C H A P T E R. Introduction 1.1

Transcription

1 C H A P T E R Introduction For y > 3/2, the slopes are negative, therefore the solutions are decreasing For y < 3/2, the slopes are positive, hence the solutions are increasing The equilibrium solution appears to be y(t) = 3/2, to which all other solutions converge

2 2 Chapter Introduction 3 For y > 3/2, the slopes are positive, therefore the solutions increase For y < 3/2, the slopes are negative, and hence the solutions decrease All solutions appear to diverge away from the equilibrium solution y(t) = 3/2 5 For y > /2, the slopes are positive, and hence the solutions increase For y < /2, the slopes are negative, and hence the solutions decrease All solutions diverge away from the equilibrium solution y(t) = /2 6 For y > 2, the slopes are positive, and hence the solutions increase For y < 2, the slopes are negative, and hence the solutions decrease All solutions diverge away from the equilibrium solution y(t) = 2

3 3 8 For all solutions to approach the equilibrium solution y(t) = 2/3, we must have y < for y > 2/3, and y > for y < 2/3 The required rates are satisfied by the differential equation y = 2 3y For solutions other than y(t) = /3 to diverge from y = /3, we must have y < for y < /3, and y > for y > /3 The required rates are satisfied by the differential equation y = 3y 2 Note that y = for y = and y = 5 The two equilibrium solutions are y(t) = and y(t) = 5 Based on the direction field, y > for y > 5; thus solutions with initial values greater than 5 diverge from the solution y(t) = 5 For < y < 5, the slopes are negative, and hence solutions with initial values between and 5 all decrease toward the solution y(t) = For y <, the slopes are all positive; thus solutions with initial values less than approach the solution y(t) = 4 Observe that y = for y = and y = 2 The two equilibrium solutions are y(t) = and y(t) = 2 Based on the direction field, y > for y > 2; thus solutions with initial values greater than 2 diverge from y(t) = 2 For < y < 2, the slopes are also positive, and hence solutions with initial values between and 2 all increase toward the solution y(t) = 2 For y <, the slopes are all negative; thus solutions with initial values less than diverge from the solution y(t) =

4 4 Chapter Introduction 5 -(j) y = 2 y 7 -(g) y = 2 y 8 -(b) y = 2 + y 2 -(e) y = y (y 3) 23 The difference between the temperature of the object and the ambient temperature is u 7 (u in F) Since the object is cooling when u > 7, and the rate constant is k = 5 min, the governing differential equation for the temperature of the object is du/dt = 5 (u 7) 24(a) Let M(t) be the total amount of the drug (in milligrams) in the patient s body at any given time t (hr) The drug is administered into the body at a constant rate of 5 mg/hr The rate at which the drug leaves the bloodstream is given by 4M(t) Hence the accumulation rate of the drug is described by the differential equation dm/dt = 5 4 M (mg/hr) (b) Based on the direction field, the amount of drug in the bloodstream approaches the equilibrium level of 25 mg (within a few hours) 26 All solutions appear to approach a linear asymptote (with slope equal to ) It is

5 5 easy to verify that y(t) = t 3 is a solution 27 All solutions appear to approach y = 3 All solutions (except y() = 5/2) appear to diverge from the sinusoid y(t) = 3 sin(t + π/4)/ 2, which is also a solution corresponding to the initial value y() = 5/2 32 All solutions appear to converge to y(t) = Solutions above the line y = 2t (but below the t-axis) have positive slope and increase rapidly to meet the t axis

6 6 Chapter Introduction Solutions that begin below the line y = 2t eventually cross it and have positive slope 33 The direction field is rather complicated Nevertheless, the collection of points at which the slope field is zero, is given by the implicit equation y 3 6y = 2t 2 The graph of these points is shown below: The y-intercepts of these curves are at y =, ± 6 It follows that for solutions with initial values y > 6, all solutions increase without bound For solutions with initial values in the range y < 6 or < y < 6, the slopes remain negative, and hence these solutions decrease without bound Solutions with initial conditions in the range 6 < y < initially increase Once the solutions reach the critical value, given by the equation y 3 6y = 2t 2, the slopes become negative and remain negative These solutions eventually decrease without bound 2 4(a) The equilibrium solution satisfies the differential equation dy e /dt = Setting ay e b =, we obtain y e (t) = b/a (b) Since dy/dt = dy/dt, it follows that dy/dt = a(y + y e ) b = a Y

7 2 7 6(a) Consider the simpler equation dy /dt = ay As in the previous solutions, rewrite the equation as (/y )dy = a dt Integrating both sides results in y (t) = c e at (b) Now set y(t) = y (t) + k, and substitute into the original differential equation We find that ay + = a(y + k) + b That is, ak + b =, and hence k = b/a (c) The general solution of the differential equation is y(t) = c e at + b/a This is exactly the form given by Eq(7) in the text Invoking an initial condition y() = y, the solution may also be expressed as y(t) = b/a + (y b/a)e at 8(a) The general solution is p(t) = p e rt Based on the discussion in the text, time t is measured in months Assuming month= 3 days, the hypothesis can be expressed as p e r = 2p Solving for the rate constant, r = ln(2), with units of per month (b) N days= N/3 months The hypothesis is stated mathematically as p e rn/3 = 2p It follows that rn/3 = ln(2), and hence the rate constant is given by r = 3 ln(2)/n The units are understood to be per month (a) Assuming no air resistance, with the positive direction taken as downward, Newton s Second Law can be expressed as mdv/dt = mg, in which g is the gravitational constant measured in appropriate units The equation can be written as dv/dt = g, with solution v(t) = gt + v The object is released with an initial velocity v (b) Suppose that the object is released from a height of h units above the ground Using the fact that v = dx/dt, in which x is the downward displacement of the object, we obtain the differential equation for the displacement as dx/dt = gt + v With the origin placed at the point of release, direct integration results in x(t) = gt 2 /2 + v t Based on the chosen coordinate system, the object reaches the ground when x(t) = h Let t = T be the time that it takes the object to reach the ground Then gt 2 /2 + v T = h Using the quadratic formula to solve for T, we obtain T = ( v ± v + 2gh )/g The positive answer corresponds to the time it takes for the object to fall to the ground The negative answer represents a previous instant at which the object could have been launched upward (with the same impact speed), only to ultimately fall downward with speed v, from a height of h units above the ground The numerical value is T = / s (c) The impact speed is calculated by substituting t = T into v(t) in part (a) That is, v(t ) = v + 2gh The numerical value is v = m/s 3 The general solution of the differential equation dq/dt = r Q is Q(t) = Q e rt, in which Q = Q() is the initial amount of the substance Let τ be the time that it takes the substance to decay to one-half of its original amount, Q Setting t = τ in the solution, we have 5 Q = Q e rτ Taking the natural logarithm of both sides, it follows that rτ = ln(5) or rτ = ln 2

8 8 Chapter Introduction 4 The differential equation governing the amount of radium-226 is dq/dt = r Q, with solution Q(t) = Q()e rt Using the result in Problem 3, and the fact that the half-life τ = 62 years, the decay rate is given by r = ln(2)/62 per year The amount of radium-226, after t years, is therefore Q(t) = Q()e 42786t Let T be the time that it takes the isotope to decay to 3/4 of its original amount Then setting t = T, and Q(T ) = (3/4)Q(), we obtain (3/4)Q() = Q()e 42786T Solving for the decay time, it follows that T = ln(3/4) or T years 6 Based on Problem 5, the governing differential equation for the temperature in the room is du/dt = 5 (u ) Setting t = at the instant that the heating system fail, the initial condition is u() = 7 F Using separation of variables, the general solution of the differential equation is u(t) = + C e 5 t Invoking the given initial condition, the temperature in the room is given by u(t) = + 6 e 5 t Setting u(t) = 32, we obtain t = 669 hr 8(a) The accumulation rate of the chemical is ()(3) grams per hour At any given time t, the concentration of the chemical in the pond is Q(t)/ 6 grams per gallon Consequently, the chemical leaves the pond at a rate of (3 4 )Q(t) grams per hour Hence, the rate of change of the chemical is given by dq = 3 3 Q(t) g/hr dt Since the pond is initially free of the chemical, Q() = (b) The differential equation can be rewritten as dq/( Q) = 3 dt Integrating both sides of the equation results in ln Q = 3t + C Taking the exponential of both sides gives Q = c e 3t Since Q() =, the value of the constant is c = Hence the amount of chemical in the pond at any time is Q(t) = ( e 3t ) grams Note that year= 876 hours Setting t = 876, the amount of chemical present after one year is Q(876) grams, that is, kilograms (c) With the accumulation rate now equal to zero, the governing equation becomes dq/dt = 3 Q(t) g/hr Resetting the time variable, we now assign the new initial value as Q() = grams (d) The solution of the differential equation in part (c) is Q(t) = e 3t Hence, one year after the source is removed, the amount of chemical in the pond is Q(876) 67 grams (e) Letting t be the amount of time after the source is removed, we obtain the equation = e 3t Taking the natural logarithm of both sides, 3 t = ln(/927777) or t 22, 776 hours 26 years

9 3 9 (f) 9(a) It is assumed that dye is no longer entering the pool In fact, the rate at which the dye leaves the pool is 2 [q(t)/6] g/min Hence the equation that governs the amount of dye in the pool is dq/dt = q/3 (g/min) The initial amount of dye in the pool is q() = 5 grams (b) The solution of the governing differential equation, with the specified initial value, is q(t) = 5 e t/3 (c) The amount of dye in the pool after four hours is obtained by setting t = 24 That is, q(4) = 5 e 8 = grams Since the size of the pool is 6, gallons, the concentration of the dye is 374 grams/gallon, and the answer is no (d) Let T be the time that it takes to reduce the concentration level of the dye to 2 grams/gallon At that time, the amount of dye in the pool is, 2 grams Using the answer in part (b), we have 5 e T/3 = 2 Taking the natural logarithm of both sides of the equation results in the required time T 74 hours (e) Consider the differential equation dq/dt = (r/6, ) q Here the parameter r corresponds to the flow rate, measured in gallons per minute Using the same initial value, the solution is given by q(t) = 5 e r t/6, In order to determine the appropriate flow rate, set t = 24 and q = 2 (Recall that 2 grams of dye has a concentration of 2 g/gal) We obtain the equation 2 = 5 e r /25 Taking the natural logarithm of both sides of the equation results in the required flow rate r 357 gallons per minute 3 The differential equation is second order, since the highest derivative in the equation is of order two The equation is linear, since the left hand side is a linear function of y and its derivatives 3 The differential equation is fourth order, since the highest derivative of the function y is of order four The equation is also linear, since the terms containing

10 Chapter Introduction the dependent variable is linear in y and its derivatives 4 The differential equation is first order, since the only derivative is of order one The dependent variable is squared, hence the equation is nonlinear 5 The differential equation is second order Furthermore, the equation is nonlinear, since the dependent variable y is an argument of the sine function, which is not a linear function 7 y (t) = e t, so y (t) = y (t) = e t Hence y y = Also, y 2 (t) = cosh t, so y (t) = sinh t and y 2 (t) = cosh t Thus y 2 y 2 = 9 y(t) = 3t + t 2, so y (t) = 3 + 2t Substituting into the differential equation, we have t(3 + 2t) (3t + t 2 ) = 3t + 2t 2 3t t 2 = t 2 Hence the given function is a solution y (t) = t/3, so y (t) = /3 and y (t) = y (t) = y (t) = Clearly, y (t) is a solution Likewise, y 2 (t) = e t + t/3, so y 2(t) = e t + /3, y 2 (t) = e t, y 2 (t) = e t, y 2 (t) = e t Substituting into the left hand side of the equation, we find that e t + 4( e t ) + 3(e t + t/3) = e t 4e t + 3e t + t = t Hence both functions are solutions of the differential equation 2 y (t) = t 2, so y (t) = 2t 3 and y (t) = 6 t 4 Substituting into the left hand side of the differential equation, we have t 2 (6 t 4 ) + 5t( 2t 3 ) + 4 t 2 = 6 t 2 t t 2 = Likewise, y 2 (t) = t 2 ln t, so y 2(t) = t 3 2t 3 ln t and y 2 (t) = 5 t t 4 ln t Substituting into the left hand side of the equation, we have t 2 ( 5 t t 4 ln t) + 5t(t 3 2t 3 ln t) + 4(t 2 ln t) = = 5 t t 2 ln t + 5 t 2 t 2 ln t + 4 t 2 ln t = Hence both functions are solutions of the differential equation 3 y(t) = (cos t) ln cos t + t sin t, so y (t) = (sin t) ln cos t + t cos t and y (t) = (cos t) ln cos t t sin t + sec t Substituting into the left hand side of the differential equation, we have ( (cos t) ln cos t t sin t + sec t) + (cos t) ln cos t + t sin t = (cos t) ln cos t t sin t + sec t + (cos t) ln cos t + t sin t = sec t Hence the function y(t) is a solution of the differential equation 5 Let y(t) = e rt Then y (t) = re rt, and substitution into the differential equation results in re rt + 2e rt = Since e rt, we obtain the algebraic equation r + 2 = The root of this equation is r = 2 7 y(t) = e rt, so y (t) = r e rt and y (t) = r 2 e rt Substituting into the differential equation, we have r 2 e rt + re rt 6 e rt = Since e rt, we obtain the algebraic equation r 2 + r 6 =, that is, (r 2)(r + 3) = The roots are r = 2 and r 2 = 3

11 3 8 Let y(t) = e rt Then y (t) = re rt, y (t) = r 2 e rt and y (t) = r 3 e rt Substituting the derivatives into the differential equation, we have r 3 e rt 3r 2 e rt + 2re rt = Since e rt, we obtain the algebraic equation r 3 3r 2 + 2r = By inspection, it follows that r(r )(r 2) = Clearly, the roots are r =, r 2 = and r 3 = 2 2 y(t) = t r, so y (t) = r t r and y (t) = r(r )t r 2 Substituting the derivatives into the differential equation, we have t 2 [ r(r )t r 2] 4t(r t r ) + 4 t r = After some algebra, it follows that r(r )t r 4r t r + 4 t r = For t, we obtain the algebraic equation r 2 5r + 4 = The roots of this equation are r = and r 2 = 4 2 The order of the partial differential equation is two, since the highest derivative, in fact each one of the derivatives, is of second order The equation is linear, since the left hand side is a linear function of the partial derivatives 23 The partial differential equation is fourth order, since the highest derivative, and in fact each of the derivatives, is of order four The equation is linear, since the left hand side is a linear function of the partial derivatives 24 The partial differential equation is second order, since the highest derivative of the function u(x, y) is of order two The equation is nonlinear, due to the product u u x on the left hand side of the equation 25 If u (x, y) = cos x cosh y, then 2 u / x 2 = cos x cosh y and 2 u / y 2 = cos x cosh y It is evident that 2 u / x u / y 2 = Also, when u 2 (x, y) = ln(x 2 + y 2 ), the second derivatives are 2 u 2 x 2 = 2 x 2 + y 2 4x 2 (x 2 + y 2 ) 2 and Adding the partial derivatives, 2 u 2 y 2 = 2 x 2 + y 2 4y 2 (x 2 + y 2 ) 2 2 u 2 x u 2 y 2 = 2 x 2 + y 2 4x 2 (x 2 + y 2 ) x 2 + y 2 4y 2 (x 2 + y 2 ) 2 = = 4 x 2 + y 2 4(x2 + y 2 ) (x 2 + y 2 ) 2 = Hence u 2 (x, y) is also a solution of the differential equation 27 Let u (x, t) = sin (λx) sin (λat) Then the second derivatives are 2 u x 2 = λ2 sin λx sin λat and 2 u t 2 = λ2 a 2 sin λx sin λat It is easy to see that a 2 2 u / x 2 = 2 u / t 2 Likewise, given u 2 (x, t) = sin(x at), we have 2 u 2 x 2 = sin(x at) and 2 u 2 t 2 = a2 sin(x at) Clearly, u 2 (x, t) is also a solution of the partial differential equation

12 2 Chapter Introduction 28 Given the function u(x, t) = π/t e x2 /4α 2t, the partial derivatives are π/t e x 2 /4α 2 t π/t x 2 e x2 /4α 2 t u xx = 2α 2 + t 4α 4 t 2 πt e x 2 /4α 2 t π x 2 e x2 /4α 2 t u t = 2t 2 + 4α 2 t 2 t It follows that π (2α α 2 2 t x 2 )e x2 /4α 2 t u xx = u t = 4α 2 t 2 t Hence u(x, t) is a solution of the partial differential equation 3(a) The kinetic energy of a particle of mass m is given by T = m v 2 /2, in which v is its speed A particle in motion on a circle of radius L has speed L (dθ/dt), where θ is its angular position and dθ/dt is its angular speed (b) Gravitational potential energy is given by V = mgh, where h is the height above a certain datum Choosing the lowest point of the swing as the datum, it follows from trigonometry that h = L( cos θ) (c) From parts (a) and (b), E = 2 ml2 ( dθ dt )2 + mgl( cos θ) Applying the chain rule for differentiation, de dt = ml2 dθ dt d 2 θ dθ + mgl sin θ dt2 dt Setting de/dt = and dividing both sides of the equation by dθ/dt results in ml 2 d2 θ + mgl sin θ =, dt2 which leads to Equation (2) 3(a) The angular momentum is the moment of the linear momentum about a given point The linear momentum is given by mv = mldθ/dt Taking the moment about the point of support, the angular momentum is M = mvl = ml 2 dθ dt (b) The moment of the gravitational force is mgl sin θ The negative sign is included since positive moments are counterclockwise Setting dm/dt equal to the moment of the gravitational force gives dm dt which leads to Equation (2) = ml 2 d2 θ = mgl sin θ, dt2

13 C H A P T E R 2 First Order Differential Equations 2 5(a) (b) If y() > 3, solutions eventually have positive slopes, and hence increase without bound If y() 3, solutions have negative slopes and decrease without bound (c) The integrating factor is µ(t) = e 2dt = e 2t The differential equation can be written as e 2t y 2e 2t y = 3e t, that is, (e 2t y) = 3e t Integration of both sides of the equation results in the general solution y(t) = 3e t + c e 2t It follows that all solutions will increase exponentially if c > and will decrease exponentially 3

14 4 Chapter 2 First Order Differential Equations if c Letting c = and then t =, we see that the boundary of these behaviors is at y() = 3 9(a) (b) All solutions eventually have positive slopes, and hence increase without bound (c) The integrating factor is µ(t) = e (/2) dt = e t/2 The differential equation can be written as e t/2 y + e t/2 y/2 = 3t e t/2 /2, that is, (e t/2 y/2) = 3t e t/2 /2 Integration of both sides of the equation results in the general solution y(t) = 3t 6 + c e t/2 All solutions approach the specific solution y (t) = 3t 6 (a) (b) For y >, the slopes are all positive, and hence the corresponding solutions increase without bound For y <, almost all solutions have negative slopes, and hence solutions tend to decrease without bound (c) First divide both sides of the equation by t (t > ) From the resulting standard form, the integrating factor is µ(t) = e (/t) dt = /t The differential equation can be written as y /t y/t 2 = t e t, that is, ( y/t) = t e t Integration leads to the general solution y(t) = te t + c t For c, solutions diverge, as implied by the direction field For the case c =, the specific solution is y(t) = te t, which evidently approaches zero as t

15 2 5 2(a) (b) All solutions eventually have positive slopes, and hence increase without bound (c) The integrating factor is µ(t) = e t/2 The differential equation can be written as e t/2 y + e t/2 y/2 = 3t 2 /2, that is, (e t/2 y/2) = 3t 2 /2 Integration of both sides of the equation results in the general solution y(t) = 3t 2 2t c e t/2 It follows that all solutions converge to the specific solution 3t 2 2t The integrating factor is µ(t) = e 2t After multiplying both sides by µ(t), the equation can be written as (e 2t y) = t Integrating both sides of the equation results in the general solution y(t) = t 2 e 2t /2 + c e 2t Invoking the specified condition, we require that e 2 /2 + c e 2 = Hence c = /2, and the solution to the initial value problem is y(t) = (t 2 )e 2t /2 6 The integrating factor is µ(t) = e (2/t) dt = t 2 Multiplying both sides by µ(t), the equation can be written as (t 2 y) = cos t Integrating both sides of the equation results in the general solution y(t) = sin t/t 2 + c t 2 Substituting t = π and setting the value equal to zero gives c = Hence the specific solution is y(t) = sin t/t 2 7 The integrating factor is µ(t) = e 2t, and the differential equation can be written as (e 2t y) = Integrating, we obtain e 2t y(t) = t + c Invoking the specified initial condition results in the solution y(t) = (t + 2)e 2t 9 After writing the equation in standard form, we find that the integrating factor is µ(t) = e (4/t) dt = t 4 Multiplying both sides by µ(t), the equation can be written as (t 4 y) = t e t Integrating both sides results in t 4 y(t) = (t + )e t + c Letting t = and setting the value equal to zero gives c = Hence the specific solution of the initial value problem is y(t) = (t 3 + t 4 )e t

16 6 Chapter 2 First Order Differential Equations 22(a) The solutions eventually increase or decrease, depending on the initial value a The critical value seems to be a = 2 (b) The integrating factor is µ(t) = e t/2, and the general solution of the differential equation is y(t) = 3e t/3 + c e t/2 Invoking the initial condition y() = a, the solution may also be expressed as y(t) = 3e t/3 + (a + 3) e t/2 The critical value is a = 3 (c) For a = 3, the solution is y(t) = 3e t/3, which diverges to as t 23(a) Solutions appear to grow infinitely large in absolute value, with signs depending on the initial value y() = a The direction field appears horizontal for a /8 (b) Dividing both sides of the given equation by 3, the integrating factor is µ(t) = e 2t/3 Multiplying both sides of the original differential equation by µ(t) and integrating results in y(t) = (2 e 2t/3 2 e πt/2 + a(4 + 3π) e 2t/3 )/(4 + 3π) The qualitative behavior of the solution is determined by the terms containing e 2t/3 : 2 e 2t/3 + a(4 + 3π) e 2t/3 The nature of the solutions will change when 2 + a(4 + 3π) = Thus the critical initial value is a = 2/(4 + 3π) (c) In addition to the behavior described in part (a), when y() = 2/(4 + 3π), the solution is y(t) = ( 2 e πt/2 )/(4 + 3π), and that specific solution will converge to y =

17 2 7 24(a) As t, solutions increase without bound if y() = a > 4, and solutions decrease without bound if y() = a < 4 (b) The integrating factor is µ(t) = e (t+)/t dt = t e t The general solution of the differential equation is y(t) = t e t + c e t /t Since y() = a, we have that + c = ae That is, c = ae Hence the solution can also be expressed as y(t) = t e t + (ae ) e t /t For small values of t, the second term is dominant Setting ae =, the critical value of the parameter is a = /e (c) When a = /e, the solution is y(t) = t e t, which approaches as t 27 The integrating factor is µ(t) = e (/2) dt = e t/2 Therefore the general solution is y(t) = (4 cos t + 8 sin t)/5 + c e t/2 Invoking the initial condition, the specific solution is y(t) = (4 cos t + 8 sin t 9 e t/2 )/5 Differentiating, it follows that y (t) = ( 4 sin t + 8 cos t + 45 e t/2 )/5 and y (t) = ( 4 cos t 8 sin t 225 e t/2 )/5 Setting y (t) =, the first solution is t = 3643, which gives the location of the first stationary point Since y (t ) <, the first stationary point in a local maximum The coordinates of the point are (3643, 828) 28 The integrating factor is µ(t) = e (2/3) dt = e 2t/3, and the differential equation can be written as (e 2t/3 y) = e 2t/3 t e 2t/3 /2 The general solution is y(t) = (2 6t)/8 + c e 2t/3 Imposing the initial condition, we have y(t) = (2 6t)/8 + (y 2/8)e 2t/3 Since the solution is smooth, the desired intersection will be a point of tangency Taking the derivative, y (t) = 3/4 (2y 2/4)e 2t/3 /3 Setting y (t) =, the solution is t = (3/2) ln [(2 8y )/9] Substituting into the solution, the respective value at the stationary point is y(t ) = 3/2 + (9/4) ln 3 (9/8) ln(2 8y ) Setting this result equal to zero, we obtain the required initial value y = (2 9 e 4/3 )/ (a) The integrating factor is µ(t) = e t/4, and the differential equation can be written as (e t/4 y) = 3 e t/4 + 2 e t/4 cos 2t After integration, we get that the general solution is y(t) = 2 + (8 cos 2t + 64 sin 2t)/65 + ce t/4 Invoking the initial condition, y() =, the specific solution is y(t) = 2 + (8 cos 2t + 64 sin 2t 788 e t/4 )/65 As t, the exponential term will decay, and the solution will oscillate about

18 8 Chapter 2 First Order Differential Equations an average value of 2, with an amplitude of 8/ 65 (b) Solving y(t) = 2, we obtain the desired value t The integrating factor is µ(t) = e 3t/2, and the differential equation can be written as (e 3t/2 y) = 3t e 3t/2 + 2 e t/2 The general solution is y(t) = 2t 4/3 4 e t + c e 3t/2 Imposing the initial condition, y(t) = 2t 4/3 4 e t + (y + 6/3) e 3t/2 Now as t, the term containing e 3t/2 will dominate the solution Its sign will determine the divergence properties Hence the critical value of the initial condition is y = 6/3 The corresponding solution, y(t) = 2t 4/3 4 e t, will also decrease without bound Note on Problems : Let g(t) be given, and consider the function y(t) = y (t) + g(t), in which y (t) as t Differentiating, y (t) = y (t) + g (t) Letting a be a constant, it follows that y (t) + ay(t) = y (t) + ay (t) + g (t) + ag(t) Note that the hypothesis on the function y (t) will be satisfied, if y (t) + ay (t) = That is, y (t) = c e at Hence y(t) = c e at + g(t), which is a solution of the equation y + ay = g (t) + ag(t) For convenience, choose a = 34 Here g(t) = 3, and we consider the linear equation y + y = 3 The integrating factor is µ(t) = e t, and the differential equation can be written as (e t y) = 3e t The general solution is y(t) = 3 + c e t 36 Here g(t) = 2t 5 Consider the linear equation y + y = 2 + 2t 5 The integrating factor is µ(t) = e t, and the differential equation can be written as (e t y) = (2t 3)e t The general solution is y(t) = 2t 5 + c e t 37 g(t) = 4 t 2 Consider the linear equation y + y = 4 2t t 2 The integrating factor is µ(t) = e t, and the equation can be written as (e t y) = (4 2t t 2 )e t The general solution is y(t) = 4 t 2 + c e t 38(a) Differentiating y and using the fundamental theorem of calculus we obtain that y = Ae p(t)dt ( p(t)), and then y + p(t)y = (b) Differentiating y we obtain that y = A (t)e p(t)dt + A(t)e p(t)dt ( p(t)) If this satisfies the differential equation then and the required condition follows y + p(t)y = A (t)e p(t)dt = g(t) (c) Let us denote µ(t) = e p(t)dt Then clearly A(t) = µ(t)g(t)dt, and after substitution y = µ(t)g(t)dt (/µ(t)), which is just Eq (33)

19 We assume a solution of the form y = A(t)e (/t) dt = A(t)e ln t = A(t)t, where A(t) satisfies A (t) = 3t cos 2t This implies that and the solution is A(t) = y = 3 cos 2t 4 3 cos 2t 4t 4 First rewrite the differential equation as + + 3t sin 2t 2 3 sin 2t 2 y + 2 t y = sin t t + c + c t Assume a solution of the form y = A(t)e (2/t) dt = A(t)t 2, where A(t) satisfies the ODE A (t) = t sin t It follows that A(t) = sin t t cos t + c and thus y = (sin t t cos t + c)/t 2 22 Problems through 2 follow the pattern of the examples worked in this section The first eight problems, however, do not have an initial condition, so the integration constant c cannot be found 2 For x, the differential equation may be written as y dy = [ x 2 /( + x 3 ) ] dx Integrating both sides, with respect to the appropriate variables, we obtain the relation y 2 /2 = (/3) ln + x 3 + c That is, y(x) = ± (2/3) ln + x3 + c 3 The differential equation may be written as y 2 dy = sin x dx Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation y = cos x + c That is, (c cos x)y =, in which c is an arbitrary constant Solving for the dependent variable, explicitly, y(x) = /(c cos x) 5 Write the differential equation as cos 2 2y dy = cos 2 x dx, which also can be written as sec 2 2y dy = cos 2 x dx Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation tan 2y = sin x cos x + x + c 7 The differential equation may be written as (y + e y )dy = (x e x )dx Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation y e y = x e x + c 8 Write the differential equation as ( + y 2 )dy = x 2 dx Integrating both sides of the equation, we obtain the relation y + y 3 /3 = x 3 /3 + c 9(a) The differential equation is separable, with y 2 dy = ( 2x)dx Integration yields y = x x 2 + c Substituting x = and y = /6, we find that c = 6 Hence the specific solution is y = /(x 2 x 6)

20 2 Chapter 2 First Order Differential Equations (b) (c) Note that x 2 x 6 = (x + 2)(x 3) Hence the solution becomes singular at x = 2 and x = 3, so the interval of existence is ( 2, 3) (a) Rewrite the differential equation as x e x dx = y dy Integrating both sides of the equation results in x e x e x = y 2 /2 + c Invoking the initial condition, we obtain c = /2 Hence y 2 = 2e x 2x e x The explicit form of the solution is y(x) = 2e x 2x e x The positive sign is chosen, since y() = (b) (c) The function under the radical becomes negative near x 7 and x 77 2(a) Write the differential equation as r 2 dr = θ dθ Integrating both sides of the equation results in the relation r = ln θ + c Imposing the condition r() = 2, we obtain c = /2 The explicit form of the solution is r = 2/( 2 ln θ)

21 22 2 (b) (c) Clearly, the solution makes sense only if θ > becomes singular when ln θ = /2, that is, θ = e Furthermore, the solution 4(a) Write the differential equation as y 3 dy = x( + x 2 ) /2 dx Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation y 2 /2 = + x 2 + c Imposing the initial condition, we obtain c = 3/2 Hence the specific solution can be expressed as y 2 = x 2 The explicit form of the solution is y(x) = / x 2 The positive sign is chosen to satisfy the initial condition (b) (c) The solution becomes singular when 2 + x 2 = 3 That is, at x = ± 5 /2 6(a) Rewrite the differential equation as 4y 3 dy = x(x 2 + )dx Integrating both sides of the equation results in y 4 = (x 2 + ) 2 /4 + c Imposing the initial condition, we obtain c = Hence the solution may be expressed as (x 2 + ) 2 4y 4 = The explicit form of the solution is y(x) = (x 2 + )/2 The sign is chosen based on y() = / 2

22 22 Chapter 2 First Order Differential Equations (b) (c) The solution is valid for all x R 8(a) Write the differential equation as (3 + 4y)dy = (e x e x )dx Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation 3y + 2y 2 = (e x + e x ) + c Imposing the initial condition, y() =, we obtain c = 7 Thus, the solution can be expressed as 3y + 2y 2 = (e x + e x ) + 7 Now by completing the square on the left hand side, 2(y + 3/4) 2 = (e x + e x ) + 65/8 Hence the explicit form of the solution is y(x) = 3/4 + 65/6 cosh x (b) (c) Note the 65 6 cosh x as long as x > 2 (approximately) Hence the solution is valid on the interval 2 < x < 2 2(a) Rewrite the differential equation as y 2 dy = arcsin x/ x 2 dx Integrating both sides of the equation results in y 3 /3 = (arcsin x) 2 /2 + c Imposing the condition y() =, we obtain c = /3 The explicit form of the solution is y(x) = (3(arcsin x) 2 /2 + ) /3

23 22 23 (b) (c) Since arcsin x is defined for x, this is the interval of existence 22 The differential equation can be written as (3y 2 4)dy = 3x 2 dx Integrating both sides, we obtain y 3 4y = x 3 + c Imposing the initial condition, the specific solution is y 3 4y = x 3 Referring back to the differential equation, we find that y as y ±2/ 3 The respective values of the abscissas are x 276, 598 Hence the solution is valid for 276 < x < Write the differential equation as (3 + 2y)dy = (2 e x )dx Integrating both sides, we obtain 3y + y 2 = 2x e x + c Based on the specified initial condition, the solution can be written as 3y + y 2 = 2x e x + Completing the square, it follows that y(x) = 3/2 + 2x e x + 3/4 The solution is defined if 2x e x + 3/4, that is, 5 x 2 (approximately) In that interval, y = for x = ln 2 It can be verified that y (ln 2) < In fact, y (x) < on the interval of definition Hence the solution attains a global maximum at x = ln 2 26 The differential equation can be written as ( + y 2 ) dy = 2( + x)dx Integrating both sides of the equation, we obtain arctan y = 2x + x 2 + c Imposing the given initial condition, the specific solution is arctan y = 2x + x 2 Therefore, y = tan(2x + x 2 ) Observe that the solution is defined as long as π/2 < 2x + x 2 < π/2 It is easy to see that 2x + x 2 Furthermore, 2x + x 2 = π/2 for x 26 and 6 Hence the solution is valid on the interval 26 < x < 6 Referring back to the differential equation, the solution is stationary at x = Since y ( ) >, the solution attains a global minimum at x = 28(a) Write the differential equation as y (4 y) dy = t( + t) dt Integrating both sides of the equation, we obtain ln y ln y 4 = 4t 4 ln + t + c Taking the exponential of both sides y/(y 4) = c e 4t /( + t) 4 It follows that as t, y/(y 4) = + 4/(y 4) That is, y(t) 4 (b) Setting y() = 2, we obtain that c = Based on the initial condition, the solution may be expressed as y/(y 4) = e 4t /( + t) 4 Note that y/(y 4) <, for all t Hence y < 4 for all t Referring back to the differential equation, it follows that y is always positive This means that the solution is monotone

24 24 Chapter 2 First Order Differential Equations increasing We find that the root of the equation e 4t /( + t) 4 = 399 is near t = 2844 (c) Note the y(t) = 4 is an equilibrium solution Examining the local direction field we see that if y() >, then the corresponding solutions converge to y = 4 Referring back to part (a), we have y/(y 4) = [y /(y 4)] e 4t /( + t) 4, for y 4 Setting t = 2, we obtain y /(y 4) = (3/e 2 ) 4 y(2)/(y(2) 4) Now since the function f(y) = y/(y 4) is monotone for y < 4 and y > 4, we need only solve the equations y /(y 4) = 399(3/e 2 ) 4 and y /(y 4) = 4(3/e 2 ) 4 The respective solutions are y = and y = (a) Observe that (x 2 + 3y 2 )/2xy = (/2)(y/x) + (3/2)(y/x) Hence the differential equation is homogeneous (b) The substitution y = x v results in v + x v = (x 2 + 3x 2 v 2 )/2x 2 v The transformed equation is v = ( + v 2 )/2xv This equation is separable, with general solution v 2 + = c x In terms of the original dependent variable, the solution is x 2 + y 2 = c x 3 (c) The integral curves are symmetric with respect to the origin 34(a) Observe that (4x + 3y)/(2x + y) = 2 (y/x) [2 + (y/x)] differential equation is homogeneous Hence the (b) The substitution y = x v results in v + x v = 2 v/(2 + v) The transformed equation is v = (v 2 + 5v + 4)/(2 + v)x This equation is separable, with general solution (v + 4) 2 v + = c/x 3 In terms of the original dependent variable, the solution is (4x + y) 2 x + y = c

25 22 25 (c) The integral curves are symmetric with respect to the origin 36(a) Divide by x 2 to see that the equation is homogeneous Substituting y = x v, we obtain x v = ( + v) 2 The resulting differential equation is separable (b) Write the equation as ( + v) 2 dv = x dx Integrating both sides of the equation, we obtain the general solution /( + v) = ln x + c In terms of the original dependent variable, the solution is y = x (c ln x ) x (c) The integral curves are symmetric with respect to the origin 37(a) The differential equation can be expressed as y = (/2)(y/x) (3/2)(y/x) Hence the equation is homogeneous The substitution y = xv results in xv = ( 5v 2 )/2v Separating variables, we have 2vdv/( 5v 2 ) = dx/x (b) Integrating both sides of the transformed equation yields (ln 5v 2 )/5 = ln x + c, that is, 5v 2 = c/ x 5 In terms of the original dependent variable, the general solution is 5y 2 = x 2 c/ x 3

26 26 Chapter 2 First Order Differential Equations (c) The integral curves are symmetric with respect to the origin 38(a) The differential equation can be expressed as y = (3/2)(y/x) (/2)(y/x) Hence the equation is homogeneous The substitution y = x v results in x v = (v 2 )/2v, that is, 2vdv/(v 2 ) = dx/x (b) Integrating both sides of the transformed equation yields ln v 2 = ln x + c, that is, v 2 = c x In terms of the original dependent variable, the general solution is y 2 = c x 2 x + x 2 (c) The integral curves are symmetric with respect to the origin 23 Let Q(t) be the amount of dye in the tank at time t Clearly, Q() = 2 g The differential equation governing the amount of dye is Q (t) = 2Q(t)/2 The solution of this separable equation is Q(t) = Q()e t/ = 2e t/ We need the time T such that Q(T ) = 2 g This means we have to solve 2 = 2e T/ and we obtain that T = ln(/) = ln 465 min 5(a) Let Q be the amount of salt in the tank Salt enters the tank of water at a rate of 2 (/4)( + (/2) sin t) = /2 + (/4) sin t oz/min It leaves the tank at a

27 23 27 rate of 2 Q/ oz/min Hence the differential equation governing the amount of salt at any time is dq dt = sin t Q 5 The initial amount of salt is Q = 5 oz The governing differential equation is linear, with integrating factor µ(t) = e t/5 Write the equation as (e t/5 Q) = e t/5 (/2 + (/4) sin t) The specific solution is Q(t) = 25 + (25 sin t 625 cos t e t/5 )/25 oz (b) (c) The amount of salt approaches a steady state, which is an oscillation of approximate amplitude /4 about a level of 25 oz 6(a) Using the Principle of Conservation of Energy, the speed v of a particle falling from a height h is given by 2 mv2 = mgh (b) The outflow rate is (outflow cross-section area) (outflow velocity): α a 2gh At any instant, the volume of water in the tank is V (h) = h A(u)du The time rate of change of the volume is given by dv/dt = (dv/dh)(dh/dt) = A(h)dh/dt Since the volume is decreasing, dv/dt = α a 2gh (c) With A(h) = π, a = π, α = 6, the differential equation for the water level h is π(dh/dt) = 6 π 2gh, with solution h(t) = 8gt 2 6 2gh() t + h() Setting h() = 3 and g = 98, h(t) = 764 t 2 46 t + 3, resulting in h(t) = for t 34 s 7(a) The equation governing the value of the investment is ds/dt = r S The value of the investment, at any time, is given by S(t) = S e rt Setting S(T ) = 2S, the required time is T = ln(2)/r (b) For the case r = 7, T 99 yr

28 28 Chapter 2 First Order Differential Equations (c) Referring to part (a), r = ln(2)/t Setting T = 8, the required interest rate is to be approximately r = 866% 2(a) Using Eq(5) we have ds/dt 5S = (8 + t), S() = 5, Using an integrating factor and integration by parts we obtain that S(t) = 56, 4, e 5t + 2t Setting S(t) = and solving numerically for t yields t = 4654 months (b) The solution we obtained in part (a) with a general initial condition S() = S is S(t) = 56, 56, e 5t + S e 5t + 2t Solving the equation S(24) = yields S = 246, 758 3(a) Let Q = r Q The general solution is Q(t) = Q e rt Based on the definition of half-life, consider the equation Q /2 = Q e 573 r It follows that 573 r = ln(/2), that is, r = per year (b) The amount of carbon-4 is given by Q(t) = Q e 297 4t (c) Given that Q(T ) = Q /5, we have the equation /5 = e 297 4T Solving for the decay time, the apparent age of the remains is approximately T = 3, 35 years 5(a) The differential equation dy/dt = r(t) y k is linear, with integrating factor µ(t) = e r(t)dt Write the equation as (µ y) = k µ(t) Integration of both sides yields the general solution y = [ k µ(τ)dτ + y µ() ] /µ(t) In this problem, the integrating factor is µ(t) = e (cos t t)/5 (b) The population becomes extinct, if y(t ) =, for some t = t Referring to part (a), we find that y(t ) = when t e (cos τ τ)/5 dτ = 5 e /5 y c It can be shown that the integral on the left hand side increases monotonically, from zero to a limiting value of approximately 5893 Hence extinction can happen only if 5 e /5 y < 5893 Solving 5e /5 y c = 5893 yields y c = 8333

29 23 29 (c) Repeating the argument in part (b), it follows that y(t ) = when t e (cos τ τ)/5 dτ = k e/5 y c Hence extinction can happen only if e /5 y /k < 5893, so y c = 4667 k (d) Evidently, y c is a linear function of the parameter k 7(a) The solution of the governing equation satisfies u 3 = u 3 /( 3 α u 3 t + ) With the given data, it follows that u(t) = 2/ 3 6 t/25 + (b) (c) Numerical evaluation results in u(t) = 6 for t 7577 s 22(a) The differential equation for the upward motion is mdv/dt = µv 2 mg, in which µ = /325 This equation is separable, with m/(µ v 2 + mg) dv = dt Integrating both sides and invoking the initial condition, v(t) = 4433 tan( t) Setting v(t ) =, the ball reaches the maximum height at t = 96 s Integrating v(t), the position is given by x(t) = 9875 ln [cos(222 t 425)] Therefore the maximum height is x(t ) = 4856 m (b) The differential equation for the downward motion is m dv/dt = +µv 2 mg This equation is also separable, with m/(mg µ v 2 ) dv = dt For convenience, set t = at the top of the trajectory The new initial condition becomes v() = Integrating both sides and invoking the initial condition, we obtain ln((443 v)/(443 + v)) = t/225 Solving for the velocity, v(t) = 443( e t/225 )/( + e t/225 ) Integrating v(t), we obtain x(t) = 9929 ln(e t/225 /( + e t/225 ) 2 ) To estimate the duration of the downward motion, set x(t 2 ) =, resulting in t 2 = 3276 s Hence the total time that the ball spends in the air is t + t 2 = 592 s

30 3 Chapter 2 First Order Differential Equations (c) 24(a) Setting µv 2 = v(dv/dx), we obtain dv/dx = µv (b) The speed v of the sled satisfies ln(v/v ) = µx Noting that the unit conversion factors cancel, solution of ln(5/5) = 2 µ results in µ = ln()/2 ft 5 ft 6788 mi (c) Solution of dv/dt = µv 2 can be expressed as /v /v = µt Noting that mi/hr = 528/36 ft/s, the elapsed time is t = (/5 /5)/((528/36)(ln()/2)) 3553 s 25(a) Measure the positive direction of motion upward The equation of motion is given by mdv/dt = k v mg The initial value problem is dv/dt = kv/m g, with v() = v The solution is v(t) = mg/k + (v + mg/k)e kt/m Setting v(t m ) =, the maximum height is reached at time t m = (m/k) ln [(mg + k v )/mg] Integrating the velocity, the position of the body is [ x(t) = mg t/k + ( m k )2 g + m v k ] ( e kt/m ) Hence the maximum height reached is x m = x(t m ) = m v g( m [ ] mg + k v k k )2 ln mg (b) Recall that for δ, ln( + δ) = δ δ 2 /2 + δ 3 /3 δ 4 /4 + (c) The dimensions of the quantities involved are [k] = MT, [v ] = LT, [m] = M and [g] = LT 2 This implies that kv /mg is dimensionless 3(a) Both equations are linear and separable Initial conditions: v() = u cos A and w() = u sin A We obtain the solutions v(t) = (u cos A)e rt and w(t) = g/r + (u sin A + g/r)e rt (b) Integrating the solutions in part (a), and invoking the initial conditions, the coordinates are x(t) = u cos A( e rt )/r and y(t) = gt r + g + ur sin A + hr2 r 2 ( u r sin A + g r 2 )e rt

31 23 3 (c) (d) Let T be the time that it takes the ball to go 35 ft horizontally Then from above, e T/5 = (u cos A 7)/u cos A At the same time, the height of the ball is given by y(t ) = 6T u sin A (8 + 5u sin A)(u cos A 7) u cos A Hence A and u must satisfy the equality [ ] u cos A 7 (8 + 5u sin A)(u cos A 7) 8 ln u sin A = u cos A u cos A for the ball to touch the top of the wall To find the optimal values for u and A, consider u as a function of A and use implicit differentiation in the above equation to find that du da = cos A 7u 2 sin A) u(u2 2 cos A Solving this equation simultaneously with the above equation yields optimal values for u and A: u 453 ft/s, A 644 rad 32(a) Solving equation (i), y (x) = [ (k 2 y)/y ] /2 The positive answer is chosen, since y is an increasing function of x (b) Let y = k 2 sin 2 t Then dy = 2k 2 sin t cos tdt Substituting into the equation in part (a), we find that Hence 2k 2 sin 2 tdt = dx 2k 2 sin t cos tdt dx = cos t sin t (c) Setting θ = 2t, we further obtain k 2 sin 2 (θ/2) dθ = dx Integrating both sides of the equation and noting that t = θ = corresponds to the origin, we obtain the solutions x(θ) = k 2 (θ sin θ)/2 and (from part (b)) y(θ) = k 2 ( cos θ)/2

32 32 Chapter 2 First Order Differential Equations (d) Note that y/x = ( cos θ)/(θ sin θ) Setting x =, y = 2, the solution of the equation ( cos θ)/(θ sin θ) = 2 is θ 4 Substitution into either of the expressions yields k Rewrite the differential equation as y + /(t(t 4)) y = It is evident that the coefficient /t(t 4) is continuous everywhere except at t =, 4 Since the initial condition is specified at t = 2, Theorem 24 assures the existence of a unique solution on the interval < t < 4 3 The function tan t is discontinuous at odd multiples of π/2 Since π/2 < π < 3π/2, the initial value problem has a unique solution on the interval (π/2, 3π/2) 5 p(t) = 2t/(4 t 2 ) and g(t) = 3t 2 /(4 t 2 ) These functions are discontinuous at x = ±2 The initial value problem has a unique solution on the interval ( 2, 2) 6 The function ln t is defined and continuous on the interval (, ) At t =, ln t =, so the normal form of the differential equation has a singularity there Also, cot t is not defined at integer multiples of π, so the initial value problem will have a solution on the interval (, π) 7 The function f(t, y) is continuous everywhere on the plane, except along the straight line y = 2t/5 The partial derivative f/ y = 7t/(2t + 5y) 2 has the same region of continuity 9 The function f(t, y) is discontinuous along the coordinate axes, and on the hyperbola t 2 y 2 = Furthermore, has the same points of discontinuity f y = ± y( t 2 + y 2 ) 2 y ln ty ( t 2 + y 2 ) 2 f(t, y) is continuous everywhere on the plane The partial derivative f/ y is also continuous everywhere 2 The function f(t, y) is discontinuous along the lines t = ±k π for k =,, 2, and y = The partial derivative f/ y = cot t/( + y) 2 has the same region of continuity 4 The equation is separable, with dy/y 2 = 2t dt Integrating both sides, the solution is given by y(t) = y /( y t 2 ) For y >, solutions exist as long as t 2 < /y For y, solutions are defined for all t 5 The equation is separable, with dy/y 3 = dt Integrating both sides and invoking the initial condition, y(t) = y / 2y 2t + Solutions exist as long as

33 y 2 t + >, that is, 2y 2 t > If y, solutions exist for t > /2y 2 If y =, then the solution y(t) = exists for all t 6 The function f(t, y) is discontinuous along the straight lines t = and y = The partial derivative f/ y is discontinuous along the same lines The equation is separable, with y dy = t 2 dt/( + t 3 ) Integrating and invoking the initial condition, the solution is y(t) = [ (2/3) ln + t 3 + y 2 ] /2 Solutions exist as long as (2/3) ln + t 3 + y 2, that is, y 2 (2/3) ln + t 3 For all y (it can be verified that y = yields a valid solution, even though Theorem 242 does not guarantee one), solutions exist as long as + t 3 e 3y 2 /2 From above, we must have t > Hence the inequality may be written as t 3 e 3y2 /2 It follows that the solutions are valid for (e 3y2 /2 ) /3 < t < 8 Based on the direction field, and the differential equation, for y <, the slopes eventually become negative, and hence solutions tend to For y >, solutions increase without bound if t < Otherwise, the slopes eventually become negative, and solutions tend to zero Furthermore, y = is an equilibrium solution Note that slopes are zero along the curves y = and ty = 3 9 For initial conditions (t, y ) satisfying ty < 3, the respective solutions all tend to zero For y 9, the solutions tend to ; for y > 9, the solutions tend to Also, y = is an equilibrium solution

34 34 Chapter 2 First Order Differential Equations 2 Solutions with t < all tend to Solutions with initial conditions (t, y ) to the right of the parabola t = + y 2 asymptotically approach the parabola as t Integral curves with initial conditions above the parabola (and y > ) also approach the curve The slopes for solutions with initial conditions below the parabola (and y < ) are all negative These solutions tend to 2(a) No There is no value of t for which (2/3)(t t ) 2/3 satisfies the condition y() = (b) Yes Let t = /2 in Eq(9) (c) For t >, y(2) (4/3) 3/ The assumption is φ (t) + p(t)φ(t) = But then cφ (t) + p(t)cφ(t) = as well 26(a) Recalling Eq(33) in Section 2, y = µ(t) t t µ(s)g(s) ds + c µ(t) It is evident that y (t) = /µ(t) and y 2 (t) = (/µ(t)) t t µ(s)g(s) ds (b) By definition, /µ(t) = e p(t)dt Hence y = p(t)/µ(t) = p(t)y That is, y + p(t)y = (c) y 2 = ( p(t)/µ(t)) t µ(s)g(s) ds + µ(t)g(t)/µ(t) = p(t)y 2 + g(t) This implies that y 2 + p(t)y 2 = g(t) 3 Since n = 3, set v = y 2 It follows that v = 2y 3 y and y = (y 3 /2)v Substitution into the differential equation yields (y 3 /2)v εy = σy 3, which further results in v + 2εv = 2σ The latter differential equation is linear, and can be written as (ve 2εt ) = 2σe 2εt The solution is given by v(t) = σ/ε + ce 2εt Converting back to the original dependent variable, y = ±v /2 = ±(σ/ε + ce 2εt ) /2 3 Since n = 3, set v = y 2 It follows that v = 2y 3 y and y = (y 3 /2)v The differential equation is written as (y 3 /2)v (Γ cos t + T )y = σy 3, which upon

35 25 35 further substitution is v + 2(Γ cos t + T )v = 2 This ODE is linear, with integrating factor µ(t) = e 2 (Γ cos t+t )dt = e 2Γ sin t+2t t The solution is t v(t) = 2e (2Γ sin t+2t t) e 2Γ sin τ+2t τ dτ + ce (2Γ sin t+2t t) Converting back to the original dependent variable, y = ±v /2 33 The solution of the initial value problem y + 2y =, y () = is y (t) = e 2t Therefore y( ) = y () = e 2 On the interval (, ), the differential equation is y 2 + y 2 =, with y 2 (t) = ce t Therefore y( + ) = y 2 () = ce Equating the limits y( ) = y( + ), we require that c = e Hence the global solution of the initial value problem is y(t) = { e 2t, t e t, t > Note the discontinuity of the derivative y (t) = { 2e 2t, < t < e t, t > 25

36 36 Chapter 2 First Order Differential Equations For y, the only equilibrium point is y =, and y = ay + by 2 > when y >, hence the equilibrium solution y = is unstable 2 -a/b The equilibrium points are y = a/b and y =, and y > when y > or y < a/b, and y < when a/b < y <, therefore the equilibrium solution y = a/b is asymptotically stable and the equilibrium solution y = is unstable

37 The only equilibrium point is y =, and y > when y >, y < when y <, hence the equilibrium solution y = is unstable 6

38 38 Chapter 2 First Order Differential Equations The only equilibrium point is y =, and y > when y <, y < when y >, hence the equilibrium solution y = is asymptotically stable 8 The only equilibrium point is y =, and y < for y As long as y, the corresponding solution is monotone decreasing Hence the equilibrium solution y = is semistable

39 The equilibrium points are y =, ±, and y > for y < or < y < and y < for < y < or y > The equilibrium solution y = is unstable, and the remaining two are asymptotically stable 2

40 4 Chapter 2 First Order Differential Equations 2-2 The equilibrium points are y =, ±2, and y < when y < 2 or y > 2, and y > for 2 < y < or < y < 2 The equilibrium solutions y = 2 and y = 2 are unstable and asymptotically stable, respectively The equilibrium solution y = is semistable 3 The equilibrium points are y =, y > for all y except y = and y = Both equilibrium solutions are semistable 5(a) Inverting Eq(), Eq(3) shows t as a function of the population y and the

41 25 4 carrying capacity K With y = K/3, t = r ln (/3) [ (y/k)] (y/k) [ (/3)] Setting y = 2y, τ = r ln (/3) [ (2/3)] (2/3) [ (/3)] That is, τ = (ln 4)/r If r = 25 per year, τ 5545 years (b) In Eq(3), set y /K = α and y/k = β As a result, we obtain T = r ln α [ β] β [ α] Given α =, β = 9 and r = 25 per year, τ 7578 years 9(a) The rate of increase of the volume is given by rate of flow in rate of flow out That is, dv/dt = k αa 2gh Since the cross section is constant, dv/dt = Adh/dt Hence the governing equation is dh/dt = (k αa 2gh )/A (b) Setting dh/dt =, the equilibrium height is h e = (/2g)(k/αa) 2 Furthermore, since dh/dt < for h > h e and dh/dt > for h < h e, it follows that the equilibrium height is asymptotically stable 22(a) The equilibrium points are at y = and y = Since f (y) = α 2αy, the equilibrium solution y = is unstable and the equilibrium solution y = is asymptotically stable (b) The differential equation is separable, with [y( y)] dy = α dt Integrating both sides and invoking the initial condition, the solution is y(t) = y e αt y y + y e αt = y + ( y )e αt It is evident that (independent of y ) lim t y(t) = and lim t y(t) = 23(a) y(t) = y e βt (b) From part (a), dx/dt = αxy e βt Separating variables, dx/x = αy e βt dt Integrating both sides, the solution is x(t) = x e α y( e βt )/β (c) As t, y(t) and x(t) x e α y/β Over a long period of time, the proportion of carriers vanishes Therefore the proportion of the population that escapes the epidemic is the proportion of susceptibles left at that time, x e α y/β 26(a) For a <, the only critical point is at y =, which is asymptotically stable For a =, the only critical point is at y =, which is asymptotically stable For a >, the three critical points are at y =, ± a The critical point at y = is unstable, whereas the other two are asymptotically stable

42 42 Chapter 2 First Order Differential Equations (b) Below, we graph solutions in the case a =, a = and a = respectively

43 25 43 (c) 27(a) f(y) = y(a y); f (y) = a 2y For a <, the critical points are at y = a and y = Observe that f (a) > and f () < Hence y = a is unstable and y = asymptotically stable For a =, the only critical point is at y =, which is semistable since f(y) = y 2 is concave down For a >, the critical points are at y = and y = a Observe that f () > and f (a) < Hence y = is unstable and y = a asymptotically stable (b) Below, we graph solutions in the case a =, a = and a = respectively

44 44 Chapter 2 First Order Differential Equations (c) 26 M(x, y) = 2x + 3 and N(x, y) = 2y 2 Since M y = N x =, the equation is exact Integrating M with respect to x, while holding y constant, yields ψ(x, y) = x 2 + 3x + h(y) Now ψ y = h (y), and equating with N results in the possible function h(y) = y 2 2y Hence ψ(x, y) = x 2 + 3x + y 2 2y, and the solution is defined implicitly as x 2 + 3x + y 2 2y = c 2 M(x, y) = 2x + 4y and N(x, y) = 2x 2y Note that M y N x, and hence the differential equation is not exact 4 First divide both sides by (2xy + 2) We now have M(x, y) = y and N(x, y) = x Since M y = N x =, the resulting equation is exact Integrating M with respect to x, while holding y constant, results in ψ(x, y) = xy + h(y) Differentiating with respect to y, ψ y = x + h (y) Setting ψ y = N, we find that h (y) =, and hence h(y) = is acceptable Therefore the solution is defined implicitly as xy = c Note that if xy + =, the equation is trivially satisfied 6 Write the equation as (ax by)dx + (bx cy)dy = Now M(x, y) = ax by and N(x, y) = bx cy Since M y N x, the differential equation is not exact

45 M(x, y) = e x sin y + 3y and N(x, y) = 3x + e x sin y Note that M y N x, and hence the differential equation is not exact M(x, y) = y/x + 6x and N(x, y) = ln x 2 Since M y = N x = /x, the given equation is exact Integrating N with respect to y, while holding x constant, results in ψ(x, y) = y ln x 2y + h(x) Differentiating with respect to x, ψ x = y/x + h (x) Setting ψ x = M, we find that h (x) = 6x, and hence h(x) = 3x 2 Therefore the solution is defined implicitly as 3x 2 + y ln x 2y = c M(x, y) = x ln y + xy and N(x, y) = y ln x + xy Note that M y N x, and hence the differential equation is not exact 3 M(x, y) = 2x y and N(x, y) = 2y x Since M y = N x =, the equation is exact Integrating M with respect to x, while holding y constant, yields ψ(x, y) = x 2 xy + h(y) Now ψ y = x + h (y) Equating ψ y with N results in h (y) = 2y, and hence h(y) = y 2 Thus ψ(x, y) = x 2 xy + y 2, and the solution is given implicitly as x 2 xy + y 2 = c Invoking the initial condition y() = 3, the specific solution is x 2 xy + y 2 = 7 The explicit form of the solution is y(x) = (x x 2 )/2 Hence the solution is valid as long as 3x M(x, y) = y e 2xy + x and N(x, y) = bx e 2xy Note that M y = e 2xy + 2xy e 2xy, and N x = b e 2xy + 2bxy e 2xy The given equation is exact, as long as b = Integrating N with respect to y, while holding x constant, results in ψ(x, y) = e 2xy /2 + h(x) Now differentiating with respect to x, ψ x = y e 2xy + h (x) Setting ψ x = M, we find that h (x) = x, and hence h(x) = x 2 /2 We conclude that ψ(x, y) = e 2xy /2 + x 2 /2 Hence the solution is given implicitly as e 2xy + x 2 = c 7 Note that ψ is of the form ψ(x, y) = f(x) + g(y), since each of the integrands is a function of a single variable It follows that ψ x = f (x) and ψ y = g (y) That is, ψ x = M(x, y ) and ψ y = N(x, y) Furthermore, 2 ψ x y (x, y ) = M y (x, y ) and 2 ψ y x (x, y ) = N x (x, y ), based on the hypothesis and the fact that the point (x, y ) is arbitrary, ψ xy = ψ yx and M y (x, y) = N x (x, y) 8 Observe that (M(x)) y = (N(y)) x = 2 M y = y cos y y 2 sin y and N x = 2 e x (cos x + sin x)/y Multiplying both sides by the integrating factor µ(x, y) = y e x, the given equation can be written as (e x sin y 2y sin x)dx + (e x cos y + 2 cos x)dy = Let M = µm and Ñ = µn Observe that M y = Ñx, and hence the latter ODE is exact Integrating Ñ with respect to y, while holding x constant, results in ψ(x, y) = e x sin y + 2y cos x + h(x) Now differentiating with respect to x, ψ x = e x sin y 2y sin x + h (x) Setting ψ x = M, we find that h (x) =, and hence h(x) = is feasible Hence the solution of the given equation is defined implicitly by e x sin y + 2y cos x = c

46 46 Chapter 2 First Order Differential Equations 2 M y = and N x = 2 Multiply both sides by the integrating factor µ(x, y) = y to obtain y 2 dx + (2xy y 2 e y )dy = Let M = ym and Ñ = yn It is easy to see that My = Ñx, and hence the latter ODE is exact Integrating M with respect to x yields ψ(x, y) = xy 2 + h(y) Equating ψ y with Ñ results in h (y) = y 2 e y, and hence h(y) = e y (y 2 2y + 2) Thus ψ(x, y) = xy 2 e y (y 2 2y + 2), and the solution is defined implicitly by xy 2 e y (y 2 2y + 2) = c 24 The equation µm + µny = has an integrating factor if (µm) y = (µn) x, that is, µ y M µ x N = µn x µm y Suppose that N x M y = R (xm yn), in which R is some function depending only on the quantity z = xy It follows that the modified form of the equation is exact, if µ y M µ x N = µr (xm yn) = R (µ xm µ yn) This relation is satisfied if µ y = (µ x)r and µ x = (µ y)r Now consider µ = µ(xy) Then the partial derivatives are µ x = µ y and µ y = µ x Note that µ = dµ/dz Thus µ must satisfy µ (z) = R(z) The latter equation is separable, with dµ = R(z)dz, and µ(z) = R(z)dz Therefore, given R = R(xy), it is possible to determine µ = µ(xy) which becomes an integrating factor of the differential equation 28 The equation is not exact, since N x M y = 2y However, (N x M y )/M = (2y )/y is a function of y alone Hence there exists µ = µ(y), which is a solution of the differential equation µ = (2 /y)µ The latter equation is separable, with dµ/µ = 2 /y One solution is µ(y) = e 2y ln y = e 2y /y Now rewrite the given ODE as e 2y dx + (2x e 2y /y)dy = This equation is exact, and it is easy to see that ψ(x, y) = x e 2y ln y Therefore the solution of the given equation is defined implicitly by x e 2y ln y = c 3 The given equation is not exact, since N x M y = 8x 3 /y 3 + 6/y 2 But note that (N x M y )/M = 2/y is a function of y alone, and hence there is an integrating factor µ = µ(y) Solving the equation µ = (2/y)µ, an integrating factor is µ(y) = y 2 Now rewrite the differential equation as (4x 3 + 3y)dx + (3x + 4y 3 )dy = By inspection, ψ(x, y) = x 4 + 3xy + y 4, and the solution of the given equation is defined implicitly by x 4 + 3xy + y 4 = c 32 Multiplying both sides of the ODE by µ = [xy(2x + y)], the given equation is equivalent to [ (3x + y)/(2x 2 + xy) ] dx + [ (x + y)/(2xy + y 2 ) ] dy = Rewrite the differential equation as [ 2 x + 2 ] [ dx + 2x + y y + ] dy = 2x + y It is easy to see that M y = N x Integrating M with respect to x, while keeping y constant, results in ψ(x, y) = 2 ln x + ln 2x + y + h(y) Now taking the partial derivative with respect to y, ψ y = (2x + y) + h (y) Setting ψ y = N, we find that h (y) = /y, and hence h(y) = ln y Therefore ψ(x, y) = 2 ln x + ln 2x + y + ln y, and the solution of the given equation is defined implicitly by 2x 3 y + x 2 y 2 = c

47 The Euler formula is given by y n+ = y n + h(2y n ) = ( + 2h)y n h (a), 22, 364, 5368 (b) 5, 2325, 38578, 5779 (c) 775, 23873, 39793, 5944 (d) The differential equation is linear with solution y(t) = ( + e 2t )/2 The values are 7, 2459, 46, All solutions seem to converge to y = 25/9 7 All solutions seem to converge to a specific function

48 48 Chapter 2 First Order Differential Equations 8 Solutions with initial conditions y() > 25 seem to diverge On the other hand, solutions with initial conditions y() < 25 seem to converge to zero Also, y = is an equilibrium solution Solutions with positive initial conditions increase without bound Solutions with negative initial conditions decrease without bound Note that y = is an equilibrium solution The Euler formula is y n+ = y n 3h y n + 5h The initial value is y = 2 (a) 238, 2496, 2623, , 27939, (b) 2367, , , , 2759, (c) , 24793, 25924, , 273, (d) , 24769, 25883, , 2785, The Euler formula is y n+ = ( + 3h)y n ht n y 2 n The initial value is (t, y ) = (, 5) (a) 738, 3665, 2443, 7724, 37348, 925

49 27 49 (b) 79548, 365, , 7787, 37795, 29 (c) 84579, 35769, 24295, 7874, 387, 2328 (d) 87734, 3567, , 78224, 385, 24 4 The Euler formula is y n+ = ( ht n )y n + hy 3 n/, with (t, y ) = (, ) (a) 9557, 68755, 36988, 4599, 42429, (b) , 67245, 36264, 47659, 454, 493 (c) , , , 4846, 46954, 3722 (d) , 66463, , 48848, , The Euler formula is y n+ = y n + h(y 2 n + 2t n y n )/(3 + t 2 n) The initial point is (t, y ) = (, 2) Using this iteration formula with the specified h values, the value of the solution at t = 25 is somewhere between 8 and 9 At t = 3 there is no reliable estimate 9(a) (b) The iteration formula is y n+ = y n + h y 2 n h t 2 n The critical value α appears to be between 67 and 68 For y > α, the iterations diverge 2(a) The ODE is linear, with general solution y(t) = t + ce t Invoking the specified initial condition, y(t ) = y, we have y = t + ce t Hence c = (y t )e t Thus the solution is given by φ(t) = (y t )e t t + t (b) The Euler formula is y n+ = ( + h)y n + h h t n Now set k = n + (c) We have y = ( + h)y + h ht = ( + h)y + (t t ) ht Rearranging the terms, y = ( + h)(y t ) + t Now suppose that y k = ( + h) k (y t ) + t k, for some k Then y k+ = ( + h)y k + h ht k Substituting for y k, we find

50 5 Chapter 2 First Order Differential Equations that y k+ = ( + h) k+ (y t ) + ( + h)t k + h ht k = ( + h) k+ (y t ) + t k + h Noting that t k+ = t k + h, the result is verified (d) Substituting h = (t t )/n, with t n = t, y n = ( + (t t )/n) n (y t ) + t Taking the limit of both sides, and using the fact that lim n ( + a/n) n = e a, pointwise convergence is proved 2 The exact solution is y(t) = e t The Euler formula is y n+ = ( + h)y n It is easy to see that y n = ( + h) n y = ( + h) n Given t >, set h = t/n Taking the limit, we find that lim n y n = lim n ( + t/n) n = e t 23 The exact solution is y(t) = t/2 + e 2t The Euler formula is y n+ = ( + 2h)y n + h/2 h t n Since y =, y = ( + 2h) + h/2 = ( + 2h) + t /2 It is easy to show by mathematical induction, that y n = ( + 2h) n + t n /2 For t >, set h = t/n and thus t n = t Taking the limit, we find that lim n y n = lim n [( + 2t/n) n + t/2] = e 2t + t/2 Hence pointwise convergence is proved 28 2 Let z = y 3 and τ = t + It follows that dz/dτ = (dz/dt)(dt/dτ) = dz/dt Furthermore, dz/dt = dy/dt = y 3 Hence dz/dτ = (z + 3) 3 The new initial condition is z() = 3(a) The approximating functions are defined recursively by φ n+ (t) = t 2 [φ n (s) + ] ds Setting φ (t) =, φ (t) = 2t Continuing, φ 2 (t) = 2t 2 + 2t, φ 3 (t) = 4t 3 /3 + 2t 2 + 2t, φ 4 (t) = 2t 4 /3 + 4t 3 /3 + 2t 2 + 2t, Based upon these we conjecture that φ n (t) = n k= 2k t k /k! and use mathematical induction to verify this form for φ n (t) First, let n =, then φ n (t) = 2t, so it is certainly true for n = Then, using Eq(7) again we have t [ t n ] 2 k n+ 2 k φ n+ (t) = 2 [φ n (s) + ] ds = 2 k! sk + ds = k! tk, k= and we have verified our conjecture k=

51 28 5 (b) (c) Recall from calculus that e at = + k= ak t k /k! Thus 2 k φ(t) = k! tk = e 2t (d) k= From the plot it appears that φ 4 is a good estimate for t < /2 5(a) The approximating functions are defined recursively by φ n+ (t) = t [ φ n (s)/2 + s] ds Setting φ (t) =, φ (t) = t 2 /2 Continuing, φ 2 (t) = t 2 /2 t 3 /2, φ 3 (t) = t 2 /2 t 3 /2 + t 4 /96, φ 4 (t) = t 2 /2 t 3 /2 + t 4 /96 t 5 /96, Based upon these we conjecture that φ n (t) = n k= 4( /2)k+ t k+ /(k + )! and use mathematical induction to verify this form for φ n (t)

52 52 Chapter 2 First Order Differential Equations (b) (c) Recall from calculus that e at = + k= ak t k /k! Thus φ(t) = 4 ( /2)k+ t k+ = 4e t/2 + 2t 4 k +! (d) k= From the plot it appears that φ 4 is a good estimate for t < 2 6(a) The approximating functions are defined recursively by φ n+ (t) = t [φ n (s) + s] ds Setting φ (t) =, φ (t) = t t 2 /2, φ 2 (t) = t t 3 /6, φ 3 (t) = t t 4 /24, φ 4 (t) = t t 5 /2, Based upon these we conjecture that φ n (t) = t t n+ /(n + )! and use mathematical induction to verify this form for φ n (t)

53 28 53 (b) (c) Clearly φ(t) = t (d) From the plot it appears that φ 4 is a good estimate for t < 8(a) The approximating functions are defined recursively by φ n+ (t) = t [ s 2 φ n (s) s ] ds Set φ (t) = The iterates are given by φ (t) = t 2 /2, φ 2 (t) = t 2 /2 t 5 /, φ 3 (t) = t 2 /2 t 5 / t 8 /8, φ 4 (t) = t 2 /2 t 5 / t 8 /8 t /88, Upon inspection, it becomes apparent that φ n (t) = t 2 [ 2 + t t (t 3 ) n [2 + 3(n )] = t 2 n k= (t 3 ) k [2 + 3(k )] ] =

54 54 Chapter 2 First Order Differential Equations (b) (c) Using the identity φ n (t) = φ (t) + [φ 2 (t) φ (t)] + [φ 3 (t) φ 2 (t)] + + [φ n (t) φ n (t)], consider the series φ (t) + k= [φ k+(t) φ k (t)] Fix any t value now We use the Ratio Test to prove the convergence of this series: φ k+ (t) φ k (t) φ k (t) φ k (t) = ( t 2 )(t 3 ) k 2 5 (2+3k) ( t 2 )(t 3 ) k 2 5 (2+3(k )) = t k The limit of this quantity is for any fixed t as k, and we obtain that φ n (t) is convergent for any t 9(a) The approximating functions are defined recursively by φ n+ (t) = t [ s 2 + φ 2 n(s) ] ds Set φ (t) = The first three iterates are given by φ (t) = t 3 /3, φ 2 (t) = t 3 /3 + t 7 /63, φ 3 (t) = t 3 /3 + t 7 /63 + 2t /279 + t 5 /59535 (b) The iterates appear to be converging

55 (a) The approximating functions are defined recursively by t [ 3s 2 ] + 4s + 2 φ n+ (t) = ds 2(φ n (s) ) Note that /(2y 2) = (/2) 6 k= yk + O(y 7 ) For computational purposes, use the geometric series sum to replace the above iteration formula by φ n+ (t) = [ ] t 6 (3s 2 + 4s + 2) φ k 2 n(s) ds Set φ (t) = The first four approximations are given by φ (t) = t t 2 t 3 /2, φ 2 (t) = t t 2 /2 + t 3 /6 + t 4 /4 t 5 /5 t 6 /24 +, φ 3 (t) = t t 2 /2 + t 4 /2 3t 5 /2 + 4t 6 /45 +, φ 4 (t) = t t 2 /2 + t 4 /8 7t 5 /6 + t 6 /5 + (b) k= The approximations appear to be converging to the exact solution, which can be found by separating the variables: φ(t) = + 2t + 2t 2 + t 3 4(a) φ n () =, for every n Let a (, ] Then φ n (a) = 2na e na2 = 2na/e na2 Using l Hospital s rule, lim z 2az/e az2 = lim z /ze az2 = Hence lim n φ n (a) = (b) 2nx dx = e e nx2 nx2 = e n Therefore, lim n φ n (x)dx lim φ n(x)dx n 5 Let t be fixed, such that (t, y ), (t, y 2 ) D Without loss of generality, assume that y < y 2 Since f is differentiable with respect to y, the mean value theorem asserts that there exists ξ (y, y 2 ) such that f(t, y ) f(t, y 2 ) = f y (t, ξ)(y y 2 ) This means that f(t, y ) f(t, y 2 ) = f y (t, ξ) y y 2 Since, by assumption, f/ y is continuous in D, f y attains a maximum K on any closed and bounded subset of D Hence f(t, y ) f(t, y 2 ) K y y 2 6 For a sufficiently small interval of t, φ n (t), φ n (t) D Since f satisfies a Lipschitz condition, f(t, φ n (t)) f(t, φ n (t)) K φ n (t) φ n (t) Here K = max f y

56 56 Chapter 2 First Order Differential Equations 7(a) φ (t) = t f(s, )ds Hence φ (t) t f(s, ) ds t Mds = M t, in which M is the maximum value of f(t, y) on D (b) By definition, φ 2 (t) φ (t) = t [f(s, φ (s)) f(s, )] ds Taking the absolute value of both sides, φ 2 (t) φ (t) t [f(s, φ (s)) f(s, )] ds Based on the results in Problems 6 and 7, φ 2 (t) φ (t) t K φ (s) ds KM t s ds Evaluating the last integral, we obtain that φ 2 (t) φ (t) MK t 2 /2 (c) Suppose that for some i By definition, It follows that φ i+ (t) φ i (t) = φ i+ (t) φ i (t) t φ i (t) φ i (t) MKi t i t t K φ i (s) φ i (s) ds = MK i t i+ (i + )! i! [f(s, φ i (s)) f(s, φ i (s))] ds f(s, φ i (s)) f(s, φ i (s)) ds t MK i h i+ (i + )! Hence, by mathematical induction, the assertion is true 8(a) Use the triangle inequality, a + b a + b K MK i s i ds = i! (b) For t h, φ (t) Mh, and φ n (t) φ n (t) MK n h n /(n!) Hence n K i h i φ n (t) M = M n (Kh) i i! K i! i= (c) The sequence of partial sums in (b) converges to M(e Kh )/K By the comparison test, the sums in (a) also converge Since individual terms of a convergent series must tend to zero, φ n (t) φ n (t), and it follows that the sequence φ n (t) is convergent 9(a) Let φ(t) = t f(s, φ(s))ds and ψ(t) = t f(s, ψ(s))ds Then by linearity of the integral, φ(t) ψ(t) = t [f(s, φ(s)) f(s, ψ(s))] ds (b) It follows that φ(t) ψ(t) t f(s, φ(s)) f(s, ψ(s)) ds i=

57 29 57 (c) We know that f satisfies a Lipschitz condition, f(t, y ) f(t, y 2 ) K y y 2, based on f/ y K in D Therefore, φ(t) ψ(t) t f(s, φ(s)) f(s, ψ(s)) ds t K φ(s) ψ(s) ds 29 Writing the equation for each n, y = 9 y, y 2 = 9 y = ( 9 2 )y, y 3 = 9 y 2 = ( 9) 3 y and so on, it is apparent that y n = ( 9) n y The terms constitute an alternating series, which converge to zero, regardless of y 3 Write the equation for each n, y = 3 y, y 2 = 4/2 y, y 3 = 5/3 y 2, Upon substitution, we find that y 2 = (4 3)/2 y, y 3 = (5 4 3)/(3 2) y, It can be proved by mathematical induction, that y n = 2 (n + 2)! n! This sequence is divergent, except for y = y = 2 (n + )(n + 2) y 4 Writing the equation for each n, y = y, y 2 = y, y 3 = y 2, y 4 = y 3, and so on It can be shown that { y, for n = 4k or n = 4k y n = y, for n = 4k 2 or n = 4k 3 The sequence is convergent only for y = 6 Writing the equation for each n, y = 5 y + 6 y 2 = 5 y + 6 = 5( 5 y + 6) + 6 = ( 5) 2 y ( 5)6 y 3 = 5 y = 5( 5 y + 6) + 6 = ( 5) 3 y + 6 [ + ( 5) + ( 5) 2] y n = ( 5) n y + 4 [ ( 5) n ] which follows from Eq(3) and (4) The sequence is convergent for all y, and in fact y n 4 8 Let y n be the balance at the end of the nth month Then y n+ = ( + r/2)y n + 25 We have y n = ρ n [y 25/( ρ)] + 25/( ρ), in which ρ = ( + r/2) Here r is the annual interest rate, given as 8% Thus y 36 = (66) 36 [ /r] 2 25/r = $2, Let y n be the balance due at the end of the nth month The appropriate difference equation is y n+ = ( + r/2) y n P Here r is the annual interest rate

58 58 Chapter 2 First Order Differential Equations and P is the monthly payment The solution, in terms of the amount borrowed, is given by y n = ρ n [y + P/( ρ)] P/( ρ), in which ρ = ( + r/2) and y = 8, To figure out the monthly payment P, we require that y 36 = That is, ρ 36 [y + P/( ρ)] = P/( ρ) After the specified amounts are substituted, we find that P = $2584 Let y n be the balance due at the end of the nth month The appropriate difference equation is y n+ = ( + r/2) y n P, in which r = 9 and P is the monthly payment The initial value of the mortgage is y = $, Then the balance due at the end of the n-th month is y n = ρ n [y + P/( ρ)] P/( ρ), where ρ = ( + r/2) In terms of the specified values, y n = (75) n [ 5 2P/r] + 2P/r Setting n = 3 2 = 36, and y 36 =, we find that P = $8462 For the monthly payment corresponding to a 2 year mortgage, set n = 24 and y 24 = to find that P = $89973 The total amount paid during the term of the loan is = $289, 6632 for the 3-year loan and is = $25, 9352 for the 2-year loan 2 Let y n be the balance due at the end of the nth month, with y the initial value of the mortgage The appropriate difference equation is y n+ = ( + r/2) y n P, in which r = and P = $ is the maximum monthly payment Given that the life of the mortgage is 2 years, we require that y 24 = The balance due at the end of the n-th month is y n = ρ n [y + P/( ρ)] P/( ρ) In terms of the specified values for the parameters, the solution of (833) 24 [y 2 /] = 2 / is y = $3, (a) δ 2 = (ρ 2 ρ )/(ρ 3 ρ 2 ) = (3449 3)/( ) = (b) diff= ( δ δ 2 /δ) = ( /46692) 22% (c) Assuming (ρ 3 ρ 2 )/(ρ 4 ρ 3 ) = δ, ρ (d) A period 6 solution appears near ρ 3565 (e) Note that (ρ n+ ρ n ) = δ n (ρ n ρ n ) With the assumption that δ n = δ, we have (ρ n+ ρ n ) = δ (ρ n ρ n ), which is of the form y n+ = α y n, n 3 It

59 29 59 follows that (ρ k ρ k ) = δ 3 k (ρ 3 ρ 2 ) for k 4 Then ρ n = ρ + (ρ 2 ρ ) + (ρ 3 ρ 2 ) + (ρ 4 ρ 3 ) + + (ρ n ρ n ) = ρ + (ρ 2 ρ ) + (ρ 3 ρ 2 ) [ + δ + δ δ 3 n] [ ] δ 4 n = ρ + (ρ 2 ρ ) + (ρ 3 ρ 2 ) δ [ ] δ Hence lim n ρ n = ρ 2 + (ρ 3 ρ 2 ) δ Substitution of the appropriate values yields lim ρ n = n PROBLEMS The equation is linear It can be written in the form y + 2y/x = x 2, and the integrating factor is µ(x) = e (2/x) dx = e 2 ln x = x 2 Multiplication by µ(x) yields x 2 y + 2yx = (yx 2 ) = x 4 Integration with respect to x and division by x 2 gives that y = x 3 /5 + c/x 2 5 The equation is exact Algebraic manipulations give the symmetric form of the equation, (2xy + y 2 + )dx + (x 2 + 2xy)dy = We can check that M y = 2x + 2y = N x, so the equation is really exact Integrating M with respect to x gives that ψ(x, y) = x 2 y + xy 2 + x + g(y), then ψ y = x 2 + 2xy + g (y) = x 2 + 2xy, so we get that g (y) =, so we obtain that g(y) = is acceptable Therefore the solution is defined implicitly as x 2 y + xy 2 + x = c 6 The equation is linear It can be written in the form y + ( + (/x))y = /x and the integrating factor is µ(x) = e +(/x) dx = e x+ln x = xe x Multiplication by µ(x) yields xe x y + (xe x + e x )y = (xe x y) = e x Integration with respect to x and division by xe x shows that the general solution of the equation is y = /x + c/(xe x ) The initial condition implies that = + c/e, which means that c = e and the solution is y = /x e/(xe x ) = x ( e x ) 7 The equation is separable Separation of variables gives the differential equation y(2 + 3y)dy = (4x 3 + )dx, and then after integration we obtain that the solution is x 4 + x y 2 y 3 = c 8 The equation is linear It can be written in the form y + 2y/x = sin x/x 2 and the integrating factor is µ(x) = e (2/x) dx = e 2 ln x = x 2 Multiplication by µ(x) gives x 2 y + 2xy = (x 2 y) = sin x, and after integration with respect to x and division by x 2 we obtain the general solution y = (c cos x)/x 2 The initial condition implies that c = 4 + cos 2 and the solution becomes y = (4 + cos 2 cos x)/x 2 The equation is exact It is easy to check that M y = = N x Integrating M with respect to x gives that ψ(x, y) = x 3 /3 + xy + g(y), then ψ y = x + g (y) =

60 6 Chapter 2 First Order Differential Equations x + e y, which means that g (y) = e y, so we obtain that g(y) = e y Therefore the solution is defined implicitly as x 3 /3 + xy + e y = c 3 The equation is separable Factoring the right hand side leads to the equation y = ( + y 2 )( + 2x) We separate the variables to obtain dy/( + y 2 ) = ( + 2x)dx, then integration gives us arctan y = x + x 2 + c The solution is y = tan(x + x 2 + c) 4 The equation is exact We can check that M y = = N x Integrating M with respect to x gives that ψ(x, y) = x 2 /2 + xy + g(y), then ψ y = x + g (y) = x + 2y, which means that g (y) = 2y, so we obtain that g(y) = y 2 Therefore the general solution is defined implicitly as x 2 /2 + xy + y 2 = c The initial condition gives us c = 7, so the solution is x 2 + 2xy + 2y 2 = 34 5 The equation is separable Separation of variables leads us to the equation dy y = ex + e x dx Note that + e x 2e x = e x We obtain that e x ln y = + e x dx = 2ex + e x dx = x 2 ln( + ex ) + c This means that y = ce x ( + e x ) 2, which also can be written as y = c/ cosh 2 (x/2) after some algebraic manipulations 6 The equation is exact The symmetric form is ( e x cos y + e 2y cos x)dx + ( e x sin y + 2e 2y sin x)dy = We can check that M y = e x sin y + 2e 2y cos x = N x Integrating M with respect to x gives that ψ(x, y) = e x cos y + e 2y sin x + g(y), then ψ y = e x sin y + 2e 2y sin x + g (y) = e x sin y + 2e 2y sin x, so we get that g (y) =, so we obtain that g(y) = is acceptable Therefore the solution is defined implicitly as e x cos y + e 2y sin x = c 7 The equation is linear The integrating factor is µ(x) = e 3 dx = e 3x, which turns the equation into e 3x y 3e 3x y = (e 3x y) = e x We integrate with respect to x to obtain e 3x y = e x + c, and the solution is y = ce 3x e 2x after multiplication by e 3x 8 The equation is linear The integrating factor is µ(x) = e 2 dx = e 2x, which gives us e 2x y + 2e 2x y = (e 2x y) = e x2 The antiderivative of the function on the right hand side can not be expressed in a closed form using elementary functions, so we have to express the solution using integrals Let us integrate both sides of this equation from to x We obtain that the left hand side turns into x (e 2s y(s)) ds = e 2x y(x) e y() = e 2x y 3 The right hand side gives us x e s2 ds So we found that x y = e 2x e s2 ds + 3e 2x

61 The equation is exact Algebraic manipulations give us the symmetric form (y 3 + 2y 3x 2 )dx + (2x + 3xy 2 )dy = We can check that M y = 3y = N x Integrating M with respect to x gives that ψ(x, y) = xy 3 + 2xy x 3 + g(y), then ψ y = 3xy 2 + 2x + g (y) = 2x + 3xy 2, which means that g (y) =, so we obtain that g(y) = is acceptable Therefore the solution is xy 3 + 2xy x 3 = c 2 The equation is separable, because y = e x+y = e x e y Separation of variables yields the equation e y dy = e x dx, which turns into e y = e x + c after integration and we obtain the implicitly defined solution e x + e y = c 22 The equation is separable Separation of variables turns the equation into (y 2 + )dy = (x 2 )dx, which, after integration, gives y 3 /3 + y = x 3 /3 x + c The initial condition yields c = 2/3, and the solution is y 3 + 3y x 3 + 3x = 2 23 The equation is linear Division by t gives y + ( + (/t))y = e 2t /t, so the integrating factor is µ(t) = e (+(/t))dt = e t+ln t = te t The equation turns into te t y + (te t + e t )y = (te t y) = e 3t Integration therefore leads to te t y = e 3t /3 + c and the solution is y = e 2t /(3t) + ce t /t 24 The equation is exact We can check that M y = 2 cos y sin x cos x = N x Integrating M with respect to x gives that ψ(x, y) = sin y sin 2 x + g(y), then ψ y = cos y sin 2 x + g (y) = cos y sin 2 x, which means that g (y) =, so we obtain that g(y) = is acceptable Therefore the solution is defined implicitly as sin y sin 2 x = c 25 The equation is exact We can check that M y = 2x y 2 x2 y 2 (x 2 + y 2 ) 2 = N x Integrating M with respect to x gives that ψ(x, y) = x 2 /y + arctan(y/x) + g(y), then ψ y = x 2 /y 2 + x/(x 2 + y 2 ) + g (y) = x/(x 2 + y 2 ) x 2 /y 2, which means that g (y) =, so we obtain that g(y) = is acceptable Therefore the solution is defined implicitly as x 2 /y + arctan(y/x) = c 28 The equation can be made exact by choosing an appropriate integrating factor We can check that (M y N x )/N = (2 )/x = /x depends only on x, so µ(x) = e (/x)dx = e ln x = x is an integrating factor After multiplication, the equation becomes (2yx + 3x 2 )dx + x 2 dy = This equation is exact now, because M y = 2x = N x Integrating M with respect to x gives that ψ(x, y) = yx 2 + x 3 + g(y), then ψ y = x 2 + g (y) = x 2, which means that g (y) =, so we obtain that g(y) = is acceptable Therefore the solution is defined implicitly as x 3 + x 2 y = c 29 The equation is homogeneous (See Section 22, Problem 3) We can see that y = x + y x y = + (y/x) (y/x) We substitute u = y/x, which means also that y = ux and then y = u x + u =

62 62 Chapter 2 First Order Differential Equations ( + u)/( u), which implies that u x = + u u u = + u2 u, a separable equation Separating the variables yields u + u 2 du = dx x, and then integration gives arctan u ln( + u 2 )/2 = ln x + c Substituting u = y/x back into this expression and using that ln( + (y/x) 2 )/2 ln x = ln( x + (y/x) 2 ) = ln( x 2 + y 2 ) we obtain that the solution is arctan(y/x) ln( x 2 + y 2 ) = c 3 The equation is homogeneous (See Section 22, Problem 3) Algebraic manipulations show that it can be written in the form y = 3y2 + 2xy 2xy + x 2 Substituting u = y/x gives that y = ux and then which implies that = 3(y/x)2 + 2(y/x) 2(y/x) + y = u x + u = 3u2 + 2u 2u +, u x = 3u2 + 2u 2u + u = u2 + u 2u +, a separable equation We obtain that (2u + )du/(u 2 + u) = dx/x, which in turn means that ln(u 2 + u) = ln x + c Therefore, u 2 + u = cx and then substituting u = y/x gives us the solution (y 2 /x 3 ) + (y/x 2 ) = c 3 The equation can be made exact by choosing an appropriate integrating factor We can check that (M y N x )/M = (3x 2 + y)/(y(3x 2 + y)) = /y depends only on y, so µ(y) = e (/y)dy = e ln y = y is an integrating factor After multiplication, the equation becomes (3x 2 y 2 + y 3 )dx + (2x 3 y + 3xy 2 )dy = This equation is exact now, because M y = 6x 2 y + 3y 2 = N x Integrating M with respect to x gives that ψ(x, y) = x 3 y 2 + y 3 x + g(y), then ψ y = 2x 3 y + 3y 2 x + g (y) = 2x 3 y + 3xy 2, which means that g (y) =, so we obtain that g(y) = is acceptable Therefore the general solution is defined implicitly as x 3 y 2 + xy 3 = c The initial condition gives us 4 8 = c = 4, and the solution is x 3 y 2 + xy 3 = 4 33 Let y be a solution, ie y = q + q 2 y + q 3 y 2 Now let y = y + (/v) also be a solution Differentiating this expression with respect to t and using that y is also a solution we obtain y = y (/v 2 )v = q + q 2 y + q 3 y 2 = q + q 2 (y + (/v)) + q 3 (y + (/v)) 2 Now using that y was also a solution we get that (/v 2 )v = q 2 (/v) + 2q 3 (y /v) + q 3 (/v 2 ), which, after some simple algebraic manipulations turns into v = (q 2 + 2q 3 y )v q 3

63 (a) The equation is y = ( y)(x + by) = x + (b x)y by 2 We set y = + (/v) and differentiate: y = v 2 v = x + (b x)( + (/v)) b( + (/v)) 2, which, after simplification, turns into v = (b + x)v + b (b) When x = at, the equation is v (b + at)v = b, so the integrating factor is µ(t) = e bt at2 /2 This turns the equation into (vµ(t)) = bµ(t), so vµ(t) = bµ(t)dt, and then v = (b µ(t)dt)/µ(t) 36 Substitute v = y, then v = y The equation turns into t 2 v + 2tv = (t 2 v) =, which yields t 2 v = t + c, so y = v = (/t) + (c /t 2 ) Integrating this expression gives us the solution y = ln t (c /t) + c 2 37 Set v = y, then v = y The equation with this substitution is tv + v = (tv) =, which gives tv = t + c, so y = v = + (c /t) Integrating this expression yields the solution y = t + c ln t + c 2 38 Set v = y, so v = y The equation is v + tv 2 =, which is a separable equation Separating the variables we obtain dv/v 2 = tdt, so /v = t 2 /2 + c, and then y = v = 2/(t 2 + c ) Now depending on the value of c, we have the following possibilities: when c =, then y = 2/t + c 2, when < c = k 2, then y = (2/k) arctan(t/k) + c 2, and when > c = k 2 then y = (/k) ln (t k)/(t + k) + c 2 We also divided by v = y when we separated the variables, and v = (which is y = c) is also a solution 39 Substitute v = y and v = y The equation is 2t 2 v + v 3 = 2tv This is a Bernoulli equation (See Section 24, Problem 27), so the substitution z = v 2 yields z = 2v 3 v, and the equation turns into 2t 2 v v 3 + = 2t/v 2, ie into 2t 2 z /2 + = 2tz, which in turn simplifies to t 2 z + 2tz = (t 2 z) = Integration yields t 2 z = t + c, which means that z = (/t) + (c/t 2 ) Now y = v = ± /z = ±t/ t + c and another integration gives y = ± 2 3 (t 2c ) t + c + c 2 The substitution also loses the solution v =, ie y = c 4 Set v = y, then v = y The equation reads v + v = e t, which is a linear equation with integrating factor µ(t) = e t This turns the equation into e t v + e t v = (e t v) =, which means that e t v = t + c and then y = v = te t + ce t Another integration yields the solution y = te t + c e t + c 2 4 Let v = y and v = y The equation is t 2 v = v 2, which is a separable equation Separating the variables we obtain dv/v 2 = dt/t 2, which gives us /v = (/t) + c, and then y = v = t/( + c t) Now when c =, then y = t 2 /2 + c 2, and when c, then y = t/c (ln + c t )/c 2 + c 2 Also, at the separation we divided by v =, which also gives us the solution y = c

64 64 Chapter 2 First Order Differential Equations 43 Set y = v(y) Then y = v (y)(dy/dt) = v (y)v(y) We obtain the equation v v + y =, where the differentiation is with respect to y This is a separable equation which simplifies to vdv = ydy We obtain that v 2 /2 = y 2 /2 + c, so y = v(y) = ± c y 2 We separate the variables again to get dy/ c y 2 = ±dt, so arcsin(y/ c) = t + d, which means that y = c sin(±t + d) = c sin(t + c 2 ) 44 Set y = v(y) Then y = v (y)(dy/dt) = v (y)v(y) We obtain the equation v v + yv 3 =, where the differentiation is with respect to y Separation of variables turns this into dv/v 2 = ydy, which gives us y = v = 2/(c + y 2 ) This implies that (c + y 2 )dy = 2dt and then the solution is defined implicitly as c y + y 3 /3 = 2t + c 2 Also, y = c is a solution which we lost when divided by y = v = 46 Set y = v(y) Then y = v (y)(dy/dt) = v (y)v(y) We obtain the equation yv v v 3 =, where the differentiation is with respect to y This separable equation gives us dv/v 2 = dy/y, which means that /v = ln y + c, and then y = v = /(c ln y ) We separate variables again to obtain (c ln y )dy = dt, and then integration yields the implicitly defined solution cy (y ln y y) = t + d Also, y = c is a solution which we lost when we divided by v = 49 Set y = v(y) Then y = v (y)(dy/dt) = v (y)v(y) We obtain the equation v v 3y 2 =, where the differentiation is with respect to y Separation of variables gives vdv = 3y 2 dy, and after integration this turns into v 2 /2 = y 3 + c The initial conditions imply that c = here, so (y ) 2 = v 2 = 2y 3 This implies that y = 2y 3/2 (the sign is determined by the initial conditions again), and this separable equation now turns into y 3/2 dy = 2dt Integration yields 2y /2 = 2t + d, and the initial conditions at this point give that d = 2 Algebraic manipulations find that y = 2( t) 2 5 Set v = y, then v = y The equation with this substitution turns into the equation ( + t 2 )v + 2tv = (( + t 2 )v) = 3t 2 Integrating this we get that ( + t 2 )v = 3t + c, and c = 5 from the initial conditions This means that y = v = 3/(t( + t 2 )) 5/( + t 2 ) The partial fraction decomposition of the first expression shows that y = 3/t 3t/( + t 2 ) 5/( + t 2 ) and then another integration here gives us that y = 3 ln t (3/2) ln( + t 2 ) 5 arctan t + d The initial conditions identify d = 2 + (3/2) ln 2 + 5π/4, and we obtained the solution

65 C H A P T E R 3 Second Order Linear Equations 3 Let y = e rt, so that y = r e rt and y = r 2 e rt Direct substitution into the differential equation yields (r 2 + 2r 3)e rt = Canceling the exponential, the characteristic equation is r 2 + 2r 3 = The roots of the equation are r = 3, Hence the general solution is y = c e t + c 2 e 3t 2 Let y = e rt Substitution of the assumed solution results in the characteristic equation r 2 + 3r + 2 = The roots of the equation are r = 2, Hence the general solution is y = c e t + c 2 e 2t 4 Substitution of the assumed solution y = e rt results in the characteristic equation 2r 2 3r + = The roots of the equation are r = /2, Hence the general solution is y = c e t/2 + c 2 e t 6 The characteristic equation is 4r 2 9 =, with roots r = ±3/2 Therefore the general solution is y = c e 3t/2 + c 2 e 3t/2 8 The characteristic equation is r 2 2r 2 =, with roots r = ± 3 Hence the general solution is y = c e ( 3)t + c 2 e (+ 3)t 9 Substitution of the assumed solution y = e rt results in the characteristic equation r 2 + r 2 = The roots of the equation are r = 2, Hence the general solution is y = c e 2t + c 2 e t Its derivative is y = 2c e 2t + c 2 e t Based on the 65

66 66 Chapter 3 Second Order Linear Equations first condition, y() =, we require that c + c 2 = In order to satisfy y () =, we find that 2c + c 2 = Solving for the constants, c = and c 2 = Hence the specific solution is y(t) = e t It clearly increases without bound as t Substitution of the assumed solution y = e rt results in the characteristic equation 6r 2 5r + = The roots of the equation are r = /3, /2 Hence the general solution is y = c e t/3 + c 2 e t/2 Its derivative is y = c e t/3 /3 + c 2 e t/2 /2 Based on the first condition, y() =, we require that c + c 2 = 4 In order to satisfy the condition y () =, we find that c /3 + c 2 /2 = Solving for the constants, c = 2 and c 2 = 8 Hence the specific solution is y(t) = 2 e t/3 8 e t/2 It clearly decreases without bound as t 2 The characteristic equation is r 2 + 3r =, with roots r = 3, Therefore the general solution is y = c + c 2 e 3t, with derivative y = 3 c 2 e 3t In order to satisfy the initial conditions, we find that c + c 2 = 2, and 3 c 2 = 3 Hence the specific solution is y(t) = e 3t This converges to as t

67 The characteristic equation is r 2 + 5r + 3 =, with roots r = ( 5 ± 3 )/2 The general solution is y = c e ( 5 3)t/2 + c 2 e ( 5+ 3)t/2, with derivative y = c e ( 5 3)t/ In order to satisfy the initial conditions, we require that c + c 2 = and c c 2 e ( 5+ 3)t/2 c 2 = Solving for the coefficients, c = ( 5/ 3 )/2 and c 2 = ( + 5/ 3 )/2 The solution clearly converges to as t 4 The characteristic equation is 2r 2 + r 4 =, with roots r = ( ± 33 )/4 The general solution is y = c e ( 33)t/4 + c 2 e ( + 33)t/4, with derivative y = 33 4 c e ( 33)t/ In order to satisfy the initial conditions, we require that c + c 2 = and 33 4 c c 2 e ( + 33)t/4 c 2 = Solving for the coefficients, c = 2/ 33 and c 2 = 2/ 33 The specific solution is [ ] y(t) = 2 e ( 33)t/4 e ( + 33)t/4 / 33

68 68 Chapter 3 Second Order Linear Equations It clearly increases without bound as t 6 The characteristic equation is 4r 2 =, with roots r = ±/2 Therefore the general solution is y = c e t/2 + c 2 e t/2 Since the initial conditions are specified at t = 2, is more convenient to write y = d e (t+2)/2 + d 2 e (t+2)/2 The derivative is given by y = [ d e (t+2)/2] /2 + [ d 2 e (t+2)/2] /2 In order to satisfy the initial conditions, we find that d + d 2 =, and d /2 + d 2 /2 = Solving for the coefficients, d = 3/2, and d 2 = /2 The specific solution is y(t) = 3 2 e (t+2)/2 2 e(t+2)/2 = 3 2e e t/2 e 2 et/2 It clearly decreases without bound as t 8 An algebraic equation with roots 2 and /2 is 2r 2 + 5r + 2 = This is the characteristic equation for the differential equation 2y + 5y + 2 y = 2 The characteristic equation is 2r 2 3r + =, with roots r = /2, Therefore the general solution is y = c e t/2 + c 2 e t, with derivative y = c e t/2 /2 + c 2 e t In order to satisfy the initial conditions, we require c + c 2 = 2 and c /2 + c 2 = /2 Solving for the coefficients, c = 3, and c 2 = The specific solution is y(t) = 3e t/2 e t To find the stationary point, set y = 3e t/2 /2 e t = There is a unique solution, with t = ln(9/4) The maximum value is then y(t ) = 9/4 To find the x-intercept, solve the equation 3e t/2 e t = The solution is readily found to be t 2 = ln

69 The characteristic equation is 4r 2 =, with roots r = ±/2 Hence the general solution is y = c e t/2 + c 2 e t/2 and y = c e t/2 /2 + c 2 e t/2 /2 Invoking the initial conditions, we require that c + c 2 = 2 and c + c 2 = 2β The specific solution is y(t) = ( β)e t/2 + ( + β)e t/2 Based on the form of the solution, it is evident that as t, y(t) as long as β = 23 The characteristic equation is r 2 (2α )r + α(α ) = Examining the coefficients, the roots are r = α, α Hence the general solution of the differential equation is y(t) = c e αt + c 2 e (α )t Assuming α R, all solutions will tend to zero as long as α < On the other hand, all solutions will become unbounded as long as α >, that is, α > 26(a) The characteristic roots are r = 3, 2 The solution of the initial value problem is y(t) = (6 + β)e 2t (4 + β)e 3t (b) The maximum point has coordinates t = ln [(3(4 + β))/(2(6 + β))], y = 4(6 + β) 3 /(27(4 + β) 2 ) (c) y = 4(6 + β) 3 /(27(4 + β) 2 ) 4, as long as β (d) lim β t = ln(3/2), lim β y = 27(a) Assuming that y is a constant, the differential equation reduces to c y = d Hence the only equilibrium solution is y = d/c (b) Setting y = Y + d/c, substitution into the differential equation results in the equation a Y + b Y + c(y + d/c ) = d The equation satisfied by Y is a Y + b Y + cy = 32 3 W (e 2t, e 3t/2 ) = W (e 2t, t e 2t ) = e2t 2e 2t e 2t 2e 2t e 3t/2 3 2 e 3t/2 = 7 2 et/2 te 2t ( 2t)e 2t = e 4t 5 W (e t sin t, e t cos t) = e t sin t e t cos t e t (sin t + cos t) e t (cos t sin t) = e2t 6 W (cos 2 θ, + cos 2θ) = cos 2 θ 2 sin θ cos θ + cos 2θ 2 sin 2θ =

70 7 Chapter 3 Second Order Linear Equations 7 Write the equation as y + (3/t)y = p(t) = 3/t is continuous for all t > Since t >, the IVP has a unique solution for all t > 9 Write the equation as y + (3/(t 4))y + (4/t(t 4))y = 2/t(t 4) The coefficients are not continuous at t = and t = 4 Since t (, 4), the largest interval is < t < 4 The coefficient 3 ln t is discontinuous at t = Since t >, the largest interval of existence is < t < Write the equation as y + (x/(x 3))y + (ln x /(x 3))y = The coefficients are discontinuous at x = and x = 3 Since x (, 3), the largest interval is < x < 3 3 y = 2 We see that t 2 (2) 2(t 2 ) = y 2 = 2 t 3, with t 2 (y 2 ) 2(y 2 ) = Let y 3 = c t 2 + c 2 t, then y 3 = 2c + 2c 2 t 3 It is evident that y 3 is also a solution 6 No Substituting y = sin(t 2 ) into the differential equation, 4t 2 sin(t 2 ) + 2 cos(t 2 ) + 2t cos(t 2 )p(t) + sin(t 2 )q(t) = At t =, this equation becomes 2 = (if we suppose that p(t) and q(t) are continuous), which is impossible 7 W (e 2t, g(t)) = e 2t g (t) 2e 2t g(t) = 3e 4t Dividing both sides by e 2t, we find that g must satisfy the ODE g 2g = 3e 2t Hence g(t) = 3t e 2t + c e 2t 9 W (f, g) = fg f g Also, W (u, v) = W (2f g, f + 2g) Upon evaluation, W (u, v) = 5fg 5f g = 5W (f, g) 2 W (f, g) = fg f g = t cos t sin t, and W (u, v) = 4fg + 4f g Hence W (u, v) = 4t cos t + 4 sin t 2 We compute W (a y + a 2 y 2, b y + b 2 y 2 ) = a y + a 2 y 2 b y + b 2 y 2 a y + a 2 y 2 b y + b 2 y 2 = = (a y + a 2 y 2 )(b y + b 2 y 2) (b y + b 2 y 2 )(a y + a 2 y 2) = = a b 2 (y y 2 y y 2 ) a 2 b (y y 2 y y 2 ) = (a b 2 a 2 b )W (y, y 2 ) This now readily shows that y 3 and y 4 form a fundamental set of solutions if and only if a b 2 a 2 b 23 The general solution is y = c e 3t + c 2 e t W (e 3t, e t ) = 2e 4t, and hence the exponentials form a fundamental set of solutions On the other hand, the fundamental solutions must also satisfy the conditions y () =, y () = ; y 2 () =, y 2() = For y, the initial conditions require c + c 2 = e, 3c c 2 = The

71 32 7 coefficients are c = e 3 /2, c 2 = 3e/2 For the solution y 2, the initial conditions require c + c 2 =, 3c c 2 = e The coefficients are c = e 3 /2, c 2 = e/2 Hence the fundamental solutions are y = 2 e 3(t ) e (t ) and y 2 = 2 e 3(t ) + 2 e (t ) 24 Yes y = 4 cos 2t ; y 2 = 4 sin 2t W (cos 2t, sin 2t) = 2 25 Clearly, y = e t is a solution y 2 = ( + t)e t, y 2 = (2 + t)e t Substitution into the ODE results in (2 + t)e t 2( + t)e t + t e t = Furthermore, W (e t, te t ) = e 2t Hence the solutions form a fundamental set of solutions 27 Clearly, y = x is a solution y 2 = cos x, y 2 = sin x Substitution into the ODE results in ( x cot x)( sin x) x(cos x) + sin x = We can compute that W (y, y 2 ) = x cos x sin x, which is nonzero for < x < π Hence {x, sin x} is a fundamental set of solutions 3 Writing the equation in standard form, we find that P (t) = sin t/ cos t Hence the Wronskian is W (t) = c e (sin t/ cos t) dt = c e ln cos t = c cos t, in which c is some constant 3 After writing the equation in standard form, we have P (x) = /x The Wronskian is W (x) = c e (/x) dx = c e ln x = c/x, in which c is some constant 32 Writing the equation in standard form, we find that P (x) = 2x/( x 2 ) The Wronskian is W (x) = c e 2x/( x 2 ) dx = c e ln x2 = c/( x 2 ), in which c is some constant 33 Rewrite the equation as p(t)y + p (t)y + q(t)y = After writing the equation in standard form, we have P (t) = p (t)/p(t) Hence the Wronskian is W (t) = c e p (t)/p(t) dt = c e ln p(t) = c/p(t) 35 The Wronskian associated with the solutions of the differential equation is given by W (t) = c e 2/t 2 dt = c e 2/t Since W (2) = 3, it follows that for the hypothesized set of solutions, c = 3 e Hence W (4) = 3 e 36 For the given differential equation, the Wronskian satisfies the first order differential equation W + p(t)w = Given that W is constant, it is necessary that p(t) 37 Direct calculation shows that W (f g, f h) = (fg)(fh) (fg) (fh) = (fg)(f h + fh ) (f g + fg )(fh) = f 2 W (g, h) 39 Since y and y 2 are solutions, they are differentiable The hypothesis can thus be restated as y (t ) = y 2(t ) = at some point t in the interval of definition

72 72 Chapter 3 Second Order Linear Equations This implies that W (y, y 2 )(t ) = But W (y, y 2 )(t ) = c e p(t)dt, which cannot be equal to zero, unless c = Hence W (y, y 2 ), which is ruled out for a fundamental set of solutions 42 P =, Q = x, R = We have P Q + R = The equation is exact Note that (y ) + (xy) = Hence y + xy = c This equation is linear, with integrating factor µ = e x2 /2 Therefore the general solution is y(x) = c e x2 /2 x x e u 2 /2 du + c 2 e x2 /2 43 P =, Q = 3x 2, R = x Note that P Q + R = 5x, and therefore the differential equation is not exact 45 P = x 2, Q = x, R = We have P Q + R = The equation is exact Write the equation as (x 2 y ) (xy) = After integration, we conclude that x 2 y xy = c Divide both sides of the differential equation by x 2 The resulting equation is linear, with integrating factor µ = /x Hence (y/x) = c x 3 The solution is y(t) = c x + c 2 x 47 P = x 2, Q = x, R = x 2 ν 2 Hence the coefficients are 2P Q = 3x and P Q + R = x 2 + ν 2 The adjoint of the original differential equation is given by x 2 µ + 3x µ + (x 2 + ν 2 )µ = 49 P =, Q =, R = x Hence the coefficients are given by 2P Q = and P Q + R = x Therefore the adjoint of the original equation is µ x µ = 33 2 e 2 3i = e 2 e 3i = e 2 (cos 3 i sin 3) 3 e iπ = cos π + i sin π = 4 e 2 (π/2) i = e 2 (cos(π/2) i sin(π/2)) = e 2 i 6 π +2i = e ( +2i) ln π = e ln π e 2 ln π i = (cos (2 ln π) + i sin (2 ln π))/π 8 The characteristic equation is r 2 2r + 6 =, with roots r = ± i 5 Hence the general solution is y = c e t cos 5 t + c 2 e t sin 5 t 9 The characteristic equation is r 2 + 2r 8 =, with roots r = 4, 2 The roots are real and different, hence the general solution is y = c e 4t + c 2 e 2t The characteristic equation is r 2 + 2r + 2 =, with roots r = ± i Hence the general solution is y = c e t cos t + c 2 e t sin t

73 The characteristic equation is 4r =, with roots r = ±(3/2) i Hence the general solution is y = c cos(3t/2) + c 2 sin(3t/2) 3 The characteristic equation is r 2 + 2r + 25 =, with roots r = ± i/2 Hence the general solution is y = c e t cos(t/2) + c 2 e t sin(t/2) 5 The characteristic equation is r 2 + r + 25 =, with roots r = (/2) ± i Hence the general solution is y = c e t/2 cos t + c 2 e t/2 sin t 6 The characteristic equation is r 2 + 4r =, with roots r = 2 ± (3/2) i Hence the general solution is y = c e 2t cos(3t/2) + c 2 e 2t sin(3t/2) 7 The characteristic equation is r =, with roots r = ± 2i Hence the general solution is y = c cos 2t + c 2 sin 2t Now y = 2c sin 2t + 2c 2 cos 2t Based on the first condition, y() =, we require that c = In order to satisfy the condition y () =, we find that 2c 2 = The constants are c = and c 2 = /2 Hence the specific solution is y(t) = sin 2t /2 The solution is periodic 9 The characteristic equation is r 2 2r + 5 =, with roots r = ± 2i Hence the general solution is y = c e t cos 2t + c 2 e t sin 2t Based on the initial condition y(π/2) =, we require that c = It follows that y = c 2 e t sin 2t, and so the first derivative is y = c 2 e t sin 2t + 2c 2 e t cos 2t In order to satisfy the condition y (π/2) = 2, we find that 2e π/2 c 2 = 2 Hence we have c 2 = e π/2 Therefore the specific solution is y(t) = e t π/2 sin 2t The solution oscillates with an exponentially growing amplitude

74 74 Chapter 3 Second Order Linear Equations 2 The characteristic equation is r 2 + =, with roots r = ± i Hence the general solution is y = c cos t + c 2 sin t Its derivative is y = c sin t + c 2 cos t Based on the first condition, y(π/3) = 2, we require that c + 3 c 2 = 4 In order to satisfy the condition y (π/3) = 4, we find that 3 c + c 2 = 8 Solving these for the constants, c = and c 2 = 3 2 Hence the specific solution is a steady oscillation, given by y(t) = ( + 2 3) cos t + ( 3 2) sin t 2 From Problem 5, the general solution is y = c e t/2 cos t + c 2 e t/2 sin t Invoking the first initial condition, y() = 3, which implies that c = 3 Substituting, it follows that y = 3e t/2 cos t + c 2 e t/2 sin t, and so the first derivative is y = 3 2 e t/2 cos t 3e t/2 sin t + c 2 e t/2 cos t c 2 2 e t/2 sin t Invoking the initial condition, y () =, we find that 3/2 + c 2 =, and so c 2 = 5/2 Hence the specific solution is y(t) = 3e t/2 cos t + (5/2) e t/2 sin t It oscillates with an exponentially decreasing amplitude 24(a) The characteristic equation is 5r 2 + 2r + 7 =, with roots r = ( ± i 34)/5 The solution is u = c e t/5 cos 34t/5 + c 2 e t/5 sin 34t/5 Invoking the given initial conditions, we obtain the equations for the coefficients : c = 2, c 2 = 5 That is, c = 2, c 2 = 7/ 34 Hence the specific solution is u(t) = 2e t/5 cos 34 5 t e t/5 sin 34 5 t

75 33 75 (b) Based on the graph of u(t), T is in the interval 4 < t < 6 solution on that interval yields T 455 A numerical 26(a) The characteristic equation is r 2 + 2ar + (a 2 + ) =, with roots r = a ± i Hence the general solution is y(t) = c e at cos t + c 2 e at sin t Based on the initial conditions, we find that c = and c 2 = a Therefore the specific solution is given by y(t) = e at cos t + a e at sin t = + a 2 e at cos (t φ), in which φ = arctan(a) (b) For estimation, note that y(t) + a 2 e at Now consider the inequality + a2 e at / The inequality holds for t (/a) ln( + a 2 ) Therefore T (/a) ln( + a 2 ) Setting a =, the numerical value is T 8763 (c) Similarly, T / , T /2 433, T 2 56 (d) Note that the estimates T a approach the graph of (/a) ln( + a 2 ) as a gets large 27 Direct calculation gives the result On the other hand, it was shown in Problem 3237 that W (f g, f h) = f 2 W (g, h) Hence W (e λt cos µt, e λt sin µt) = e 2λt W (cos µt, sin µt) = e 2λt [cos µt(sin µt) (cos µt) sin µt] = µ e 2λt 28(a) Clearly, y and y 2 are solutions Also, W (cos t, sin t) = cos 2 t + sin 2 t =

76 76 Chapter 3 Second Order Linear Equations (b) y = i e it, y = i 2 e it = e it Evidently, y is a solution and so y = c y + c 2 y 2 (c) Setting t =, = c cos + c 2 sin, and c = (d) Differentiating, i e it = c 2 cos t Setting t =, i = c 2 cos and hence c 2 = i Therefore e it = cos t + i sin t 29 Euler s formula is e it = cos t + i sin t It follows that e it = cos t i sin t Adding these equation, e it + e it = 2 cos t Subtracting the two equations results in e it e it = 2i sin t 3 Let r = λ + iµ, and r 2 = λ 2 + iµ 2 Then e (r+r2)t = e (λ+λ2)t+i(µ+µ2)t = e (λ+λ2)t [cos(µ + µ 2 )t + i sin(µ + µ 2 )t] = = e (λ+λ2)t [(cos µ t + i sin µ t)(cos µ 2 t + i sin µ 2 t)] = = e λt (cos µ t + i sin µ t) e λ2t (cos µ t + i sin µ t) = e rt e r2t Hence e (r+r2)t = e rt e r2t 32 Clearly, u = λe λt cos µt µe λt sin µt = e λt (λ cos µt µ sin µt) and then u = λe λt (λ cos µt µ sin µt) + e λt ( λµ sin µt µ 2 cos µt) Plugging these into the differential equation, dividing by e λt and arranging the sine and cosine terms we obtain that the identity to prove is (a(λ 2 µ 2 ) + bλ + c) cos µt + ( 2λµa bµ) sin µt = We know that λ ± iµ solves the characteristic equation ar 2 + br + c =, so a(λ iµ) 2 + b(λ iµ) + c = a(λ 2 µ 2 ) + bλ + c + i( 2λµa µb) = If this complex number is zero, then both the real and imaginary parts of it are zero, but those are the coefficients of cos µt and sin µt in the above identity, which proves that au + bu + cu = The solution for v is analogous 35 The equation transforms into y + y = The characteristic roots are r = ±i The solution is y = c cos(x) + c 2 sin(x) = c cos(ln t) + c 2 sin(ln t) 37 The equation transforms into y + 2y + 25y = The characteristic roots are r = ± i/2 The solution is y = c e x cos(x/2) + c 2 e x cos( 2 ln t) sin( 2 ln t) sin(x/2) = c + c 2 t t 38 The equation transforms into y 5y 6y = The characteristic roots are r =, 6 The solution is y = c e x + c 2 e 6x = c e ln t + c 2 e 6 ln t = c /t + c 2 t 6 39 The equation transforms into y 5y + 6y = The characteristic roots are r = 2, 3 The solution is y = c e 2x + c 2 e 3x = c e 2 ln t + c 2 e 3 ln t = c t 2 + c 2 t 3 4 The equation transforms into y + 2y 3y = The characteristic roots are r =, 3 The solution is y = c e x + c 2 e 3x = c e ln t + c 2 e 3 ln t = c t + c 2 /t 3

77 The equation transforms into y + 6y + y = The characteristic roots are r = 3 ± i The solution is y = c e 3x cos(x) + c 2 e 3x sin(x) = c t 3 cos(ln t) + c 2 sin(ln t) t3 43(a) By the chain rule, y (x) = (dy/dx)x In general, dz/dt = (dz/dx)(dx/dt) Setting z = (dy/dt), we have d 2 y dt 2 = dz dx dx dt = d [ ] [ dy dx dx d 2 ] dx dx dt dt = y dx dx dx 2 dt dt + dy [ ] d dx dx dx dx dt dt However, Hence d dx [ ] dx dx dt dt = d 2 y dt 2 = d2 y dx 2 [ d 2 ] x dt dt 2 dx dx dt = d2 x dt 2 [ ] 2 dx + dy d 2 x dt dx dt 2 (b) Substituting the results in part (a) into the general differential equation, y + p(t)y + q(t)y =, we find that d 2 [ ] 2 y dx dx 2 + dy d 2 x dy dx + p(t) dt dx dt2 dx dt + q(t)y = Collecting the terms, [ dx dt ] 2 d 2 [ y d 2 ] dx 2 + x dy dt 2 + p(t)dx dt dx + q(t)y = (c) Assuming (dx/dt) 2 = k q(t), and q(t) >, we find that dx/dt = k q(t), which can be integrated That is, x = u(t) = k q(t) dt = q(t) dt, since k = (d) Let k = It follows that d 2 x/dt 2 + p(t)dx/dt = du/dt + p(t)u(t) = q /2 q + p q Hence [ d 2 ] [ ] 2 x dx dt 2 + p(t)dx / = q (t) + 2p(t)q(t) dt dt 2 [q(t)] 3/2 As long as dx/dt, the differential equation can be expressed as [ ] d 2 y dx 2 + q (t) + 2p(t)q(t) dy 2 [q(t)] 3/2 dx + y = For the case q(t) <, write q(t) = [ q(t)], and set (dx/dt) 2 = q(t) 45 p(t) = 3t and q(t) = t 2 We have x = t dt = t 2 /2 Furthermore, q (t) + 2p(t)q(t) 2 [q(t)] 3/2 = + 3t2 t 2 The ratio is not constant, and therefore the equation cannot be transformed

78 78 Chapter 3 Second Order Linear Equations 46 p(t) = t /t and q(t) = t 2 We have x = t dt = t 2 /2 Furthermore, q (t) + 2p(t)q(t) 2 [q(t)] 3/2 = The ratio is constant, and therefore the equation can be transformed From Problem 43, the transformed equation is d 2 y dx 2 + dy dx + y = Based on the methods in this section, the characteristic equation is r 2 + r + =, with roots r = ( ± i 3)/2 The general solution is y(x) = c e x/2 cos 3 x/2 + c 2 e x/2 sin 3 x/2 Since x = t 2 /2, the solution in the original variable t is [ y(t) = e t2 /4 c cos ( 3 t 2 /4) + c 2 sin ( ] 3 t 2 /4) 34 2 The characteristic equation is 9r 2 + 6r + =, with the double root r = /3 The general solution is y(t) = c e t/3 + c 2 t e t/3 3 The characteristic equation is 4r 2 4r 3 =, with roots r = /2, 3/2 The general solution is y(t) = c e t/2 + c 2 e 3t/2 4 The characteristic equation is 4r 2 + 2r + 9 =, with double root r = 3/2 The general solution is y(t) = (c + c 2 t)e 3t/2 6 The characteristic equation is r 2 6r + 9 =, with the double root r = 3 The general solution is y(t) = c e 3t + c 2 t e 3t 7 The characteristic equation is 4r 2 + 7r + 4 =, with roots r = /4, 4 The general solution is y(t) = c e t/4 + c 2 e 4t 8 The characteristic equation is 6r r + 9 =, with double root r = 3/4 The general solution is y(t) = c e 3t/4 + c 2 t e 3t/4 The characteristic equation is 2r 2 + 2r + = We obtain the complex roots r = ( ± i)/2 The general solution is y(t) = c e t/2 cos(t/2) + c 2 e t/2 sin(t/2) The characteristic equation is 9r 2 2r + 4 =, with the double root r = 2/3 The general solution is y(t) = c e 2t/3 + c 2 t e 2t/3 Invoking the first initial condition, it follows that c = 2 Now y (t) = (4/3 + c 2 )e 2t/3 + 2c 2 t e 2t/3 /3 Invoking the second initial condition, 4/3 + c 2 =, or c 2 = 7/3 Hence we obtain the solution y(t) = 2e 2t/3 (7/3)te 2t/3 Since the second term dominates for large t, y(t)

79 The characteristic equation is 9r 2 + 6r + 82 = We obtain the complex roots r = /3 ± 3 i The general solution is y(t) = c e t/3 cos 3t + c 2 e t/3 sin 3t Based on the first initial condition, c = Invoking the second initial condition, we conclude that /3 + 3c 2 = 2, or c 2 = 5/9 Hence y(t) = e t/3 cos 3t + (5/9)e t/3 sin 3t The solution oscillates with an exponentially decreasing amplitude 5(a) The characteristic equation is 4r 2 + 2r + 9 =, with double root r = 3/2 The general solution is y(t) = c e 3t/2 + c 2 t e 3t/2 Invoking the first initial condition, it follows that c = Now y (t) = ( 3/2 + c 2 )e 3t/2 (3/2)c 2 t e 3t/2 The second initial condition requires that 3/2 + c 2 = 4, or c 2 = 5/2 Hence the specific solution is y(t) = e 3t/2 (5/2)t e 3t/2

80 8 Chapter 3 Second Order Linear Equations (b) The solution crosses the x-axis at t = 2/5 (c) The solution has a minimum at the point (6/5, 5e 8/5 /3) (d) Given that y () = b, we have 3/2 + c 2 = b, or c 2 = b + 3/2 Hence the solution is y(t) = e 3t/2 + (b + 3/2)t e 3t/2 Since the second term dominates, the long-term solution depends on the sign of the coefficient b + 3/2 The critical value is b = 3/2 6 The characteristic roots are r = r 2 = /2 Hence the general solution is given by y(t) = c e t/2 + c 2 t e t/2 Invoking the initial conditions, we require that c = 2, and that + c 2 = b The specific solution is y(t) = 2e t/2 + (b )t e t/2 Since the second term dominates, the long-term solution depends on the sign of the coefficient b The critical value is b = 8(a) The characteristic roots are r = r 2 = 2/3 Therefore the general solution is given by y(t) = c e 2t/3 + c 2 t e 2t/3 Invoking the initial conditions, we require that c = a, and that 2a/3 + c 2 = After solving for the coefficients, the specific solution is y(t) = ae 2t/3 + (2a/3 )t e 2t/3 (b) Since the second term dominates, the long-term solution depends on the sign of the coefficient 2a/3 The critical value is a = 3/2 2(a) The characteristic equation is r 2 + 2a r + a 2 = (r + a) 2 = (b) With p(t) = 2a, Abel s Formula becomes W (y, y 2 ) = c e 2a dt = c e 2at (c) y (t) = e at is a solution From part (b), with c =, e at y 2(t) + a e at y 2 (t) = e 2at, which can be written as (e at y 2 (t)) =, resulting in e at y 2 (t) = t 22(a) If the characteristic equation ar 2 + br + c has equal roots r, then ar 2 + br + c = a(r r ) 2 = Then clearly L[e rt ] = (ar 2 + br + c)e rt = a(r r ) 2 e rt This gives immediately that L[e rt ] = (b) Differentiating the identity in part (a) with respect to r we get (2ar + b)e rt + (ar 2 + br + c)te rt = 2a(r r )e rt + a(r r ) 2 te rt Again, this gives L[te rt ] = 23 Set y 2 (t) = t 2 v(t) Substitution into the differential equation results in t 2 (t 2 v + 4tv + 2v) 4t(t 2 v + 2tv) + 6t 2 v = After collecting terms, we end up with t 4 v = Hence v(t) = c + c 2 t, and thus y 2 (t) = c t 2 + c 2 t 3 Setting c = and c 2 =, we obtain y 2 (t) = t 3 24 Set y 2 (t) = t v(t) Substitution into the differential equation results in t 2 (tv + 2v ) + 2t(tv + v) 2tv = After collecting terms, we end up with t 3 v + 4t 2 v = This equation is linear in the variable w = v It follows that v (t) = c t 4, and v(t) = c t 3 + c 2 Thus

81 34 8 y 2 (t) = c t 2 + c 2 t Setting c = and c 2 =, we obtain y 2 (t) = t 2 26 Set y 2 (t) = tv(t) Substitution into the differential equation results in v v = This equation is linear in the variable w = v It follows that v (t) = c e t, and v(t) = c e t + c 2 Thus y 2 (t) = c te t + c 2 t Setting c = and c 2 =, we obtain y 2 (t) = te t 28 Set y 2 (x) = e x v(x) Substitution into the differential equation results in v + (x 2)/(x ) v = This equation is linear in the variable w = v An integrating factor is µ = e (x 2)/(x ) dx = e x /(x ) Rewrite the equation as [e x v /(x )] =, from which it follows that v (x) = c(x )e x Hence v(x) = c xe x + c 2 and y 2 (x) = c x + c 2 e x Setting c = and c 2 =, we obtain y 2 (x) = x 29 Set y 2 (x) = y (x) v(x), in which y (x) = x /4 e 2 x It can be verified that y is a solution of the differential equation, that is, x 2 y (x 875)y = Substitution of the given form of y 2 results in the differential equation 2x 9/4 v + (4x 7/4 + x 5/4 )v = This equation is linear in the variable w = v An integrating factor is µ = e [2x /2 +/(2x)]dx = x e 4 x Rewrite the equation as [ x e 4 x v ] =, from which it follows that v (x) = c e 4 x / x Integrating, v(x) = c e 4 x + c 2 and as a result, y 2 (x) = c x /4 e 2 x + c 2 x /4 e 2 x Setting c = and c 2 =, we obtain y 2 (x) = x /4 e 2 x 3 Direct substitution verifies that y (t) = e δx2 /2 is a solution of the differential equation Now set y 2 (x) = y (x) v(x) Substitution of y 2 into the equation results in v δxv = This equation is linear in the variable w = v An integrating factor is µ = e δx2 /2 Rewrite the equation as [ e δx2 /2 v ] =, from which it follows that v (x) = c e δx2 /2 Integrating, we obtain x v(x) = c e δu2 /2 du + v() Hence y 2 (x) = c e δx2 /2 x e δu2 /2 du + c 2 e δx2 /2 Setting c 2 =, we obtain a second independent solution 33 After writing the differential equation in standard form, we have p(t) = 3/t Based on Abel s identity, W (y, y 2 ) = c e 3/t dt = c t 3 As shown in Problem 32, two solutions of a second order linear equation satisfy (y 2 /y ) = W (y, y 2 )/y 2 In the given problem, y (t) = t Hence (t y 2 ) = c t Integrating both sides of the equation, y 2 (t) = c t ln t + c 2 t Setting c = and c 2 = we obtain y 2 (t) = t ln t 35 After writing the differential equation in standard form, we have p(x) = x/(x ) Based on Abel s identity, W (y, y 2 ) = c e x/(x ) dx = c e x (x ) Two solutions of a second order linear equation satisfy (y 2 /y ) = W (y, y 2 )/y 2 In the given problem, y (x) = e x Hence (e x y 2 ) = c e x (x ) Integrating both

82 82 Chapter 3 Second Order Linear Equations sides of the equation, y 2 (x) = c x + c 2 e x Setting c = and c 2 =, we obtain y 2 (x) = x 36 Write the differential equation in standard form to find p(x) = /x Based on Abel s identity, W (y, y 2 ) = c e /x dx = c x Two solutions of a second order linear differential equation satisfy (y 2 /y ) = W (y, y 2 )/y 2 In the given problem, y (x) = x /2 sin x Hence x ( sin x y 2) = c sin 2 x Integrating both sides of the equation, y 2 (x) = c x /2 cos x + c 2 x /2 sin x Setting c = and c 2 =, we obtain y 2 (x) = x /2 cos x 38(a) The characteristic equation is ar 2 + c = If a, c >, then the roots are r = ± i c/a The general solution is c c y(t) = c cos a t + c 2 sin a t, which is bounded (b) The characteristic equation is ar 2 + br = The roots are r =, b/a, and hence the general solution is y(t) = c + c 2 e bt/a Clearly, y(t) c With the given initial conditions, c = y + (a/b)y 39 Note that 2 cos t sin t = sin 2t Then k cos t sin t = (k/2) sin 2t Now if < k < 2, then (k/2) sin 2t < sin 2t and (k/2) sin 2t > sin 2t Hence k cos t sin t = k sin 2t > sin 2t 2 4 The equation transforms into y 4y + 4y = We obtain a double root r = 2 The solution is y = c e 2x + c 2 xe 2x = c e 2 ln t + c 2 ln te 2 ln t = c t 2 + c 2 t 2 ln t 42 The equation transforms into y 7y /2 + 5y/2 = The characteristic roots are r =, 5/2, so the solution is y = c e x + c 2 e 5x/2 = c e ln t + c 2 e 5 ln t/2 = c t + c 2 t 5/2 43 The equation transforms into y + 2y + y = We get a double root r = The solution is y = c e x + c 2 xe x = c e ln t + c 2 ln te ln t = c t + c 2 t ln t 44 The equation transforms into y 3y + 9y/4 = We obtain the double root r = 3/2 The solution is y = c e 3x/2 + c 2 xe 3x/2 = c e 3 ln t/2 + c 2 ln te 3 ln t/2 = c t 3/2 + c 2 t 3/2 ln t

83 The characteristic equation for the homogeneous problem is r 2 + 2r + 5 =, with complex roots r = ± 2i Hence y c (t) = c e t cos 2t + c 2 e t sin 2t Since the function g(t) = 3 sin 2t is not proportional to the solutions of the homogeneous equation, set Y = A cos 2t + B sin 2t Substitution into the given differential equation, and comparing the coefficients, results in the system of equations B 4A = 3 and A + 4B = Hence Y = (2/7) cos 2t + (3/7) sin 2t The general solution is y(t) = y c (t) + Y 3 The characteristic equation for the homogeneous problem is r 2 r 2 =, with roots r =, 2 Hence y c (t) = c e t + c 2 e 2t Set Y = At 2 + Bt + C Substitution into the given differential equation, and comparing the coefficients, results in the system of equations 2A = 4, 2A 2B = 2 and 2A B 2C = Hence Y = 2t 2 + 3t 7/2 The general solution is y(t) = y c (t) + Y 4 The characteristic equation for the homogeneous problem is r 2 + r 6 =, with roots r = 3, 2 Hence y c (t) = c e 3t + c 2 e 2t Set Y = Ae 3t + Be 2t Substitution into the given differential equation, and comparing the coefficients, results in the system of equations 6A = 2 and 4B = 2 Hence Y = 2e 3t 3e 2t The general solution is y(t) = y c (t) + Y 5 The characteristic equation for the homogeneous problem is r 2 2r 3 =, with roots r =, 3 Hence y c (t) = c e t + c 2 e 3t Note that the assignment Y = Ate t is not sufficient to match the coefficients Try Y = Ate t + Bt 2 e t Substitution into the differential equation, and comparing the coefficients, results in the system of equations 4A + 2B = and 8B = 3 This implies that Y = (3/6)te t + (3/8)t 2 e t The general solution is y(t) = y c (t) + Y 7 The characteristic equation for the homogeneous problem is r =, with complex roots r = ±3i Hence y c (t) = c cos 3t + c 2 sin 3t To simplify the analysis, set g (t) = 6 and g 2 (t) = t 2 e 3t By inspection, we have Y = 2/3 Based on the form of g 2, set Y 2 = Ae 3t + Bte 3t + Ct 2 e 3t Substitution into the differential equation, and comparing the coefficients, results in the system of equations 8A + 6B + 2C =, 8B + 2C =, and 8C = Hence Y 2 = 62 e3t 27 te3t + 8 t2 e 3t The general solution is y(t) = y c (t) + Y + Y 2 9 The characteristic equation for the homogeneous problem is 2r 2 + 3r + =, with roots r =, /2 Hence y c (t) = c e t + c 2 e t/2 To simplify the analysis, set g (t) = t 2 and g 2 (t) = 3 sin t Based on the form of g, set Y = A + Bt + Ct 2 Substitution into the differential equation, and comparing the coefficients, results in the system of equations A + 3B + 4C =, B + 6C =, and C = Hence we obtain Y = 4 6t + t 2 On the other hand, set Y 2 = D cos t + E sin t After substitution into the ODE, we find that D = 9/ and E = 3/ The general solution is y(t) = y c (t) + Y + Y 2

84 84 Chapter 3 Second Order Linear Equations The characteristic equation for the homogeneous problem is r 2 + ω 2 =, with complex roots r = ± ω i Hence y c (t) = c cos ω t + c 2 sin ω t Since ω ω, set Y = A cos ωt + B sin ωt Substitution into the ODE and comparing the coefficients results in the system of equations (ω 2 ω 2 )A = and (ω 2 ω 2 )B = Hence Y = cos ωt ω2 ω 2 The general solution is y(t) = y c (t) + Y 2 From Problem, y c (t) is known Since cos ω t is a solution of the homogeneous problem, set Y = At cos ω t + Bt sin ω t Substitution into the given ODE and comparing the coefficients results in A = and B = /2ω Hence the general solution is y(t) = c cos ω t + c 2 sin ω t + t sin ω t/(2ω ) 4 The characteristic equation for the homogeneous problem is r 2 r 2 =, with roots r =, 2 Hence y c (t) = c e t + c 2 e 2t Based on the form of the right hand side, that is, cosh(2t) = (e 2t + e 2t )/2, set Y = At e 2t + Be 2t Substitution into the given ODE and comparing the coefficients results in A = /6 and B = /8 Hence the general solution is y(t) = c e t + c 2 e 2t + t e 2t /6 + e 2t /8 6 The characteristic equation for the homogeneous problem is r =, with roots r = ± 2i Hence y c (t) = c cos 2t + c 2 sin 2t Set Y = A + Bt + Ct 2 Comparing the coefficients of the respective terms, we find that A = /8, B =, C = /4 Now set Y 2 = D e t, and obtain D = 3/5 Hence the general solution is y(t) = c cos 2t + c 2 sin 2t /8 + t 2 /4 + 3 e t /5 Invoking the initial conditions, we require that 9/4 + c = and 3/5 + 2c 2 = 2 Hence c = 9/4 and c 2 = 7/ 7 The characteristic equation for the homogeneous problem is r 2 2r + =, with a double root r = Hence y c (t) = c e t + c 2 t e t Consider g (t) = t e t Note that g is a solution of the homogeneous problem Set Y = At 2 e t + Bt 3 e t (the first term is not sufficient for a match) Upon substitution, we obtain Y = t 3 e t /6 By inspection, Y 2 = 4 Hence the general solution is y(t) = c e t + c 2 t e t + t 3 e t /6 + 4 Invoking the initial conditions, we require that c + 4 = and c + c 2 = Hence c = 3 and c 2 = 4 9 The characteristic equation for the homogeneous problem is r =, with roots r = ±2i Hence y c (t) = c cos 2t + c 2 sin 2t Since the function sin 2t is a solution of the homogeneous problem, set Y = At cos 2t + Bt sin 2t Upon substitution, we obtain Y = 3t cos 2t /4 Hence the general solution is y(t) = c cos 2t + c 2 sin 2t 3t cos 2t /4 Invoking the initial conditions, we require that c = 2 and 2c 2 (3/4) = Hence c = 2 and c 2 = /8 2 The characteristic equation for the homogeneous problem is r 2 + 2r + 5 =, with complex roots r = ± 2i Hence y c (t) = c e t cos 2t + c 2 e t sin 2t Based on the form of g(t), set Y = At e t cos 2t + Bt e t sin 2t After comparing coefficients, we obtain Y = t e t sin 2t Hence the general solution is y(t) = c e t cos 2t + c 2 e t sin 2t + t e t sin 2t Invoking the initial conditions, we require

85 35 85 that c = and c + 2c 2 = Hence c = and c 2 = /2 22(a) The characteristic equation for the homogeneous problem is r 2 + =, with complex roots r = ± i Hence y c (t) = c cos t + c 2 sin t Let g (t) = t sin t and g 2 (t) = t By inspection, it is easy to see that Y 2 (t) = t Based on the form of g (t), set Y (t) = At cos t + Bt sin t + Ct 2 cos t + Dt 2 sin t (b) Substitution into the equation and comparing the coefficients results in A =, B = /4, C = /4, and D = Hence Y (t) = t + t sin t /4 t 2 cos t /4 23(a) The characteristic equation for the homogeneous problem is r 2 5r + 6 =, with roots r = 2, 3 Hence y c (t) = c e 2t + c 2 e 3t Consider g (t) = e 2t (3t + 4) sin t, and g 2 (t) = e t cos 2t Based on the form of these functions on the right hand side of the ODE, set Y 2 (t) = e t (A cos 2t + A 2 sin 2t) and Y (t) = (B + B 2 t)e 2t sin t + (C + C 2 t)e 2t cos t (b) Substitution into the equation and comparing the coefficients results in Y (t) = 2 (et cos 2t + 3e t sin 2t) te2t (cos t sin t) + e 2t ( cos t 5 sin t) 2 25(a) We obtain the double characteristic root r = 2 Hence y c (t) = c e 2t + c 2 te 2t Consider the functions g (t) = 2t 2, g 2 (t) = 4te 2t, and g 3 (t) = t sin 2t The corresponding forms of the respective parts of the particular solution are Y (t) = A + A t + A 2 t 2, Y 2 (t) = e 2t (B 2 t 2 + B 3 t 3 ), and Y 3 (t) = t(c cos 2t + C 2 sin 2t) + (D cos 2t + D 2 sin 2t) (b) Substitution into the equation and comparing the coefficients results in Y (t) = 4 (3 + 4t + 2t2 ) t3 e 2t + t cos 2t + (cos 2t sin 2t) (a) The homogeneous solution is y c (t) = c cos 2t + c 2 sin 2t Since cos 2t and sin 2t are both solutions of the homogeneous equation, set Y (t) = t(a + A t + A 2 t 2 ) cos 2t + t(b + B t + B 2 t 2 ) sin 2t (b) Substitution into the equation and comparing the coefficients results in Y (t) = ( 3 32 t 2 t3 ) cos 2t + 6 (28t + 3t2 ) sin 2t 27(a) The homogeneous solution is y c (t) = c e t + c 2 te 2t None of the functions on the right hand side are solutions of the homogenous equation In order to include all possible combinations of the derivatives, consider Y (t) = e t (A + A t + A 2 t 2 ) cos 2t + e t (B + B t + B 2 t 2 ) sin 2t + + e t (C cos t + C 2 sin t) + De t

86 86 Chapter 3 Second Order Linear Equations (b) Substitution into the differential equation and comparing the coefficients results in Y (t) = e t (A + A t + A 2 t 2 ) cos 2t + e t (B + B t + B 2 t 2 ) sin 2t + e t ( 3 2 cos t sin t) + 2et /3, in which A = 45/3552, A = 73/676, A 2 = 5/52, B = 233/3552, B = /69, B 2 = /52 28(a) The homogeneous solution is y c (t) = c e t cos 2t + c 2 e t sin 2t None of the terms on the right hand side are solutions of the homogenous equation In order to include the appropriate combinations of derivatives, consider Y (t) = e t (A t + A 2 t 2 ) cos 2t + e t (B t + B 2 t 2 ) sin 2t + + e 2t (C + C t) cos 2t + e 2t (D + D t) sin 2t (b) Substitution into the differential equation and comparing the coefficients results in Y (t) = 3 6 te t cos 2t t2 e t sin 2t 25 e 2t (7 + t) cos 2t + 25 e 2t ( + 5t) sin 2t 3 The homogeneous solution is y c (t) = c cos λt + c 2 sin λt Since the differential operator does not contain a first derivative (and λ mπ), we can set Y (t) = N C m sin mπt m= Substitution into the differential equation yields N m 2 π 2 C m sin mπt + λ 2 m= N m= C m sin mπt = Equating coefficients of the individual terms, we obtain C m = N a m sin mπt m= a m λ 2 m 2, m =, 2 N π2 32 The homogeneous solution is y c (t) = c e t cos 2t + c 2 e t sin 2t The input function is independent of the homogeneous solutions, on any interval Since the right hand side is piecewise constant, it follows by inspection that { /5, t π/2 Y (t) =, t > π/2

87 35 87 For t π/2, the general solution is y(t) = c e t cos 2t + c 2 e t sin 2t + /5 Invoking the initial conditions y() = y () =, we require that c = /5, and that c 2 = / Hence y(t) = 5 (2e t cos 2t + e t sin 2t) on the interval t π/2 We now have the values y(π/2) = ( + e π/2 )/5, and y (π/2) = For t > π/2, the general solution is y(t) = d e t cos 2t + d 2 e t sin 2t It follows that y(π/2) = e π/2 d and y (π/2) = e π/2 d 2e π/2 d 2 Since the solution is continuously differentiable, we require that e π/2 d = ( + e π/2 )/5 and e π/2 d 2e π/2 d 2 = Solving for the coefficients, d = 2d 2 = (e π/2 + )/5 34 Since a, b, c >, the roots of the characteristic equation have negative real parts That is, r = α ± β i, where α < Hence the homogeneous solution is If g(t) = d, then the general solution is y c (t) = c e αt cos βt + c 2 e αt sin βt y(t) = d/c + c e αt cos βt + c 2 e αt sin βt Since α <, y(t) d/c as t If c =, then the characteristic roots are r = and r = b/a The ODE becomes ay + by = d Integrating both sides, we find that ay + by = d t + c The general solution can be expressed as y(t) = d t/b + c + c 2 e bt/a In this case, the solution grows without bound If b =, also, then the differential equation can be written as y = d/a, which has general solution y(t) = d t 2 /2a + c + c 2 Hence the assertion is true only if the coefficients are positive 35(a) Since D is a linear operator, D 2 y + bdy + cy = D 2 y (r + r 2 )Dy + r r 2 y = D 2 y r 2 Dy r Dy + r r 2 y = D(Dy r 2 y) r (Dy r 2 y) = (D r )(D r 2 )y (b) Let u = (D r 2 )y Then the ODE (i) can be written as (D r )u = g(t), that is, u r u = g(t) The latter is a linear first order equation in u Its general solution is t u(t) = e rt e rτ g(τ)dτ + c e rt t

88 88 Chapter 3 Second Order Linear Equations From above, we have y r 2 y = u(t) This equation is also a first order ODE Hence the general solution of the original second order equation is y(t) = e r2t t t e r2τ u(τ)dτ + c 2 e r2t Note that the solution y(t) contains two arbitrary constants 37 Note that (2D 2 + 3D + )y = (2D + )(D + )y Let u = (D + )y, and solve the ODE 2u + u = t sin t This equation is a linear first order ODE, with solution t u(t) = e t/2 e [τ τ/2 2 / ] sin τ dτ + c e t/2 = t = t 2 4t cos t sin t + c e t/2 Now consider the ODE y + y = u(t) The general solution of this first order ODE is y(t) = e t t t e τ u(τ)dτ + c 2 e t, in which u(t) is given above Substituting for u(t) and performing the integration, y(t) = t 2 6t cos t 3 sin t + c e t/2 + c 2 e t 38 We have (D 2 + 2D + )y = (D + )(D + )y Let u = (D + )y, and consider the ODE u + u = 2e t The general solution is u(t) = 2t e t + c e t We therefore have the first order equation u + u = 2t e t + c e t The general solution of the latter differential equation is y(t) = e t t t [2τ + c ] dτ + c 2 e t = e t (t 2 + c t + c 2 ) 39 We have (D 2 + 2D)y = D(D + 2)y Let u = (D + 2)y, and consider the equation u = sin 2t Direct integration results in u(t) = 3t 2 cos 2t + c The problem is reduced to solving the ODE y + 2y = 3t 2 cos 2t + c The general solution of this first order differential equation is y(t) = e 2t t t e 2τ [3τ 2 cos 2τ + c ] dτ + c 2 e 2t = = 3 2 t 2 (cos 2t + sin 2t) + c + c 2 e 2t 36 The solution of the homogeneous equation is y c (t) = c e 2t + c 2 e 3t The functions y (t) = e 2t and y 2 (t) = e 3t form a fundamental set of solutions The Wronskian

89 36 89 of these functions is W (y, y 2 ) = e 5t Using the method of variation of parameters, the particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which e 3t (2e t ) e 2t u (t) = W (t) dt = (2e t ) 2e t and u 2 (t) = W (t) dt = e 2t Hence the particular solution is Y (t) = 2e t e t = e t 3 The solution of the homogeneous equation is y c (t) = c e t + c 2 te t The functions y (t) = e t and y 2 (t) = te t form a fundamental set of solutions The Wronskian of these functions is W (y, y 2 ) = e 2t Using the method of variation of parameters, the particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which te t (3e t ) e u (t) = dt = 3t 2 t (3e t ) /2 and u 2 (t) = dt = 3t W (t) W (t) Hence the particular solution is Y (t) = 3t 2 e t /2 + 3t 2 e t = 3t 2 e t /2 4 The functions y (t) = e t/2 and y 2 (t) = te t/2 form a fundamental set of solutions The Wronskian of these functions is W (y, y 2 ) = e t First write the equation in standard form, so that g(t) = 4e t/2 Using the method of variation of parameters, the particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which te t/2 (4e t/2 ) e u (t) = dt = 2t 2 t/2 (4e t/2 ) and u 2 (t) = dt = 4t W (t) W (t) Hence the particular solution is Y (t) = 2t 2 e t/2 + 4t 2 e t/2 = 2t 2 e t/2 6 The solution of the homogeneous equation is y c (t) = c cos 3t + c 2 sin 3t The two functions y (t) = cos 3t and y 2 (t) = sin 3t form a fundamental set of solutions, with W (y, y 2 ) = 3 The particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which sin 3t(9 sec 2 3t) u (t) = dt = csc 3t W (t) cos 3t(9 sec 2 3t) u 2 (t) = dt = ln(sec 3t + tan 3t), W (t) since < t < π/6 Hence Y (t) = + (sin 3t) ln(sec 3t + tan 3t) The general solution is given by y(t) = c cos 3t + c 2 sin 3t + (sin 3t) ln(sec 3t + tan 3t) 7 The functions y (t) = e 2t and y 2 (t) = te 2t form a fundamental set of solutions The Wronskian of these functions is W (y, y 2 ) = e 4t The particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which te 2t (t 2 e 2t ) e 2t (t 2 e 2t ) u (t) = dt = ln t and u 2 (t) = dt = /t W (t) W (t)

90 9 Chapter 3 Second Order Linear Equations Hence the particular solution is Y (t) = e 2t ln t e 2t Since the second term is a solution of the homogeneous equation, the general solution is given by y(t) = c e 2t + c 2 te 2t e 2t ln t 8 The solution of the homogeneous equation is y c (t) = c cos 2t + c 2 sin 2t The two functions y (t) = cos 2t and y 2 (t) = sin 2t form a fundamental set of solutions, with W (y, y 2 ) = 2 The particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which sin 2t(3 csc 2t) u (t) = dt = 3t/2 W (t) cos 2t(3 csc 2t) u 2 (t) = dt = 3 ln(sin 2t), W (t) 4 since < t < π/2 Hence Y (t) = (3/2)t cos 2t + (3/4)(sin 2t) ln(sin 2t) The general solution is given by y(t) = c cos 2t + c 2 sin 2t 3 2 t cos 2t + 3 (sin 2t) ln(sin 2t) 4 9 The functions y (t) = cos (t/2) and y 2 (t) = sin(t/2) form a fundamental set of solutions The Wronskian of these functions is W (y, y 2 ) = /2 First write the ODE in standard form, so that g(t) = sec(t/2)/2 The particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which cos (t/2) [sec(t/2)] u (t) = dt = 2 ln(cos (t/2)) 2W (t) sin(t/2) [sec(t/2)] u 2 (t) = dt = t 2W (t) The particular solution is Y (t) = 2 cos(t/2) ln(cos (t/2)) + t sin(t/2) The general solution is given by y(t) = c cos (t/2) + c 2 sin(t/2) + 2 cos(t/2) ln(cos (t/2)) + t sin(t/2) The solution of the homogeneous equation is y c (t) = c e t + c 2 te t The functions y (t) = e t and y 2 (t) = te t form a fundamental set of solutions, with W (y, y 2 ) = e 2t The particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which te t (e t ) u (t) = W (t)( + t 2 ) dt = 2 ln( + t2 ) u 2 (t) = e t (e t ) W (t)( + t 2 dt = arctan t ) The particular solution is Y (t) = (/2)e t ln( + t 2 ) + te t arctan(t) general solution is given by y(t) = c e t + c 2 te t 2 et ln( + t 2 ) + te t arctan(t) Hence the

91 The functions y (t) = cos 2t and y 2 (t) = sin 2t form a fundamental set of solutions, with W (y, y 2 ) = 2 The particular solution is given by Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which u (t) = 2 t Hence the particular solution is Y (t) = 2 cos 2t t t g(s) sin 2s ds and u 2 (t) = 2 t g(s) sin 2s ds + 2 t t g(s) cos 2s ds t sin 2t g(s) cos 2s ds t Note that sin 2t cos 2s cos 2t sin 2s = sin(2t 2s) It follows that Y (t) = 2 t t g(s) sin(2t 2s)ds The general solution of the differential equation is given by y(t) = c cos 2t + c 2 sin 2t + 2 t t g(s) sin(2t 2s)ds 3 Note first that p(t) =, q(t) = 2/t 2 and g(t) = (3t 2 )/t 2 The functions y (t) and y 2 (t) are solutions of the homogeneous equation, verified by substitution The Wronskian of these two functions is W (y,y 2 ) = 3 Using the method of variation of parameters, the particular solution is Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which t (3t 2 ) u (t) = t 2 dt = t 2 /6 + ln t W (t) t 2 (3t 2 ) u 2 (t) = t 2 dt = t 3 /3 + t/3 W (t) Therefore Y (t) = /6 + t 2 ln t t 2 /3 + /3 5 Observe that g(t) = t e 2t The functions y (t) and y 2 (t) are a fundamental set of solutions The Wronskian of these two functions is W (y,y 2 ) = t e t Using the method of variation of parameters, the particular solution is Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which e t (t e 2t ) ( + t)(t e 2t u (t) = W (t) dt = ) e2t /2 and u 2 (t) = dt = t e t W (t) Therefore Y (t) = ( + t)e 2t /2 + t e 2t = e 2t /2 + t e 2t /2 6 Observe that g(t) = 2( t) e t Direct substitution of y (t) = e t and y 2 (t) = t verifies that they are solutions of the homogeneous equation The Wronskian of the two solutions is W (y,y 2 ) = ( t) e t Using the method of variation of parameters, the particular solution is Y (t) = u (t) y (t) + u 2 (t) y 2 (t), in which 2t( t)e t u (t) = dt = te 2t + e 2t /2 W (t)

92 92 Chapter 3 Second Order Linear Equations u 2 (t) = 2( t) W (t) dt = 2 e t Therefore Y (t) = te t + e t /2 2t e t = te t + e t /2 7 Note that g(x) = ln x The functions y (x) = x 2 and y 2 (x) = x 2 ln x are solutions of the homogeneous equation, as verified by substitution The Wronskian of the solutions is W (y,y 2 ) = x 3 Using the method of variation of parameters, the particular solution is Y (x) = u (x) y (x) + u 2 (x) y 2 (x), in which x 2 ln x(ln x) u (x) = dx = (ln x) 3 /3 W (x) x 2 (ln x) u 2 (x) = W (x) dx = (ln x)2 /2 Therefore Y (x) = x 2 (ln x) 3 /3 + x 2 (ln x) 3 /2 = x 2 (ln x) 3 /6 9 First write the equation in standard form Note that the forcing function becomes g(x)/( x) The functions y (x) = e x and y 2 (x) = x are a fundamental set of solutions, as verified by substitution The Wronskian of the solutions is W (y,y 2 ) = ( x)e x Using the method of variation of parameters, the particular solution is Y (x) = u (x) y (x) + u 2 (x) y 2 (x), in which Therefore u (x) = x x τ(g(τ)) ( τ)w (τ) dτ and u 2(x) = x Y (x) = e x τ(g(τ)) ( τ)w (τ) dτ + x x = x x x x (xe τ e x τ)g(τ) ( τ) 2 e τ dτ x x e τ (g(τ)) ( τ)w (τ) dτ e τ (g(τ)) ( τ)w (τ) dτ = 2 First write the equation in standard form The forcing function becomes g(x)/x 2 The functions y (x) = x /2 sin x and y 2 (x) = x /2 cos x are a fundamental set of solutions The Wronskian of the solutions is W (y,y 2 ) = /x Using the method of variation of parameters, the particular solution is Y (x) = u (x) y (x) + u 2 (x) y 2 (x), in which Therefore u (x) = x x Y (x) = sin x x x cos τ (g(τ)) τ dτ and u 2 (x) = τ x cos τ (g(τ)) τ dt cos x x τ x x = x x x sin(x τ) g(τ) τ dτ τ x x sin τ (g(τ)) τ dτ τ sin τ (g(τ)) τ dτ = τ 2 Let y (t) and y 2 (t) be a fundamental set of solutions, and W (t) = W (y, y 2 ) be the corresponding Wronskian Any solution, u(t), of the homogeneous equation is

93 36 93 a linear combination u(t) = α y (t) + α 2 y 2 (t) Invoking the initial conditions, we require that y = α y (t ) + α 2 y 2 (t ) y = α y (t ) + α 2 y 2(t ) Note that this system of equations has a unique solution, since W (t ) Now consider the nonhomogeneous problem, L [v] = g(t), with homogeneous initial conditions Using the method of variation of parameters, the particular solution is given by t y 2 (s) g(s) t y (s) g(s) Y (t) = y (t) ds + y 2 (t) ds t W (s) t W (s) The general solution of the IVP (iii) is v(t) = β y (t) + β 2 y 2 (t) + Y (t) = β y (t) + β 2 y 2 (t) + y (t)u (t) + y 2 (t)u 2 (t) in which u and u 2 that are defined above Invoking the initial conditions, we require = β y (t ) + β 2 y 2 (t ) + Y (t ) = β y (t ) + β 2 y 2(t ) + Y (t ) Based on the definition of u and u 2, Y (t ) = Furthermore, since y u + y 2 u 2 =, it follows that Y (t ) = Hence the only solution of the above system of equations is the trivial solution Therefore v(t) = Y (t) Now consider the function y = u + v Then L [y] = L [u + v] = L [u] + L [v] = g(t) That is, y(t) is a solution of the nonhomogeneous problem Further, y(t ) = u(t ) + v(t ) = y, and similarly, y (t ) = y By the uniqueness theorems, y(t) is the unique solution of the initial value problem 23(a) A fundamental set of solutions is y (t) = cos t and y 2 (t) = sin t The Wronskian W (t) = y y 2 y y 2 = By the result in Problem 22, Y (t) = = t t t cos(s) sin(t) cos(t) sin(s) g(s)ds W (s) t [cos(s) sin(t) cos(t) sin(s)] g(s)ds Finally, we have cos(s) sin(t) cos(t) sin(s) = sin(t s) (b) Using Problem 2 and part (a), the solution is y(t) = y cos t + y sin t + t sin(t s)g(s)ds 24 A fundamental set of solutions is y (t) = e at and y 2 (t) = e bt The Wronskian W (t) = y y 2 y y 2 = (b a)e (a+b)t By the result in Problem 22, Y (t) = t t e as e bt e at e bs W (s) g(s)ds = b a t t e as e bt e at e bs e (a+b)s g(s)ds

94 94 Chapter 3 Second Order Linear Equations Hence the particular solution is Y (t) = b a t t [ e b(t s) e a(t s)] g(s)ds 26 A fundamental set of solutions is y (t) = e at and y 2 (t) = te at The Wronskian W (t) = y y 2 y y 2 = e 2at By the result in Problem 22, Y (t) = t Hence the particular solution is t te as+at s e at+as g(s)ds = W (s) Y (t) = t t t t (t s)e a(t s) g(s)ds (t s)e as+at e 2as g(s)ds 27 The form of the kernel depends on the characteristic roots If the roots are real and distinct, K(t s) = eb(t s) e a(t s) b a If the roots are real and identical, If the roots are complex conjugates, K(t s) = (t s)e a(t s) K(t s) = eλ(t s) sin µ(t s) µ 28 Let y(t) = v(t)y (t), in which y (t) is a solution of the homogeneous equation Substitution into the given ODE results in v y + 2v y + vy + p(t) [v y + vy ] + q(t)vy = g(t) By assumption, y + p(t)y + q(t)y =, hence v(t) must be a solution of the ODE v y + [2y + p(t)y ] v = g(t) Setting w = v, we also have w y + [2y + p(t)y ] w = g(t) 3 First write the equation as y + 7t y + 5t 2 y = t As shown in Problem 28, the function y(t) = t v(t) is a solution of the given ODE as long as v is a solution of t v + [ 2t 2 + 7t 2] v = t, that is, v + 5t v = This ODE is linear and first order in v The integrating factor is µ = t 5 The solution is v = t/6 + c t 5 Direct integration now results in v(t) = t 2 /2 + c t 4 + c 2 Hence y(t) = t/2 + c t 5 + c 2 t 3 Write the equation as y t ( + t)y + t y = t e 2t As shown in Problem 28, the function y(t) = ( + t)v(t) is a solution of the given ODE as long as v is a solution of ( + t) v + [ 2 t ( + t) 2] v = t e 2t,

95 37 95 that is, v + t2 t(t + ) v = t t + e2t This equation is first order linear in v, with integrating factor µ = t ( + t) 2 e t The solution is v = (t 2 e 2t + c te t )/( + t) 2 Integrating, we obtain v(t) = e 2t /2 e 2t /(t + ) + c e t /(t + ) + c 2 Hence the solution of the original ODE is y(t) = (t )e 2t /2 + c e t + c 2 (t + ) 32 Write the equation as y + t( t) y ( t) y = 2( t) e t The function y(t) = e t v(t) is a solution to the given ODE as long as v is a solution of e t v + [ 2e t + t( t) e t] v = 2( t) e t, that is, v + [(2 t)/( t)] v = 2( t) e 2t This equation is first order linear in v, with integrating factor µ = e t /(t ) The solution is v = (t )(2e 2t + c e t ) Integrating, we obtain v(t) = (/2 t)e 2t c te t + c 2 Hence the solution of the original ODE is y(t) = (/2 t)e t c t + c 2 e t 37 R cos δ = 3 and R sin δ = 4, so R = 25 = 5 and δ = arctan(4/3) We obtain that u = 5 cos(2t arctan(4/3)) 3 R cos δ = 4 and R sin δ = 2, so R = 2 = 2 5 and δ = arctan(/2) We obtain that u = 2 5 cos(3t + arctan(/2)) 4 R cos δ = 2 and R sin δ = 3, so R = 3 and δ = π + arctan(3/2) We obtain that u = 3 cos(πt π arctan(3/2)) 5 The spring constant is k = 2/(/2) = 4 lb/ft Mass m = 2/32 = /6 lb-s 2 /ft Since there is no damping, the equation of motion is u /6 + 4u =, that is, u + 64u = The initial conditions are u() = /4 ft, u () = ft/s The general solution is u(t) = A cos 8t + B sin 8t Invoking the initial conditions, we have u(t) = cos 8t /4 R = /4 ft, δ = rad, ω = 8 rad/s, and T = π/4 s 7 The spring constant is k = 3/(/4) = 2 lb/ft Mass m = 3/32 lb-s 2 /ft Since there is no damping, the equation of motion is 3u /32 + 2u =, that is, u + 28u = The initial conditions are u() = /2 ft, u () = 2 ft/s The general solution is u(t) = A cos 8 2 t + B sin 8 2 t Invoking the initial conditions, we have u(t) = 2 cos 8 2 t sin 8 2 t R = /288 ft, δ = π arctan(3/ 2 ) rad, ω = 8 2 rad/s, T = π/(4 2 ) s

96 96 Chapter 3 Second Order Linear Equations The spring constant is k = 6/(/4) = 64 lb/ft Mass m = /2 lb-s 2 /ft The damping coefficient is γ = 2 lb-s/ft Hence the equation of motion is u /2 + 2u + 64u =, that is, u + 4u + 28u = The initial conditions are u() = ft, u () = /4 ft/s The general solution is u(t) = A cos 2 3 t + B sin 2 3 t Invoking the initial conditions, we have u(t) = 8 3 e 2t sin 2 3 t Solving u(t) =, on the interval [2, 4], we obtain t = π/2 3 = 282 s Based on the graph, and the solution of u(t) = /2 ft, we have u(t) for t τ = 5927 The spring constant is k = 3/() = 3 N/m The damping coefficient is given as γ = 3/5 N-s/m Hence the equation of motion is 2u + 3u /5 + 3u =, that is, u + 3u + 5u = The initial conditions are u() = 5 m and u () = m/s The general solution is u(t) = A cos µt + B sin µt, in which µ = 3878 rad/s Invoking the initial conditions, we have u(t) = e 5t (5 cos µt sin µt) Also, µ/ω = 3878/ The frequency of the undamped motion is ω = The quasi frequency of the damped motion is µ = 4 γ 2 /2 Setting µ = 2ω /3, we obtain γ = 2 5 /3 4 The spring constant is k = mg/l The equation of motion for an undamped system is mu + mgu/l = Hence the natural frequency of the system is ω = g/l The period is T = 2π/ω 5 The general solution of the system is u(t) = A cos γ(t t ) + B sin γ(t t ) Invoking the initial conditions, we have u(t) = u cos γ(t t ) + (u /γ) sin γ(t t ) Clearly, the functions v = u cos γ(t t ) and w = (u /γ) sin γ(t t ) satisfy the given criteria 6 Note that r sin( ω t θ) = r sin ω t cos θ r cos ω t sin θ Comparing the given expressions, we have A = r sin θ and B = r cos θ That is, r = R = A2 + B 2, and tan θ = A/B = / tan δ The latter relation is also tan θ + cot δ = 8 The system is critically damped, when R = 2 L/C Here R = ohms

97 (a) Let u = Re γt/2m cos(µt δ) Then attains a maximum when µt k δ = 2kπ Hence T d = t k+ t k = 2π/µ (b) u(t k )/u(t k+ ) = e γt k/2m /e γt k+/2m = e (γt k+ γt k )/2m Hence u(t k )/u(t k+ ) = e γ(2π/µ)/2m = e γ T d/2m (c) = ln [u(t k )/u(t k+ )] = γ(2π/µ)/2m = πγ/µm 22 The spring constant is k = 6/(/4) = 64 lb/ft Mass m = /2 lb-s 2 /ft The damping coefficient is γ = 2 lb-s/ft The quasi frequency is µ = 2 3 rad/s Hence = 2π/ (a) The solution of the IVP is u(t) = e t/8 (2 cos 3 7 t/8 + (2 7/2) sin 3 7 t/8) Using the plot, and numerical analysis, τ 475 (b) For γ = 5, τ 242 ; for γ =, τ 968 ; for γ = 5, τ 784 (c) (d) For γ = 6, τ 728 ; for γ = 7, τ 6767 ; for γ = 8, τ 5473 ; for γ = 9, τ 646 τ steadily decreases to about τ min 4873, corresponding to the critical value γ 73 (e) We can rewrite the solution as u(t) = Re γt/2 cos(µt δ), where R = 4/ 4 γ 2

98 98 Chapter 3 Second Order Linear Equations Neglecting the cosine factor, we can approximate τ by solving Re γτ/2 = /, thus finding ( ) τ 2 γ ln(r) = 2 γ ln 4 4 γ 2 For γ = 25, τ ; for γ = 5, τ 23223; for γ =, τ 8843; for γ = 5, τ 76554; for γ = 6, τ 72643; for γ = 7, τ ; for γ = 8, τ 68965; for γ = 9, τ (a) The characteristic equation is mr 2 + γr + k = Since γ 2 < 4km, the roots are r,2 = ( γ ± i 4mk γ 2 )/2m The general solution is ] 4mk γ u(t) = e [A γt/2m 2 4mk γ 2 cos t + B sin t 2m 2m Invoking the initial conditions, A = u and B = (2mv γu )/ 4mk γ 2 (b) We can write u(t) = R e γt/2m cos(µt δ), in which [ ] R = u 2 + (2mv γu ) 2 (2mv γu ) 4mk γ 2 and δ = arctan u 4mk γ 2 (c) R = u 2 + (2mv γu ) 2 4mk γ 2 = 2 m(ku 2 + γu v + mv 2) a + bγ 4mk γ 2 = 4mk γ 2 It is evident that R increases (monotonically) without bound as γ (2 mk) 28(a) The general solution is u(t) = A cos 2 t + B sin 2 t Invoking the initial conditions, we have u(t) = 2 sin 2 t (b)

99 37 99 (c) The condition u () = 2 implies that u(t) initially increases Hence the phase point travels clockwise 3 Based on Newton s second law, with the positive direction to the right, F = mu, where F = ku γu Hence the equation of motion is mu + γu + ku = The only difference in this problem is that the equilibrium position is located at the unstretched configuration of the spring 32(a) The restoring force exerted by the spring is F s = (ku + ɛu 3 ) The opposing viscous force is F d = γu Based on Newton s second law, with the positive direction to the right, F s + F d = mu Hence the equation of motion is mu + γu + ku + ɛu 3 = (b) With the specified parameter values, the equation of motion is u + u = The general solution of this ODE is u(t) = A cos t + B sin t Invoking the initial conditions, the specific solution is u(t) = sin t Clearly, the amplitude is R =, and the period of the motion is T = 2π (c) Given ɛ =, the equation of motion is u + u + u 3 = A solution of the IVP can be generated numerically We estimate A = 98 and T = 67

100 Chapter 3 Second Order Linear Equations (d) For ɛ = 2, A = 96 and T = 59 For ɛ = 3, A = 94 and T = 574 (e) The amplitude and period both seem to decrease (f) For ɛ =, A = 3 and T = 655 For ɛ = 2, A = 6 and T = 69 For ɛ = 3, A = and T = 74 The amplitude and period both seem to increase (a) ε = (b) ε = 2 (c) ε = We have sin(α ± β) = sin α cos β ± cos α sin β Subtracting the two identities, we obtain sin(α + β) sin(α β) = 2 cos α sin β Setting α + β = 7t and α β = 6t, we get that α = 65t and β = 5t This implies that sin 7t sin 6t = 2 sin (t/2) cos (3t/2) 3 Consider the trigonometric identities cos(α ± β) = cos α cos β sin α sin β Adding the two identities, we get cos(α β) + cos(α + β) = 2 cos α cos β Comparing the expressions, set α + β = 2πt and α β = πt This means α = 3πt/2 and β = πt/2 Upon substitution, we have cos(πt) + cos(2πt) = 2 cos(3πt/2) cos(πt/2) 4 Adding the two identities sin(α ± β) = sin α cos β ± cos α sin β, it follows that sin(α β) + sin(α + β) = 2 sin α cos β Setting α + β = 4t and α β = 3t, we have α = 7t/2 and β = t/2 Hence sin 3t + sin 4t = 2 sin(7t/2) cos(t/2)

101 38 6 Using MKS units, the spring constant is k = 5(98)/ = 49 N/m, and the damping coefficient is γ = 2/4 = 5 N-s/m The equation of motion is 5u + 5u + 49u = sin(t/2) The initial conditions are u() = m and u () = 3 m/s 8(a) The homogeneous solution is u c (t) = Ae 5t cos 73 t + Be 5t sin 73 t Based on the method of undetermined coefficients, the particular solution is U(t) = [ 6 cos(t/2) sin(t/2)] 5328 Hence the general solution of the ODE is u(t) = u c (t) + U(t) Invoking the initial conditions, we find that A = 6/5328 and B = /8953 Hence the response is [ u(t) = 6 e 5t cos 73 t e 5t sin 73 t ] + U(t) (b) u c (t) is the transient part and U(t) is the steady state part of the response (c) (d) The amplitude of the forced response is given by R = 2/, in which = 25(98 ω 2 ) ω 2 The maximum amplitude is attained when is a minimum Hence the amplitude is maximum at ω = 4 3 rad/s 9 The spring constant is k = 2 lb/ft and hence the equation of motion is 6 32 u + 2u = 4 cos 7t,

102 2 Chapter 3 Second Order Linear Equations that is, u + 64u = (64/3) cos 7t The initial conditions are u() = ft, u () = ft/s The general solution is u(t) = A cos 8t + B sin 8t + (64/45) cos 7t Invoking the initial conditions, we have u(t) = (64/45) cos 8t + (64/45) cos 7t = (28/45) sin(t/2) sin(5t/2) 2 The equation of motion is 2u + u + 3u = 3 cos 3t 2 sin 3t Since the system is damped, the steady state response is equal to the particular solution Using the method of undetermined coefficients, we obtain u ss (t) = (sin 3t cos 3t)/6 Further, we find that R = 2 /6 and δ = arctan( ) = 3π/4 Hence we can write u ss (t) = ( 2 /6) cos(3t 3π/4) 3(a,b) Plug in u(t) = R cos(ωt δ) into the equation mu + γu + ku = F cos ωt, then use trigonometric identities and compare the coefficients of cos ωt and sin ωt The result follows (c) The amplitude of the steady-state response is given by R = F m2 (ω 2 ω2 ) 2 + γ 2 ω 2 Since F is constant, the amplitude is maximum when the denominator of R is minimum Let z = ω 2, and consider the function f(z) = m 2 (ω 2 z) 2 + γ 2 z Note that f(z) is a quadratic, with minimum at z = ω 2 γ 2 /2m 2 Hence the amplitude R attains a maximum at ωmax 2 = ω 2 γ 2 /2m 2 Furthermore, since ω 2 = k/m, ω 2 max = ω 2 [ γ2 2km Substituting ω 2 = ω 2 max into the expression for the amplitude, R = F γ4 /4m 2 + γ 2 (ω 2 γ2 /2m 2 ) = F ω 2 γ 2 γ 4 /4m 2 = F γω γ2 /4mk 7(a) The steady state part of the solution U(t) = A cos ωt + B sin ωt may be found by substituting this expression into the differential equation and solving for A and B We find that A = ] 32(2 ω 2 ) 64 63ω 2 + 6ω 4, B = 8ω 64 63ω 2 + 6ω 4

103 38 3 (b) The amplitude is A = ω2 + 6 ω 4 (c) (d) See Problem 3 The amplitude is maximum when the denominator of A is minimum That is, when ω = ω max = 3 4 /8 43 Hence A = 64/ 27 8(a) The homogeneous solution is u c (t) = A cos t + B sin t Based on the method of undetermined coefficients, the particular solution is U(t) = 3 cos ωt ω2 Hence the general solution of the ODE is u(t) = u c (t) + U(t) Invoking the initial conditions, we find that A = 3/(ω 2 ) and B = Hence the response is (b) u(t) = 3 [ cos ωt cos t ] ω2 (a) ω = 7 (b) ω = 8 (c) ω = 9 Note that u(t) = 6 ω 2 sin [ ( ω)t 2 ] sin [ (ω + )t 2 ]

104 4 Chapter 3 Second Order Linear Equations 9(a) The homogeneous solution is u c (t) = A cos t + B sin t Based on the method of undetermined coefficients, the particular solution is U(t) = 3 cos ωt ω2 Hence the general solution is u(t) = u c (t) + U(t) Invoking the initial conditions, we find that A = (ω 2 + 2)/(ω 2 ) and B = Hence the response is (b) u(t) = ω 2 [ 3 cos ωt (ω 2 + 2) cos t ] + sin t (a) ω = 7 (b) ω = 8 (c) ω = 9 Note that u(t) = 6 ω 2 sin [ ( ω)t 2 ] sin [ (ω + )t 2 ] + cos t + sin t 2 (a) ω = 7 (b) ω = 8 (c) ω = 9

105 38 5 2(a) (b) Phase plot - u vs u : 23(a)

106 6 Chapter 3 Second Order Linear Equations (b) Phase plot - u vs u : 25(a) (a) ω = 5 (b) ω = 75 (c) ω = (d) ω = 25 (e) ω = 5 (f) ω = 75 (g) ω = 2

107 38 7 (b) R vs ω (c) The amplitude for a similar system with a linear spring is given by R = ω2 + 25ω 4

108 8 Chapter 3 Second Order Linear Equations

109 C H A P T E R 4 Higher Order Linear Equations 4 The differential equation is in standard form Its coefficients, as well as the function g(t) = t, are continuous everywhere Hence solutions are valid on the entire real line 3 Writing the equation in standard form, the coefficients are rational functions with singularities at t = and t = Hence the solutions are valid on the intervals (,), (, ), and (, ) 5 Writing the equation in standard form, the coefficients are rational functions with a singularity at x = Furthermore, p 4 (x) = tan x/(x ) is undefined, and hence not continuous, at x k = ±(2k + )π/2, k =,, 2, Hence solutions are defined on any interval that does not contain x or x k 6 Writing the equation in standard form, the coefficients are rational functions with singularities at x = ± 2 Hence the solutions are valid on the intervals (, 2), ( 2, 2), and (2, ) 7 Evaluating the Wronskian of the three functions, W (f, f 2, f 3 ) = 4 Hence the functions are linearly independent 9 Evaluating the Wronskian of the four functions, W (f, f 2, f 3, f 4 ) = Hence the functions are linearly dependent To find a linear relation among the functions, 9

110 Chapter 4 Higher Order Linear Equations we need to find constants c, c 2, c 3, c 4, not all zero, such that Collecting the common terms, we obtain c f (t) + c 2 f 2 (t) + c 3 f 3 (t) + c 4 f 4 (t) = (c 2 + 2c 3 + c 4 )t 2 + (2c c 3 + c 4 )t + ( 3c + c 2 + c 4 ) =, which results in three equations in four unknowns Arbitrarily setting c 4 =, we can solve the equations c 2 + 2c 3 =, 2c c 3 =, 3c + c 2 =, to find that c = 2/7, c 2 = 3/7, c 3 = 3/7 Hence 2f (t) + 3f 2 (t) 3f 3 (t) 7f 4 (t) = Evaluating the Wronskian of the three functions, W (f, f 2, f 3 ) = 56 Hence the functions are linearly independent Substitution verifies that the functions are solutions of the differential equation Furthermore, we have W (, cos t, sin t) = 2 Substitution verifies that the functions are solutions of the differential equation Furthermore, we have W (, t, cos t, sin t) = 4 Substitution verifies that the functions are solutions of the differential equation Furthermore, we have W (, t, e t, t e t ) = e 2t 5 Substitution verifies that the functions are solutions of the differential equation Furthermore, we have W (, x, x 3 ) = 6x 6 Substitution verifies that the functions are solutions of the differential equation Furthermore, we have W (x, x 2, /x) = 6/x 8 The operation of taking a derivative is linear, and hence (c y + c 2 y 2 ) (k) = c y (k) + c 2 y (k) 2 It follows that L [c y + c 2 y 2 ] = c y (n) + c 2 y (n) 2 + p (c y (n ) + c 2 y (n ) 2 ) + + p n (c y + c 2 y 2 ) Rearranging the terms, we obtain L [c y + c 2 y 2 ] = c L [y ] + c 2 L [y 2 ] Since y and y 2 are solutions, L [c y + c 2 y 2 ] = The rest follows by induction 2(a) Let f(t) and g(t) be arbitrary functions Then W (f, g) = fg f g Hence W (f, g) = f g + fg f g f g = fg f g That is, W (f, g) = f g f g Now expand the 3-by-3 determinant as y 2 W (y, y 2, y 3 ) = y y 3 y 2 y y 3 + y 3 y y 2 y 2 y 3 y y 3 y y 2

111 4 Differentiating, we obtain W (y, y 2, y 3 ) = y y 2 y 3 y 2 y 3 y 2 y y 3 y y 3 + y 3 y y 2 y y 2 + y 2 + y y 3 y 2 y y 3 + y 3 y y 2 y 2 y 3 y y 3 y y 2 The second line follows from the observation above Now we find that y W y 2 y 3 (y, y 2, y 3 ) = y y 2 y 3 + y y 2 y 3 y y 2 y 3 y y 2 y 3 y y 2 y 3 Hence the assertion is true, since the first determinant is equal to zero (b) Based on the properties of determinants, p 3 y p 3 y 2 p 3 y 3 p 2 (t)p 3 (t)w = p 2 y p 2 y 2 p 2 y 3 y y 2 y 3 Adding the first two rows to the third row does not change the value of the determinant Since the functions are assumed to be solutions of the given ODE, addition of the rows results in p 3 y p 3 y 2 p 3 y 3 p 2 (t)p 3 (t)w = p 2 y p 2 y 2 p 2 y 3 p y p y 2 p y 3 It follows that p 2 (t)p 3 (t)w = p (t)p 2 (t)p 3 (t)w As long as the coefficients are not zero, we obtain W = p (t)w (c) The first order equation W = p (t)w is linear, with integrating factor µ(t) = e p (t)dt Hence W (t) = c e p (t)dt Furthermore, W (t) is zero only if c = (d) It can be shown, by mathematical induction, that y y 2 y n y n y y 2 y n y n W (y, y 2,, y n ) = y (n 2) y (n 2) 2 y (n 2) n n y (n) y (n) 2 y (n) n y n (n) Based on the reasoning in part (b), it follows that and hence W = p (t)w y (n 2) p 2 (t)p 3 (t) p n (t)w = p (t)p 2 (t)p 3 (t) p n (t)w, 2 Inspection of the coefficients reveals that p (t) = 2 Based on Problem 2, we find that W = 2W, and hence W = ce 2t

112 2 Chapter 4 Higher Order Linear Equations 22 Inspection of the coefficients reveals that p (t) = Based on Problem 2, we find that W =, and hence W = c 24 Writing the equation in standard form, we find that p (t) = /t Using Abel s formula, the Wronskian has the form W (t) = c e /t dt = ce ln t = c/t 26 Let y(t) = y (t)v(t) Then y = y v + y v, y = y v + 2y v + y v, and y = y v + 3y v + 3y v + y v Substitution into the ODE results in y v + 3y v + 3y v + y v + p [y v + 2y v + y v ] + +p 2 [y v + y v ] + p 3 y v = Since y is assumed to be a solution, all terms containing the factor v(t) vanish Hence y v + [p y + 3y ] v + [3y + 2p y + p 2 y ] v =, which is a second order ODE in the variable u = v 28 First write the equation in standard form: y 3 t + 2 t(t + 3) y + 6 t + t 2 (t + 3) y 6 t 2 (t + 3) y = Let y(t) = t 2 v(t) Substitution into the given ODE results in t 2 v t(t + 4) + 3 t + 3 v = Set w = v Then w is a solution of the first order differential equation w + 3 t + 4 t(t + 3) w = This equation is linear, with integrating factor µ(t) = t 4 /(t + 3) The general solution is w = c(t + 3)/t 4 Integrating twice, v(t) = c t + c t 2 + c 2 t + c 3 Hence y(t) = c t + c + c 2 t 3 + c 3 t 2 Finally, since y (t) = t 2 and y 2 (t) = t 3 are given solutions, the third independent solution is y 3 (t) = c t + c 42 The magnitude of + i is R = 2 and the polar angle is π/4 Hence the polar form is given by + i = 2 e iπ/4 3 The magnitude of 3 is R = 3 and the polar angle is π Hence 3 = 3 e iπ 4 The magnitude of i is R = and the polar angle is 3π/2 Hence i = e 3πi/2 5 The magnitude of 3 i is R = 2 and the polar angle is π/6 = π/6 Hence the polar form is given by 3 i = 2 e πi/6

113 The magnitude of i is R = 2 and the polar angle is 5π/4 Hence the polar form is given by i = 2 e 5πi/4 7 Writing the complex number in polar form, = e 2mπi, where m may be any integer Thus /3 = e 2mπi/3 Setting m =,, 2 successively, we obtain the three roots as /3 =, /3 = e 2πi/3, /3 = e 4πi/3 Equivalently, the roots can also be written as, cos(2π/3) + i sin(2π/3) = ( + i 3 )/2, cos(4π/3) + i sin(4π/3) = ( i 3 )/2 9 Writing the complex number in polar form, = e 2mπi, where m may be any integer Thus /4 = e 2mπi/4 Setting m =,, 2, 3 successively, we obtain the three roots as /4 =, /4 = e πi/2, /4 = e πi, /4 = e 3πi/2 Equivalently, the roots can also be written as, cos(π/2) + i sin(π/2) = i, cos(π) + i sin(π) =, cos(3π/2) + i sin(3π/2) = i In polar form, 2(cos π/3 + i sin π/3) = 2 e i(π/3+2mπ), in which m is any integer Thus [2(cos π/3 + i sin π/3)] /2 = 2 /2 e i(π/6+mπ) With m =, one square root is given by 2 /2 e iπ/6 = ( 3 + i)/ 2 With m =, the other root is given by 2 /2 e i7π/6 = ( 3 i)/ 2 The characteristic equation is r 3 r 2 r + = The roots are r =,, One root is repeated, hence the general solution is y = c e t + c 2 e t + c 3 te t 3 The characteristic equation is r 3 2r 2 r + 2 =, with roots r =,, 2 The roots are real and distinct, so the general solution is y = c e t + c 2 e t + c 3 e 2t 4 The characteristic equation can be written as r 2 (r 2 4r + 4) = The roots are r =,, 2, 2 There are two repeated roots, and hence the general solution is given by y = c + c 2 t + c 3 e 2t + c 4 te 2t 6 The characteristic equation can be written as (r 2 )(r 2 4) = The roots are given by r = ±, ± 2 The roots are real and distinct, hence the general solution is y = c e t + c 2 e t + c 3 e 2t + c 4 e 2t 7 The characteristic equation can be written as (r 2 ) 3 = The roots are given by r = ±, each with multiplicity three Hence the general solution is y = c e t + c 2 te t + c 3 t 2 e t + c 4 e t + c 5 te t + c 6 t 2 e t 8 The characteristic equation can be written as r 2 (r 4 ) = The roots are given by r =,, ±, ± i The general solution is y = c + c 2 t + c 3 e t + c 4 e t + c 5 cos t + c 6 sin t 9 The characteristic equation can be written as r(r 4 3r 3 + 3r 2 3r + 2) = Examining the coefficients, it follows that r 4 3r 3 + 3r 2 3r + 2 = (r )(r 2)(r 2 + ) Hence the roots are r =,, 2, ± i The general solution of the ODE is given by y = c + c 2 e t + c 3 e 2t + c 4 cos t + c 5 sin t 2 The characteristic equation can be written as r(r 3 8) =, with roots r =,

114 4 Chapter 4 Higher Order Linear Equations 2 e 2mπi/3, m =,, 2 That is, r =, 2, ± i 3 Hence the general solution is y = c + c 2 e 2t + e t [ c 3 cos 3 t + c 4 sin 3 t ] 2 The characteristic equation can be written as (r 4 + 4) 2 = The roots of the equation r = are r = ± i, ± i Each of these roots has multiplicity two The general solution is y = e t [c cos t + c 2 sin t ] + te t [c 3 cos t + c 4 sin t ] + e t [c 5 cos t + c 6 sin t ] + te t [c 7 cos t + c 8 sin t ] 22 The characteristic equation can be written as (r 2 + ) 2 = The roots are given by r = ± i, each with multiplicity two The general solution is y = c cos t + c 2 sin t + t [c 3 cos t + c 4 sin t ] 24 The characteristic equation is r 3 + 5r 2 + 6r + 2 = Examining the coefficients, we find that r 3 + 5r 2 + 6r + 2 = (r + )(r 2 + 4r + 2) Hence the roots are deduced as r =, 2 ± 2 The general solution is y = c e t + c 2 e ( 2+ 2)t + c 3 e ( 2 2)t 25 The characteristic equation is 8r 3 + 2r 2 + 4r + 4 = By examining the first and last coefficients, we find that 8r 3 + 2r 2 + 4r + 4 = (2r + )(9r 2 + 6r + 4) Hence the roots are r = /2, ( ± 3 i)/3 The general solution of the ODE is given by y = c e t/2 + e t/3 [ c 2 cos(t/ 3 ) + c 3 sin(t/ 3 ) ] 26 The characteristic equation is r 4 7r 3 + 6r 2 + 3r 36 = By examining the first and last coefficients, we find that r 4 7r 3 + 6r 2 + 3r 36 = (r 3)(r + 2)(r 2 6r + 6) The roots are r = 2, 3, 3 ± 3 The general solution is y = c e 2t + c 2 e 3t + c 3 e (3 3)t + c 4 e (3+ 3)t 28 The characteristic equation is r 4 + 6r 3 + 7r r + 4 = It can be shown that r 4 + 6r 3 + 7r r + 4 = (r 2 + 2r + 2)(r 2 + 4r + 7) Hence the roots are r = ± i, 2 ± i 3 The general solution of the euqation is y = e t (c cos t + c 2 sin t) + e 2t (c 3 cos 3 t + c 4 sin 3 t) 32 The characteristic equation is r 3 r 2 + r =, with roots r =, ± i Hence the general solution is y(t) = c e t + c 2 cos t + c 3 sin t Invoking the initial conditions, we obtain the system of equations c + c 2 = 2, c + c 3 =, c c 2 = 2, with solution c =, c 2 = 2, c 3 = Therefore the solution of the initial value problem is y(t) = 2 cos t sin t, which oscillates as t

115 The characteristic equation is 2r 4 r 3 9r 2 + 4r + 4 =, with roots r = /2,, ± 2 Hence the general solution is y(t) = c e t/2 + c 2 e t + c 3 e 2t + c 4 e 2t Applying the initial conditions, we obtain the system of equations c + c 2 + c 3 + c 4 = 2, c /2 + c 2 2c 3 + 2c 4 =, c /4 + c 2 + 4c 3 + 4c 4 = 2, c /8 + c 2 8c 3 + 8c 4 =, with solution c = 6/5, c 2 = 2/3, c 3 = /6, c 4 = / Therefore the solution of the initial value problem is y(t) = (6/5)e t/2 (2/3)e t e 2t /6 e 2t / The solution decreases without bound 35 The characteristic equation is 6 r 3 + 5r 2 + r =, with roots r =, /3, /2 The general solution is y(t) = c + c 2 e t/3 + c 3 e t/2 Invoking the initial conditions, we require that c + c 2 + c 3 = 2, c 2 /3 c 3 /2 = 2, c 2 /9 + c 3 /4 = The solution is c = 8, c 2 = 8, c 3 = 8 Therefore the solution of the initial value problem is y(t) = 8 8e t/3 + 8e t/2 It approaches 8 as t

116 6 Chapter 4 Higher Order Linear Equations 36 The general solution is derived in Problem 28 as y(t) = e t [c cos t + c 2 sin t ] + e 2t [ c 3 cos 3 t + c 4 sin 3 t Invoking the initial conditions, we obtain the system of equations c + c 3 = c + c 2 2c c 4 = 2 2c 2 + c c 4 = 2c + 2c 2 + c c 4 = 3 with solution c = 2/3, c 2 = 38/3, c 3 = 8/3, c 4 = 7 3 /39 ] The solution is a rapidly decaying oscillation 4(a) Suppose that c e rt + c 2 e r2t + + c n e rnt =, and each of the r k are real and different Multiplying this equation by e rt, we obtain that c + c 2 e (r2 r)t + + c n e (rn r)t = Differentiation results in c 2 (r 2 r )e (r2 r)t + + c n (r n r )e (rn r)t = (b) Now multiplying the latter equation by e (r2 r)t, and differentiating, we obtain c 3 (r 3 r 2 )(r 3 r )e (r3 r2)t + + c n (r n r 2 )(r n r )e (rn r2)t = (c) Following the above steps in a similar manner, it follows that c n (r n r n ) (r n r )e (rn rn )t = Since these equations hold for all t, and all the r k are different, we have c n = Hence c e rt + c 2 e r2t + + c n e rn t =, < t < (d) The same procedure can now be repeated, successively, to show that c = c 2 = = c n =

117 43 7 4(a) Recall the derivative formula d n ( ) n dx n (uv) = v dn u dx n + Let u = (r r ) s and Therefore and v = q(r) Note that ( ) n dv d n u dx dx n + + ( ) n d n v n dx n u d n dr n [(r r ) s ] = s (s ) (s n + )(r r ) s n d s dr s [(r r ) s ] = s! d n dr n [(r r ) s q(r)] = r=r only if n < s, since it is assumed that q(r ) (b) Differential operators commute, so that Likewise, It follows that r ( dk dt k ert ) = dk dt j r j ( dk dt k ert ) = dk r ) = dk dt k (t ert ) k ( ert e rt dt k ( j r j j r j L [ e rt] = L [ t j e rt] ) = dk dt k (tj e rt ) (c) From Eq (i), we have j r j [ e rt Z(r) ] = L [ t j e rt] Based on the product formula in part (a), j r j [ e rt Z(r) ] r=r = if j < s Therefore L [ t j e rt] = if j < s 43 2 The general solution of the homogeneous equation is y c = c e t + c 2 e t + c 3 cos t + c 4 sin t Let g (t) = 3t and g 2 (t) = cos t By inspection, we find that Y (t) = 3t Since g 2 (t) is a solution of the homogeneous equation, set Y 2 (t) = t(a cos t + B sin t) Substitution into the given ODE and comparing the coefficients of similar term results in A = and B = /4 Hence the general solution of the nonhomogeneous problem is y(t) = y c (t) 3t t sin t /4

118 8 Chapter 4 Higher Order Linear Equations 3 The characteristic equation corresponding to the homogeneous problem can be written as (r + )(r 2 + ) = The solution of the homogeneous equation is y c = c e t + c 2 cos t + c 3 sin t Let g (t) = e t and g 2 (t) = 4t Since g (t) is a solution of the homogeneous equation, set Y (t) = Ate t Substitution into the ODE results in A = /2 Now let Y 2 (t) = Bt + C We find that B = C = 4 Hence the general solution of the nonhomogeneous problem is y(t) = y c (t) + te t /2 + 4(t ) 4 The characteristic equation corresponding to the homogeneous problem can be written as r(r + )(r ) = The solution of the homogeneous equation is y c = c + c 2 e t + c 3 e t Since g(t) = 2 sin t is not a solution of the homogeneous problem, we can set Y (t) = A cos t + B sin t Substitution into the ODE results in A = and B = Thus the general solution is y(t) = c + c 2 e t + c 3 e t + cos t 6 The characteristic equation corresponding to the homogeneous problem can be written as (r 2 + ) 2 = It follows that y c = c cos t + c 2 sin t + t(c 3 cos t + c 4 sin t) Since g(t) is not a solution of the homogeneous problem, set Y (t) = A + B cos 2t + C sin 2t Substitution into the ODE results in A = 3, B = /9, C = Thus the general solution is y(t) = y c (t) cos 2t /9 7 The characteristic equation corresponding to the homogeneous problem can be written as r 3 (r 3 + ) = Thus the homogeneous [ solution is y c = c + c 2 t + c 3 t 2 + c 4 e t + e t/2 c 5 cos( 3 t/2) + c 5 sin( ] 3 t/2) Note the g(t) = t is a solution of the homogenous problem Consider a particular solution of the form Y (t) = t 3 (At + B) Substitution into the ODE gives us that A = /24 and B = Thus the general solution is y(t) = y c (t) + t 4 /24 8 The characteristic equation corresponding to the homogeneous problem can be written as r 3 (r + ) = Hence the homogeneous solution is y c = c + c 2 t + c 3 t 2 + c 4 e t Since g(t) is not a solution of the homogeneous problem, set Y (t) = A cos 2t + B sin 2t Substitution into the ODE results in A = /4 and B = /2 Thus the general solution is y(t) = y c (t) + (cos 2t + 2 sin 2t)/4 From Problem 22 in Section 42, the homogeneous solution is y c = c cos t + c 2 sin t + t [c 3 cos t + c 4 sin t ] Since g(t) is not a solution of the homogeneous problem, substitute Y (t) = At + B into the ODE to obtain A = 3 and B = 4 Thus the general solution is y(t) = y c (t) + 3t + 4 Invoking the initial conditions, we find that c = 4, c 2 = 4, c 3 =, c 4 = 3/2 Therefore the solution of the initial value problem is y(t) = (t 4) cos t (3t/2 + 4) sin t + 3t + 4

119 43 9 The characteristic equation can be written as r(r 2 3r + 2) = Hence the homogeneous solution is y c = c + c 2 e t + c 3 e 2t Let g (t) = e t and g 2 (t) = t Note that g is a solution of the homogeneous problem Set Y (t) = Ate t Substitution into the ODE results in A = Now let Y 2 (t) = Bt 2 + Ct Substitution into the ODE results in B = /4 and C = 3/4 Therefore the general solution is y(t) = c + c 2 e t + c 3 e 2t te t + (t 2 + 3t)/4 Invoking the initial conditions, we find that c =, c 2 = c 3 = The solution of the initial value problem is y(t) = te t + (t 2 + 3t)/4 2 The characteristic equation can be written as (r )(r + 3)(r 2 + 4) = Hence the homogeneous solution is y c = c e t + c 2 e 3t + c 3 cos 2t + c 4 sin 2t None of the terms in g(t) is a solution of the homogeneous problem Therefore we can assume a form Y (t) = Ae t + B cos t + C sin t Substitution into the ODE results in the values A = /2, B = 2/5, C = 4/5 Hence the general solution is y(t) = c e t + c 2 e 3t + c 3 cos 2t + c 4 sin 2t + e t /2 (2 cos t + 4 sin t)/5 Invoking the initial conditions, we find that c = 8/4, c 2 = 73/52, c 3 = 77/65, c 4 = 49/3

120 2 Chapter 4 Higher Order Linear Equations 4 From Problem 4, the homogeneous solution is y c = c + c 2 e t + c 3 e t Consider the terms g (t) = te t and g 2 (t) = 2 cos t Note that since r = is a simple root of the characteristic equation, we set Y (t) = t(at + B)e t The function 2 cos t is not a solution of the homogeneous equation We set Y 2 (t) = C cos t + D sin t Hence the particular solution has the form Y (t) = t(at + B)e t + C cos t + D sin t 5 The characteristic equation can be written as (r 2 ) 2 = The roots are given as r = ±, each with multiplicity two Hence the solution of the homogeneous problem is y c = c e t + c 2 te t + c 3 e t + c 4 te t Let g (t) = e t and g 2 (t) = sin t The function e t is a solution of the homogeneous problem Since r = has multiplicity two, we set Y (t) = At 2 e t The function sin t is not a solution of the homogeneous equation We can set Y 2 (t) = B cos t + C sin t Hence the particular solution has the form Y (t) = At 2 e t + B cos t + C sin t 6 The characteristic equation can be written as r 2 (r 2 + 4) =, and the roots are r =, ±2i The root r = has multiplicity two, hence the homogeneous solution is y c = c + c 2 t + c 3 cos 2t + c 4 sin 2t The functions g (t) = sin 2t and g 2 (t) = 4 are solutions of the homogenous equation The complex roots have multiplicity one, therefore we need to set Y (t) = At cos 2t + Bt sin 2t Now g 2 (t) = 4 is associated with the double root r =, so we set Y 2 (t) = Ct 2 Finally, g 3 (t) = te t (and its derivatives) is independent of the homogeneous solution Therefore set Y 3 (t) = (Dt + E)e t Conclude that the particular solution has the form Y (t) = At cos 2t + Bt sin 2t + Ct 2 + (Dt + E)e t 8 The characteristic equation can be written as r 2 (r 2 + 2r + 2) =, with roots r =, with multiplicity two, and r = ± i This means that the homogeneous solution is y c = c + c 2 t + c 3 e t cos t + c 4 e t sin t The function g (t) = 3e t + 2te t, and all of its derivatives, is independent of the homogeneous solution Therefore set Y (t) = Ae t + (Bt + C)e t Now g 2 (t) = e t sin t is a solution of the homogeneous equation, associated with the complex roots We need to set Y 2 (t) = t(d e t cos t + E e t sin t) It follows that the particular solution has the form Y (t) = Ae t + (Bt + C)e t + t(d e t cos t + E e t sin t)

121 Differentiating y = u(t)v(t), successively, we have y = u v + uv y = u v + 2u v + uv y (n) = n j= ( ) n u (n j) v (j) j Setting v(t) = e αt, v (j) = α j e αt So for any p =, 2,, n, p ( ) p y (p) = e αt α j u (p j) j It follows that L [ e αt u ] = e αt j= p n a n p p = j= ( ) p α j u j (p j) ( ) It is evident that the right hand side of Eq ( ) is of the form [ ] e αt k u (n) + k u (n ) + + k n u + k n u Hence the operator equation L [e αt u] = e αt (b t m + b t m + + b m t + b m ) can be written as k u (n) + k u (n ) + + k n u + k n u = b t m + b t m + + b m t + b m The coefficients k i, i =,,, n can be determined by collecting the like terms in the double summation in Eq ( ) For example, k is the coefficient of u (n) The only term that contains u (n) is when p = n and j = Hence k = a On the other hand, k n is the coefficient of u(t) The inner summation in ( ) contains terms with u, given by α p u (when j = p), for each p =,,, n Hence n k n = a n p α p p = 2(a) Clearly, e 2t is a solution of y 2y =, and te t is a solution of the differential equation y + 2y + y = The latter ODE has characteristic equation (r + ) 2 = Hence (D 2) [ 3e 2t] = 3(D 2) [ e 2t] = and (D + ) 2 [te t ] = Furthermore, we have (D 2)(D + ) 2 [te t ] = (D 2) [] =, and (D 2)(D + ) 2 [ 3e 2t] = (D + ) 2 (D 2) [ 3e 2t] = (D + ) 2 [] = (b) Based on part (a), (D 2)(D + ) 2 [ (D 2) 3 (D + )Y ] = (D 2)(D + ) 2 [ 3e 2t te t] =, since the operators are linear The implied operations are associative and commutative Hence (D 2) 4 (D + ) 3 Y = The operator equation corresponds to the solution of a linear homogeneous ODE with characteristic equation (r 2) 4 (r + ) 3 = The roots are r = 2, with multiplicity 4 and r =, with multiplicity 3 It

122 22 Chapter 4 Higher Order Linear Equations follows that the given homogeneous solution is Y (t) = c e 2t + c 2 te 2t + c 3 t 2 e 2t + c 4 t 3 e 2t + c 5 e t + c 6 te t + c 7 t 2 e t, which is a linear combination of seven independent solutions 22 (5) Observe that (D ) [e t ] = and (D 2 + ) [sin t] = Hence the operator H(D) = (D )(D 2 + ) is an annihilator of e t + sin t The operator corresponding to the left hand side of the given ODE is (D 2 ) 2 It follows that (D + ) 2 (D ) 3 (D 2 + )Y = The resulting ODE is homogeneous, with solution Y (t) = c e t + c 2 te t + c 3 e t + c 4 te t + c 5 t 2 e t + c 6 cos t + c 7 sin t After examining the homogeneous solution of Problem 5, and eliminating duplicate terms, we have Y (t) = c 5 t 2 e t + c 6 cos t + c 7 sin t 22 (6) We find that D [4] =, (D ) 2 [te t ] =, and (D 2 + 4) [sin 2t] = The operator H(D) = D(D ) 2 (D 2 + 4)is an annihilator of 4 + te t + sin 2t The operator corresponding to the left hand side of the ODE is D 2 (D 2 + 4) It follows that D 3 (D ) 2 (D 2 + 4) 2 Y = The resulting ODE is homogeneous, with solution Y (t) = c + c 2 t + c 3 t 2 + c 4 e t + c 5 te t + c 6 cos 2t + c 7 sin 2t + c 8 t cos 2t + c 9 t sin 2t After examining the homogeneous solution of Problem 6, and eliminating duplicate terms, we have Y (t) = c 3 t 2 + c 4 e t + c 5 te t + c 8 t cos 2t + c 9 t sin 2t 22 (8) Observe that (D ) [e t ] =, (D + ) 2 [te t ] = The function e t sin t is a solution of a second order ODE with characteristic roots r = ± i It follows that (D 2 + 2D + 2) [e t sin t] = Therefore the operator H(D) = (D )(D + ) 2 (D 2 + 2D + 2) is an annihilator of 3e t + 2te t + e t sin t The operator corresponding to the left hand side of the given ODE is D 2 (D 2 + 2D + 2) It follows that D 2 (D )(D + ) 2 (D 2 + 2D + 2) 2 Y = The resulting ODE is homogeneous, with solution Y (t) = c + c 2 t + c 3 e t + c 4 e t + c 5 te t + e t (c 6 cos t + c 7 sin t) + te t (c 8 cos t + c 9 sin t ) After examining the homogeneous solution of Problem 8, and eliminating duplicate terms, we have Y (t) = c 3 e t + c 4 e t + c 5 te t + te t (c 8 cos t + c 9 sin t ) 44 2 The characteristic equation is r(r 2 ) = Hence the homogeneous solution is y c (t) = c + c 2 e t + c 3 e t The Wronskian is evaluated as W (, e t, e t ) = 2 Now compute the three determinants e t e t W (t) = e t e t e t e t = 2, W e t 2(t) = e t e t = e t,

123 44 23 e t W 3 (t) = e t e t = et The solution of the system of equations () is u (t) = t W (t) W (t) = t, u 2(t) = t W 2(t) W (t) u 3(t) = t W 3(t) W (t) = te t /2 = te t /2, Hence u (t) = t 2 /2, u 2 (t) = e t (t + )/2, u 3 (t) = e t (t )/2 The particular solution becomes Y (t) = t 2 /2 (t + )/2 + (t )/2 = t 2 /2 The constant is a solution of the homogeneous equation, therefore the general solution is y(t) = c + c 2 e t + c 3 e t t 2 /2 3 From Problem 3 in Section 42, y c (t) = c e t + c 2 e t + c 3 e 2t The Wronskian is evaluated as W (e t, e t, e 2t ) = 6 e 2t Now compute the three determinants e t e 2t W (t) = e t 2e 2t e t 4e 2t = e t e 2t e3t, W 2 (t) = e t 2e 2t e t 4e 2t = 3et, e t e t W 3 (t) = e t e t e t e t = 2 Hence u (t) = e 5t /6, u 2(t) = e 3t /2, u 3(t) = e 2t /3 Therefore the particular solution can be expressed as Y (t) = e t [ e 5t /3 ] e t [ e 3t /6 ] + e 2t [ e 2t /6 ] = e 4t /3 6 From Problem 22 in Section 42, y c (t) = c cos t + c 2 sin t + t [c 3 cos t + c 4 sin t ] The Wronskian is evaluated as W (cos t, sin t, t cos t, t sin t) = 4 Now compute the four auxiliary determinants sin t t cos t t sin t W (t) = cos t cos t t sin t sin t + t cos t sin t 2 sin t t cos t 2 cos t t sin t = cos t 3 cos t + t sin t 3 sin t t cos t = 2 sin t + 2t cos t, cos t t cos t t sin t W 2 (t) = sin t cos t t sin t sin t + t cos t cos t 2 sin t t cos t 2 cos t t sin t = sin t 3 cos t + t sin t 3 sin t t cos t = 2t sin t + 2 cos t, cos t sin t t sin t W 3 (t) = sin t cos t sin t + t cos t cos t sin t 2 cos t t sin t = 2 cos t, sin t cos t 3 sin t t cos t

124 24 Chapter 4 Higher Order Linear Equations It follows that Hence cos t sin t t cos t W 4 (t) = sin t cos t cos t t sin t cos t sin t 2 sin t t cos t = 2 sin t sin t cos t 3 cos t + t sin t u (t) = [ sin 2 t + t sin t cos t ] /2, u 2(t) = [ t sin 2 t + sin t, cos t ] /2, u 3(t) = sin t cos t/2, and u 4(t) = sin 2 t/2 u (t) = (3 sin t cos t 2t cos 2 t t)/8, u 2 (t) = (sin 2 t 2 cos 2 t 2t sin t cos t + t 2 )/8, u 3 (t) = sin 2 t/4, and u 4 (t) = [cos t sin t t] /4 Therefore the particular solution can be expressed as Y (t) = u (t) cos t + u 2 (t) sin t + u 3 (t)t cos t + u 4 (t)t sin t = (sin t 3t cos t t 2 sin t)/8 Note that only the last term is not a solution of the homogeneous equation Hence the general solution is y(t) = c cos t + c 2 sin t + t [c 3 cos t + c 4 sin t ] t 2 sin t /8 8 Based on the results in Problem 2, y c (t) = c + c 2 e t + c 3 e t It was also shown that W (, e t, e t ) = 2, with W (t) = 2, W 2 (t) = e t, W 3 (t) = e t Therefore we have u (t) = csc t, u 2(t) = e t csc t /2, u 3(t) = e t csc t /2 The particular solution can be expressed as Y (t) = [u (t)] + e t [u 2 (t)] + e t [u 3 (t)] More specifically, t Y (t) = ln csc(t) + cot(t) + et e s csc(s)ds + e t 2 t 2 t = ln csc(t) + cot(t) + cosh(t s) csc(s)ds t t t e s csc(s)ds 9 Based on Problem 4, u (t) = sec t, u 2(t) =, u 3(t) = tan t The particular solution can be expressed as Y (t) = [u (t)] + cos t [u 2 (t)] + sin t [u 3 (t)] That is, Y (t) = ln sec(t) + tan(t) t cos t + sin t ln cos(t) Hence the general solution of the initial value problem is y(t) = c + c 2 cos t + c 3 sin t + ln sec(t) + tan(t) t cos t + sin t ln cos(t) Invoking the initial conditions, we require that c + c 2 = 2, c 3 =, c 2 = 2 Therefore y(t) = 2 cos t + sin t + ln sec(t) + tan(t) t cos t + sin t ln cos(t) Since π/2 < t < π/2, the absolute value signs may be removed

125 44 25 From Problem 6, y(t) = c cos t + c 2 sin t + c 3 t cos t + c 4 t sin t t 2 sin t/8 In order to satisfy the initial conditions, we require that c = 2, c 2 + c 3 =, c + 2c 4 =, 3/4 c 2 3c 3 = Therefore y(t) = (7 sin t 7t cos t + 4t sin t t 2 sin t)/8 + 2 cos t 2 From Problem 8, the general solution of the initial value problem is y(t) = c + c 2 e t + c 3 e t + ln csc(t) + cot(t) + t t + et e s csc(s)ds + e t e s csc(s)ds 2 t 2 t In this case, t = π/2 Observe that y(π/2) = y c (π/2), y (π/2) = y c(π/2), and y (π/2) = y c (π/2) Therefore we obtain the system of equations c + c 2 e π/2 + c 3 e π/2 = 2, c 2 e π/2 c 3 e π/2 =, c 2 e π/2 + c 3 e π/2 = Hence the solution of the initial value problem is t y(t) = 3 e t+π/2 + ln csc(t) + cot(t) + cosh(t s) csc(s)ds π/2

126 26 Chapter 4 Higher Order Linear Equations 3 First write the equation as y + x y 2x 2 y + 2x 3 y = 2x The Wronskian is evaluated as W (x, x 2, /x) = 6/x Now compute the three determinants x 2 /x W (x) = 2x /x 2 2 2/x 3 = 3, W x /x 2(x) = /x 2 2/x 3 = 2/x, x x 2 W 3 (x) = 2x 2 = x2 Hence u (x) = x 2, u 2(x) = 2x/3, u 3(x) = x 4 /3 Therefore the particular solution can be expressed as Y (x) = x [ x 3 /3 ] + x 2 [ x 2 /3 ] + [ x 5 /5 ] = x 4 /5 x 5 The homogeneous solution is y c (t) = c cos t + c 2 sin t + c 3 cosh t + c 4 sinh t The Wronskian is evaluated as W (cos t, sin t, cosh t, sinh t) = 4 Now the four additional determinants are given by W (t) = 2 sin t, W 2 (t) = 2 cos t, W 3 (t) = 2 sinh t, W 4 (t) = 2 cosh t If follows that u (t) = g(t) sin(t)/2, u 3(t) = g(t) sinh(t)/2, Therefore the particular solution can be expressed as Y (t) = cos(t) 2 cosh(t) 2 t g(s) sin(s) ds sin(t) t 2 t u 2(t) = g(t) cos(t)/2, u 4(t) = g(t) cosh(t)/2 t g(s) sinh(s) ds + sinh(t) t 2 t g(s) cos(s) ds t t g(s) cosh(s) ds Using the appropriate identities, the integrals can be combined to obtain Y (t) = 2 t t g(s) sinh(t s) ds 2 t t g(s) sin(t s) ds 7 First write the equation as y 3 x y + 6 x 2 y 6 x 3 y = g(x) / x 3 It can be shown that y c (x) = c x + c 2 x 2 + c 3 x 3 is a solution of the homogeneous equation The Wronskian of this fundamental set of solutions is W (x, x 2, x 3 ) = 2x 3 The three additional determinants are given by W (x) = x 4, W 2 (x) = 2x 3, W 3 (x) = x 2 Hence u (x) = g(x)/2x 2, u 2(x) = g(x) /x 3, u 3(x) = g(x)/2x 4 Now the particular solution can be expressed as Y (x) = x = 2 x x x x g(t) dt x2 2t2 x x [ x t 2 2 x2 t 3 + x3 t 4 g(t) dt + x3 t 3 ] g(t)dt x x g(t) 2t 4 dt =

127 C H A P T E R 5 Series Solutions of Second Order Linear Equations 5 Apply the ratio test: lim n (x 3) n+ (x 3) n = lim x 3 = x 3 n Hence the series converges absolutely for x 3 < The radius of convergence is ρ = The series diverges for x = 2 and x = 4, since the n-th term does not approach zero 3 Applying the ratio test, lim n n! x 2n+2 (n + )! x 2n = lim n x 2 n + = The series converges absolutely for all values of x Thus the radius of convergence is ρ = 4 Apply the ratio test: lim n 2 n+ x n+ 2 n x n = lim 2 x = 2 x n Hence the series converges absolutely for 2 x <, or x < /2 The radius of convergence is ρ = /2 The series diverges for x = ±/2, since the n-th term does not approach zero 27

128 28 Chapter 5 Series Solutions of Second Order Linear Equations 6 Applying the ratio test, lim n n(x x ) n+ (n + )(x x ) n = lim n n n + (x x ) = (x x ) Hence the series converges absolutely for (x x ) < The radius of convergence is ρ = At x = x +, we obtain the harmonic series, which is divergent At the other endpoint, x = x, we obtain ( ) n, n which is conditionally convergent 7 Apply the ratio test : lim n 3 n (n + ) 2 (x + 2) n+ 3 n+ n 2 (x + 2) n n= (n + ) 2 = lim n 3 n 2 (x + 2) = (x + 2) 3 Hence the series converges absolutely for 3 x + 2 <, or x + 2 < 3 The radius of convergence is ρ = 3 At x = 5 and x = +, the series diverges, since the n-th term does not approach zero 8 Applying the ratio test, since lim n n n (n + )! x n+ (n + ) n+ n! x n = lim n lim n n n (n + ) n x = e x, n n (n + ) n = lim n ( + n ) n = e Hence the series converges absolutely for x < e The radius of convergence is ρ = e At x = ± e, the series diverges, since the n-th term does not approach zero This follows from the fact that lim n n! e n n n 2πn = We have f(x) = e x, with f (n) (x) = e x, for n =, 2, Therefore f (n) () = Hence the Taylor expansion about x = is Applying the ratio test, lim n e x = n!x n+ (n + )! x n = The radius of convergence is ρ = n = x n n! lim n x = n +

129 5 29 We have f(x) = x, with f (x) = and f (n) (x) =, for n = 2, Clearly, f() = and f () =, with all other derivatives equal to zero Hence the Taylor expansion about x = is x = + (x ) Since the series has only a finite number of terms, it converges absolutely for all x 4 We have f(x) = /( + x), f (x) = /( + x) 2, f (x) = 2/( + x) 3, with f (n) (x) = ( ) n n!/( + x) n+, for n It follows that f (n) () = ( ) n n! for n Hence the Taylor expansion about x = is Applying the ratio test, lim n + x = ( ) n x n x n+ x n n = = lim x = x n The series converges absolutely for x <, but diverges at x = ± 5 We have f(x) = /( x), f (x) = /( x) 2, f (x) = 2/( x) 3, with f (n) (x) = n!/( x) n+, for n It follows that f (n) () = n!, for n Hence the Taylor expansion about x = is Applying the ratio test, lim n x = x n+ x n n = x n = lim x = x n The series converges absolutely for x <, but diverges at x = ± 6 We have f(x) = /( x), f (x) = /( x) 2, f (x) = 2/( x) 3, with f (n) (x) = n!/( x) n+, for n It follows that f (n) (2) = ( ) n+ n! for n Hence the Taylor expansion about x = 2 is Applying the ratio test, lim n x = (x 2) n+ (x 2) n n = ( ) n (x 2) n = lim x 2 = x 2 n The series converges absolutely for x 2 <, but diverges at x = and x = 3 7 Applying the ratio test, lim n (n + )x n+ n x n n + = lim x = x n n

130 3 Chapter 5 Series Solutions of Second Order Linear Equations The series converges absolutely for x < Term-by-term differentiation results in y = n 2 x n = + 4x + 9x x 3 + n = y = n 2 (n ) x n 2 = 4 + 8x + 48x 2 + x 3 + n = 2 Shifting the indices, we can also write y = (n + ) 2 x n and y = (n + 2) 2 (n + ) x n n = n = 2 Shifting the index in the second series, that is, setting n = k +, a k x k+ = a n x n k = n = Hence a k+ x k + a k x k+ = a k+ x k + a k x k k = k = k = = a + k = (a k+ + a k )x k+ k = 2 Shifting the index by 2, that is, setting m = n 2, n(n )a n x n 2 = (m + 2)(m + )a m+2 x m n = 2 = m = (n + 2)(n + )a n+2 x n n = 22 Shift the index down by 2, that is, set m = n + 2 It follows that a n x n+2 = a m 2 x m = a n 2 x n n = m = 2 n = 2 24 Clearly, ( x 2 ) n(n )a n x n 2 = n(n )a n x n 2 n(n )a n x n n = 2 n = 2 n = 2 Shifting the index in the first series, that is, setting k = n 2, n(n )a n x n 2 = (k + 2)(k + )a k+2 x k n = 2 = k = (n + 2)(n + )a n+2 x n n =

131 5 3 Hence ( x 2 ) n(n )a n x n 2 = n = 2 (n + 2)(n + )a n+2 x n n = n(n )a n x n Note that when n = and n =, the coefficients in the second series are zero So ( x 2 ) n(n )a n x n 2 = [(n + 2)(n + )a n+2 n(n )a n ] x n n = 2 26 Clearly, n = na n x n + x n = a n x n = n = na n x n + n = n = 2 a n x n+ n = Shifting the index in the first series, that is, setting k = n, na n x n = (k + )a k+ x k n = k = Shifting the index in the second series, that is, setting k = n +, a n x n+ = a k x k n = k = Combining the series, and starting the summation at n =, na n x n + x a n x n = a + [(n + )a n+ + a n ] x n n = n = 27 We note that x n(n )a n x n 2 + n = 2 a n x n = n = n = n(n )a n x n + n = 2 a n x n n = Shifting the index in the first series, that is, setting k = n, n(n )a n x n = k(k + )a k+ x k = k(k + )a k+ x k, n = 2 k = since the coefficient of the term associated with k = is zero Combining the series, x n(n )a n x n 2 + a n x n = [n(n + )a n+ + a n ] x n n = 2 n = n = k =

132 32 Chapter 5 Series Solutions of Second Order Linear Equations 52 (a,b,d) Let y = a + a x + a 2 x a n x n + Then y = n(n )a n x n 2 = n = 2 (n + 2)(n + )a n+2 x n n = Substitution into the ODE results in (n + 2)(n + )a n+2 x n or n = a n x n = n = [(n + 2)(n + )a n+2 a n ] x n = n = Equating all the coefficients to zero, (n + 2)(n + )a n+2 a n =, n =,, 2, We obtain the recurrence relation a n a n+2 =, n =,, 2, (n + )(n + 2) The subscripts differ by two, so for k =, 2, and Hence a 2k = a 2k+ = a 2k 2 (2k )2k = a 2k 4 (2k 3)(2k 2)(2k )2k = = a (2k)! a 2k 2k(2k + ) = a 2k 3 (2k 2)(2k )2k(2k + ) = = a (2k + )! y = a k = The linearly independent solutions are (c) The Wronskian at is x 2k (2k)! + a k = x 2k+ (2k + )! y = + x2 2! + x4 4! + x6 6! + = cosh x y 2 = x + x3 3! + x5 5! + x7 7! + = sinh x 4(a,b,d) Let y = a + a x + a 2 x a n x n + Then y = n(n )a n x n 2 = n = 2 (n + 2)(n + )a n+2 x n n =

133 52 33 Substitution into the ODE results in (n + 2)(n + )a n+2 x n + k 2 x 2 a n x n = n = Rewriting the second summation, (n + 2)(n + )a n+2 x n + that is, n = 2a a 3 x + n = k 2 a n 2 x n =, n = 2 [ (n + 2)(n + )an+2 + k 2 ] a n 2 x n = n = 2 Setting the coefficients equal to zero, we have a 2 =, a 3 =, and (n + 2)(n + )a n+2 + k 2 a n 2 =, for n = 2, 3, 4, The recurrence relation can be written as k 2 a n 2 a n+2 =, n = 2, 3, 4, (n + 2)(n + ) The indices differ by four, so a 4, a 8, a 2, are defined by a 4 = k2 a 4 3, a 8 = k2 a 4 8 7, a 2 = k2 a 8 2, Similarly, a 5, a 9, a 3, are defined by a 5 = k2 a 5 4, a 9 = k2 a 5 9 8, a 3 = k2 a 9 3 2, The remaining coefficients are zero Therefore the general solution is y = a [ k2 4 3 x4 + + a [ x k2 5 4 x5 + Note that for the even coefficients, and for the odd coefficients, k x8 k x9 k2 a 4m 4 k 6 ] x2 + + k 6 ] x3 + a 4m =, m =, 2, 3, (4m )4m k2 a 4m 3 a 4m+ =, m =, 2, 3, 4m(4m + ) Hence the linearly independent solutions are ( ) m+ (k 2 x 4 ) m+ y (x) = (4m + 3)(4m + 4) m = [ ] ( ) m+ (k 2 x 4 ) m+ y 2 (x) = x (4m + 4)(4m + 5) m =

134 34 Chapter 5 Series Solutions of Second Order Linear Equations (c) The Wronskian at is 6(a,b) Let y = a + a x + a 2 x a n x n + Then y = na n x n = (n + )a n+ x n and y = n = n(n )a n x n 2 = n = 2 Substitution into the ODE results in (2 + x 2 ) (n + 2)(n + )a n+2 x n x n = Before proceeding, write and It follows that + x 2 n = n = (n + 2)(n + )a n+2 x n n = (n + )a n+ x n + 4 n = (n + 2)(n + )a n+2 x n = x (n + )a n+ x n = n = n(n )a n x n n = 2 n a n x n n = 4a + 4a 2 + (3a + 2a 3 )x+ a n x n = n = [2(n + 2)(n + )a n+2 + n(n )a n n a n + 4a n ] x n = n = 2 Equating the coefficients to zero, we find that a 2 = a, a 3 = a /4, and a n+2 = n2 2n + 4 2(n + 2)(n + ) a n, n =,, 2, The indices differ by two, so for k =,, 2, and a 2k+2 = (2k)2 4k + 4 2(2k + 2)(2k + ) a 2k a 2k+3 = (2k + )2 4k + 2 2(2k + 3)(2k + 2) a 2k+ Hence the linearly independent solutions are (c) The Wronskian at is y (x) = x 2 + x4 6 x6 3 + y 2 (x) = x x x5 6 9x7 92 +

135 (a,b,d) Let y = a + a x + a 2 x a n x n + Then y = na n x n = (n + )a n+ x n and y = n = n(n )a n x n 2 = n = 2 Substitution into the ODE results in (n + 2)(n + )a n+2 x n + x n = First write We then obtain 2a 2 + 2a + x n = (n + 2)(n + )a n+2 x n n = (n + )a n+ x n + 2 n = (n + )a n+ x n = n = n a n x n n = a n x n = n = [(n + 2)(n + )a n+2 + n a n + 2a n ] x n = n = It follows that a 2 = a and a n+2 = a n /(n + ), n =,, 2, Note that the indices differ by two, so for k =, 2, and a 2k = a 2k 2 2k = a 2k 4 (2k 3)(2k ) = = ( ) k a 3 5 (2k ) a 2k+ = a 2k 2k Hence the linearly independent solutions are = a 2k 3 (2k 2)2k = = ( ) k a (2k) y (x) = x2 + x4 3 x = + n = y 2 (x) = x x3 2 + x5 2 4 x = x + (c) The Wronskian at is n = 9(a,b,d) Let y = a + a x + a 2 x a n x n + Then y = na n x n = (n + )a n+ x n and y = n = n(n )a n x n 2 = n = 2 n = ( ) n x 2n 3 5 (2n ) ( ) n x 2n (2n) (n + 2)(n + )a n+2 x n n =

136 36 Chapter 5 Series Solutions of Second Order Linear Equations Substitution into the ODE results in ( + x 2 ) (n + 2)(n + )a n+2 x n 4x n = Before proceeding, write and It follows that + x 2 n = (n + )a n+ x n + 6 n = (n + 2)(n + )a n+2 x n = x (n + )a n+ x n = n = n(n )a n x n n = 2 n a n x n n = 6a + 2a 2 + (2a + 6a 3 )x+ a n x n = n = [(n + 2)(n + )a n+2 + n(n )a n 4n a n + 6a n ] x n = n = 2 Setting the coefficients equal to zero, we obtain a 2 = 3a, a 3 = a /3, and (n 2)(n 3) a n+2 = (n + )(n + 2) a n, n =,, 2, Observe that for n = 2 and n = 3, we obtain a 4 = a 5 = Since the indices differ by two, we also have a n = for n 4 Therefore the general solution is a polynomial y = a + a x 3a x 2 a x 3 /3 Hence the linearly independent solutions are y (x) = 3x 2 and y 2 (x) = x x 3 /3 (c) The Wronskian is (x 2 + ) 2 At x = it is (a,b,d) Let y = a + a x + a 2 x a n x n + Then y = n(n )a n x n 2 = (n + 2)(n + )a n+2 x n n = 2 n = Substitution into the ODE results in (4 x 2 ) (n + 2)(n + )a n+2 x n + 2 a n x n = First write It follows that x 2 n = n = (n + 2)(n + )a n+2 x n = n = n = 2 2a + 8a 2 + (2a + 24a 3 )x+ n(n )a n x n

137 [4(n + 2)(n + )a n+2 n(n )a n + 2a n ] x n = n = 2 We obtain a 2 = a /4, a 3 = a /2 and 4(n + 2)a n+2 = (n 2)a n, n =,, 2, Note that for n = 2, a 4 = Since the indices differ by two, we also have a 2k = for k = 2, 3, On the other hand, for k =, 2,, a 2k+ = (2k 3)a 2k 4(2k + ) Therefore the general solution is y = a + a x a x 2 = (2k 5)(2k 3)a 2k (2k )(2k + ) 4 a n = = = x 2n+ 4 n (2n )(2n + ) Hence the linearly independent solutions are y (x) = x 2 /4 and y 2 (x) = x x3 2 x5 24 x7 224 = x (c) The Wronskian at is n = (a,b,d) Let y = a + a x + a 2 x a n x n + Then y = na n x n = (n + )a n+ x n and y = n = n(n )a n x n 2 = n = 2 Substitution into the ODE results in (3 x 2 ) (n + 2)(n + )a n+2 x n 3x n = Before proceeding, write and It follows that + x 2 n = n = a 4 k (2k )(2k + ) x 2n+ 4 n (2n )(2n + ) (n + 2)(n + )a n+2 x n n = (n + )a n+ x n n = (n + 2)(n + )a n+2 x n = x (n + )a n+ x n = n = n(n )a n x n n = 2 n a n x n n = 6a 2 a + ( 4a + 8a 3 )x+ a n x n = n = [3(n + 2)(n + )a n+2 n(n )a n 3n a n a n ] x n = n = 2

138 38 Chapter 5 Series Solutions of Second Order Linear Equations We obtain a 2 = a /6, 2a 3 = a /9, and 3(n + 2)a n+2 = (n + )a n, n =,, 2, The indices differ by two, so for k =, 2, and a 2k = (2k )a 2k 2 3(2k) = (2k 3)(2k )a 2k (2k 2)(2k) = = 3 5 (2k ) a 3 k 2 4 (2k) a 2k+ = (2k)a 2k 3(2k + ) = (2k 2)(2k)a 2k (2k )(2k + ) = = (2k) a 3 k 3 5 (2k + ) Hence the linearly independent solutions are y (x) = + x2 6 + x x = + n = y 2 (x) = x + 2x x x = x + (c) The Wronskian at is n = 2(a,b,d) Let y = a + a x + a 2 x a n x n + Then y = na n x n = (n + )a n+ x n and y = n = n(n )a n x n 2 = n = 2 Substitution into the ODE results in ( x) (n + 2)(n + )a n+2 x n + x n = n = 3 5 (2n ) x 2n 3 n 2 4 (2n) (2n) x 2n+ 3 n 3 5 (2n + ) (n + 2)(n + )a n+2 x n n = (n + )a n+ x n n = Before proceeding, write x (n + 2)(n + )a n+2 x n = and It follows that 2a 2 a + n = x (n + )a n+ x n = n = a n x n = n = (n + )n a n+ x n n = n a n x n n = [(n + 2)(n + )a n+2 (n + )n a n+ + n a n a n ] x n = n = We obtain a 2 = a /2 and (n + 2)(n + )a n+2 (n + )n a n+ + (n )a n =

139 52 39 for n =,, 2, Writing out the individual equations, 3 2 a 3 2 a 2 = 4 3 a a 3 + a 2 = 5 4 a a a 3 = 6 5 a a a 4 = The coefficients are calculated successively as a 3 = a /(2 3), a 4 = a 3 /2 a 2 /2 = a /24, a 5 = 3a 4 /5 a 3 / = a /2, We can now see that for n 2, a n is proportional to a In fact, for n 2, a n = a /(n!) Therefore the general solution is y = a + a x + a x 2 + a x 3 + a x 4 + 2! 3! 4! Hence the linearly independent solutions are y 2 (x) = x and y (x) = + n = 2 (c) The Wronskian is e x ( x) At x = it is x n n! = ex x 3(a,b,d) Let y = a + a x + a 2 x a n x n + Then and y = y = na n x n = n = n(n )a n x n 2 = n = 2 Substitution into the ODE results in 2 (n + 2)(n + )a n+2 x n + x n = First write We then obtain 4a 2 + 3a + x (n + )a n+ x n n = (n + 2)(n + )a n+2 x n n = (n + )a n+ x n + 3 n = (n + )a n+ x n = n = n a n x n n = a n x n = n = [2(n + 2)(n + )a n+2 + n a n + 3a n ] x n = n = It follows that a 2 = 3a /4 and 2(n + 2)(n + )a n+2 + (n + 3)a n =

140 4 Chapter 5 Series Solutions of Second Order Linear Equations for n =,, 2, The indices differ by two, so for k =, 2, and a 2k = (2k + )a 2k 2 2(2k )(2k) = (2k )(2k + )a 2k (2k 3)(2k 2)(2k )(2k) = = ( )k 3 5 (2k + ) 2 k (2k)! a a 2k+ = (2k + 2)a 2k 2(2k)(2k + ) = (2k)(2k + 2)a 2k (2k 2)(2k )(2k)(2k + ) = = ( )k 4 6 (2k)(2k + 2) 2 k (2k + )! Hence the linearly independent solutions are y (x) = 3 4 x x x6 + = a y 2 (x) = x 3 x3 + 2 x5 2 x7 + = x + (c) The Wronskian at is 5(a) From Problem 2, we have y (x) = n = x 2n 2 n n! n = and y 2 (x) = n = ( ) n 3 5 (2n + ) 2 n x 2n (2n)! ( ) n 4 6 (2n + 2) 2 n x 2n+ (2n + )! n = 2 n n! x 2n+ (2n + )! Since a = y() and a = y (), we have y(x) = 2 y (x) + y 2 (x) That is, y(x) = 2 + x + x x3 + 4 x4 + 5 x x6 + The four- and five-term polynomial approximations are p 4 = 2 + x + x 2 + x 3 /3, and p 5 = 2 + x + x 2 + x 3 /3 + x 4 /4 (b)

141 52 4 (c) The four-term approximation p 4 appears to be reasonably accurate (within %) on the interval x < 7 7(a) From Problem 7, the linearly independent solutions are ( ) n x 2n y (x) = (2n ) and y ( ) n x 2n+ 2(x) = x (2n) n = n = Since a = y() and a = y (), we have y(x) = 4 y (x) y 2 (x) That is, y(x) = 4 x 4x x x4 8 x5 4 5 x6 + The four- and five-term polynomial approximations are p 4 = 4 x 4x x3, and p 5 = 4 x 4x x x4 (b) (c) The four-term approximation p 4 appears to be reasonably accurate (within %) on the interval x < 5 8(a) From Problem 2, we have y (x) = + n = 2 x n n! and y 2 (x) = x Since a = y() and a = y (), we have y(x) = 3 y (x) + 2 y 2 (x) That is, y(x) = 3 + 2x 3 2 x2 2 x3 8 x4 4 x5 24 x6 + The four- and five-term polynomial approximations are p 4 = 3 + 2x 3 2 x2 2 x3, and p 5 = 3 + 2x 3 2 x2 2 x3 8 x4

142 42 Chapter 5 Series Solutions of Second Order Linear Equations (b) (c) The four-term approximation p 4 appears to be reasonably accurate (within %) on the interval x < 9 2 Two linearly independent solutions of Airy s equation (about x = ) are y (x) = + y 2 (x) = x + n = n = x 3n 2 3 (3n )(3n) x 3n+ 3 4 (3n)(3n + ) Applying the ratio test to the terms of y (x), 2 3 (3n )(3n) x 3n+3 lim n 2 3 (3n + 2)(3n + 3) x 3n = lim n (3n + )(3n + 2)(3n + 3) x 3 = Similarly, applying the ratio test to the terms of y 2 (x), 3 4 (3n)(3n + ) x 3n+4 lim n 3 4 (3n + 3)(3n + 4) x 3n+ = lim n (3n + 2)(3n + 3)(3n + 4) x 3 = Hence both series converge absolutely for all x 2 Let y = a + a x + a 2 x a n x n + Then and y = y = na n x n = n = n(n )a n x n 2 = n = 2 Substitution into the ODE results in (n + 2)(n + )a n+2 x n 2x n = (n + )a n+ x n n = (n + 2)(n + )a n+2 x n n = (n + )a n+ x n + λ n = a n x n = n =

143 52 43 First write We then obtain 2a 2 + λ a + x (n + )a n+ x n = n = n a n x n n = [(n + 2)(n + )a n+2 2n a n + λ a n ] x n = n = Setting the coefficients equal to zero, it follows that a n+2 = (2n λ) (n + )(n + 2) a n for n =,, 2, Note that the indices differ by two, so for k =, 2, and a 2k = (4k 4 λ)a 2k 2 (2k )2k = (4k 8 λ)(4k 4 λ)a 2k 4 (2k 3)(2k 2)(2k )2k k λ (λ 4k + 8)(λ 4k + 4) = ( ) a (2k)! a 2k+ = (4k 2 λ)a 2k 2k(2k + ) = (4k 6 λ)(4k 2 λ)a 2k 3 (2k 2)(2k )2k(2k + ) k (λ 2) (λ 4k + 6)(λ 4k + 2) = ( ) a (2k + )! Hence the linearly independent solutions of the Hermite equation (about x = ) are y (x) = λ λ(λ 4) 2! x2 + x 4 λ(λ 4)(λ 8) x 6 + 4! 6! y 2 (x) = x λ 2 x 3 + 3! (b) Based on the recurrence relation (λ 2)(λ 6) x 5 5! a n+2 = (2n λ) (n + )(n + 2) a n, (λ 2)(λ 6)(λ ) x 7 + 7! the series solution will terminate as long as λ is a nonnegative even integer If λ = 2m, then one or the other of the solutions in part (b) will contain at most m/2 + terms In particular, we obtain the polynomial solutions corresponding to λ =, 2, 4, 6, 8, : λ = y (x) = λ = 2 y 2 (x) = x λ = 4 y (x) = 2x 2 λ = 6 y 2 (x) = x 2x 3 /3 λ = 8 y (x) = 4x 2 + 4x 4 /3 λ = y 2 (x) = x 4x 3 /3 + 4x 5 /5

144 44 Chapter 5 Series Solutions of Second Order Linear Equations (c) Observe that if λ = 2n, and a = a =, then k 2n (2n 4k + 8)(2n 4k + 4) a 2k = ( ) (2k)! and k (2n 2) (2n 4k + 6)(2n 4k + 2) a 2k+ = ( ) (2k + )! for k =, 2, [n/2] It follows that the coefficient of x n, in y and y 2, is { ( ) k 4 k k! (2k)! for n = 2k a n = ( ) k 4k k! (2k+)! for n = 2k + Then by definition, { ( ) k 2 n (2k)! H n (x) = 4 k k! y (x) = ( ) k (2k)! k! y (x) for n = 2k ( ) k 2 n (2k+)! y 4 k k! 2 (x) = ( ) k 2 (2k+)! k! y 2 (x) for n = 2k + Therefore the first six Hermite polynomials are H (x) = H (x) = 2x 24 The series solution is given by H 2 (x) = 4x 2 2 H 3 (x) = 8x 8 2x H 4 (x) = 6x 4 48x H 5 (x) = 32x 5 6x 3 + 2x y(x) = x 2 + x4 6 x6 3 + x The series solution is given by y(x) = x x3 2 + x5 2 4 x x

145 The series solution is given by y(x) = x4 2 + x8 672 x Let y = a + a x + a 2 x a n x n + Then y = na n x n = (n + )a n+ x n n = n = and y = n(n )a n x n 2 = (n + 2)(n + )a n+2 x n n = 2 n = Substitution into the ODE results in ( x) (n + 2)(n + )a n+2 x n + x (n + )a n+ x n 2 a n x n = n = n = n = After appropriately shifting the indices, it follows that 2a 2 2a + [(n + 2)(n + )a n+2 (n + )n a n+ + n a n 2 a n ] x n = n = We find that a 2 = a and (n + 2)(n + )a n+2 (n + )n a n+ + (n 2)a n =

146 46 Chapter 5 Series Solutions of Second Order Linear Equations for n =, 2, Writing out the individual equations, 3 2 a 3 2 a 2 a = 4 3 a a 3 = 5 4 a a 4 + a 3 = 6 5 a a a 4 = Since a = and a =, the remaining coefficients satisfy the equations 3 2 a 3 = 4 3 a a 3 = 5 4 a a 4 + a 3 = 6 5 a a a 4 = That is, a 3 = /6, a 4 = /2, a 5 = /24, a 6 = /45, Hence the series solution of the initial value problem is y(x) = x + 6 x3 + 2 x x x x Let y = φ(x) be a solution of the initial value problem First note that Differentiating twice, y = (sin x)y (cos x)y y = (sin x)y 2(cos x)y + (sin x)y y (4) = (sin x)y 3(cos x)y + 3(sin x)y + (cos x)y Given that φ() = and φ () =, the first equation gives φ () = and the last two equations give φ () = 2 and φ (4) () =

147 Let y = φ(x) be a solution of the initial value problem First note that Differentiating twice, y = x 2 y (sin x)y y = x 2 y (2x + sin x)y (cos x)y y (4) = x 2 y (4x + sin x)y (2 + 2 cos x)y + (sin x)y Given that φ() = a and φ () = a, the first equation gives φ () = and the last two equations give φ () = a and φ (4) () = 4a 5 Clearly, p(x) = 4 and q(x) = 6x are analytic for all x Hence the series solutions converge everywhere 8 The only root of P (x) = x is zero Hence ρ min = 2 The Taylor series expansion of e x, about x =, is e x x n = n! n = Let y = a + a x + a 2 x a n x n + Substituting into the ODE, [ ] [ x n ] (n + 2)(n + )a n+2 x n + x a n x n = n! n = First note that x a n x n = n = n = n = a n x n = a x + a x 2 + a 2 x a n x n + n = The coefficient of x n in the product of the two series is c n = 2a 2 n! + 6a 3 (n )! + 2a 4 (n 2)! + + (n + )n a n+ + (n + 2)(n + )a n+2 Expanding the individual series, it follows that 2a 2 + (2a 2 + 6a 3 )x + (a 2 + 6a 3 + 2a 4 )x 2 + (a 2 + 6a 3 + 2a 4 + 2a 5 )x a x + a x 2 + a 2 x 3 + = Setting the coefficients equal to zero, we obtain the system 2a 2 =, 2a 2 + 6a 3 + a =, a 2 + 6a 3 + 2a 4 + a =, a 2 + 6a 3 + 2a 4 + 2a 5 + a 2 =, Hence the general solution is y(x) = a + a x a x (a a ) x4 2 + (2a a ) x5 4 + (4 3 a 2a ) x6 2 + We find that two linearly independent solutions (W (y, y 2 )() = ) are y (x) = x3 6 + x4 2 x5 4 +

148 48 Chapter 5 Series Solutions of Second Order Linear Equations y 2 (x) = x x4 2 + x5 2 x6 6 + Since p(x) = and q(x) = xe x converge everywhere, ρ = 3 The Taylor series expansion of cos x, about x =, is ( ) n x 2n cos x = (2n)! n = Let y = a + a x + a 2 x a n x n + Substituting into the ODE, [ ] [ ( ) n x 2n ] (n + 2)(n + )a n+2 x n + na n x n 2 a n x n = (2n)! n = n = n = The coefficient of x n in the product of the two series is n = c n = 2a 2 b n + 6a 3 b n + 2a 4 b n (n + )na n+ b + (n + 2)(n + )a n+2 b, in which cos x = b + b x + b 2 x b n x n + It follows that 2a 2 2a + c n x n + (n 2)a n x n = n = n = Expanding the product of the series, it follows that 2a 2 2a + 6a 3 x+( a 2 + 2a 4 )x 2 + ( 3a 3 + 2a 5 )x 3 + a x + a 3 x 3 + 2a 4 x 4 + = Setting the coefficients equal to zero, a 2 a =, 6a 3 a =, a 2 + 2a 4 =, 3a 3 + 2a 5 + a 3 =, Hence the general solution is y(x) = a + a x + a x 2 + a x 3 x 4 x 5 x 6 x a 2 + a 6 + a 2 + a 56 + We find that two linearly independent solutions (W (y, y 2 )() = ) are y (x) = + x 2 + x4 2 + x6 2 + y 2 (x) = x + x3 6 + x5 6 + x The nearest zero of P (x) = cos x is at x = ±π/2 Hence ρ min = π/2 4 The Taylor series expansion of ln( + x), about x =, is ln( + x) = n = ( ) n+ x n n Let y = a + a x + a 2 x a n x n + Substituting into the ODE, [ ] ( ) n x n (n + 2)(n + )a n+2 x n n! n = n =

149 [ n = The first product is the series ] ( ) n+ x n (n + )a n+ x n x n n = a n x n = n = 2a 2 + ( 2a 2 + 6a 3 )x + (a 2 6a 3 + 2a 4 )x 2 + ( a 2 + 6a 3 2a 4 + 2a 5 )x 3 + The second product is the series a x + (2a 2 a /2)x 2 + (3a 3 a 2 + a /3)x 3 + (4a 4 3a 3 /2 + 2a 2 /3 a /4)x 3 + Combining the series and equating the coefficients to zero, we obtain Hence the general solution is 2a 2 = 2a 2 + 6a 3 + a a = 2a 4 6a 3 + 3a 2 3a /2 = 2a 5 2a 4 + 9a 3 3a 2 + a /3 = y(x) = a + a x + (a a ) x3 6 + (2a + a ) x a 7x (5 3 a a ) x6 2 + We find that two linearly independent solutions (W (y, y 2 )() = ) are y (x) = + x3 6 + x4 2 x6 2 + y 2 (x) = x x3 6 + x x5 2 + The coefficient p(x) = e x ln( + x) is analytic at x =, but its power series has a radius of convergence ρ = 5 If y = x and y 2 = x 2 are solutions, then substituting y 2 into the ODE results in 2 P (x) + 2x Q(x) + x 2 R(x) = Setting x =, we find that P () = Similarly, substituting y into the ODE results in Q() = Therefore P (x)/q(x) and R(x)/P (x) may not be analytic If they were, Theorem 32 would guarantee that y and y 2 were the only two solutions But note that an arbitrary value of y() cannot be a linear combination of y () and y 2 () Hence x = must be a singular point 6 Let y = a + a x + a 2 x a n x n + Substituting into the ODE, (n + )a n+ x n a n x n = That is, n = n = [(n + )a n+ a n ] x n = n =

150 5 Chapter 5 Series Solutions of Second Order Linear Equations Setting the coefficients equal to zero, we obtain a n+ = a n n + for n =,, 2, It is easy to see that a n = a /(n!) Therefore the general solution is [ ] y(x) = a + x + x2 2! + x3 3! + = a e x The coefficient a = y(), which can be arbitrary 7 Let y = a + a x + a 2 x a n x n + Substituting into the ODE, (n + )a n+ x n x a n x n = That is, n = (n + )a n+ x n n = n = a n x n = n = Combining the series, we have a + [(n + )a n+ a n ] x n = n = Setting the coefficient equal to zero, a = and a n+ = a n /(n + ) for n =, 2, Note that the indices differ by two, so for k =, 2, a 2k = a 2k 2 (2k) = a 2k 4 (2k 2)(2k) = = a 2 4 (2k) and a 2k+ = Hence the general solution is [ ] y(x) = a + x2 2 + x ! + x ! + + x2n 2 n n! + = a e x2 /2 The coefficient a = y(), which can be arbitrary 9 Let y = a + a x + a 2 x a n x n + Substituting into the ODE, ( x) (n + )a n+ x n a n x n = That is, n = (n + )a n+ x n n = n = n a n x n n = a n x n = n = Combining the series, we have a a + [(n + )a n+ n a n a n ] x n = n =

151 53 5 Setting the coefficients equal to zero, a = a and a n+ = a n for n =,, 2, Hence the general solution is y(x) = a [ + x + x 2 + x x n + ] = a x The coefficient a = y(), which can be arbitrary 2 Let y = a + a x + a 2 x a n x n + Substituting into the ODE, That is, (n + )a n+ x n + x n = (n + )a n+ x n + n = a n x n = + x n = a n x n = + x n = Combining the series, and the nonhomogeneous terms, we have (a ) + (2a 2 + a )x + [(n + )a n+ + a n ] x n = n = 2 Setting the coefficients equal to zero, we obtain a =, 2a 2 + a =, and a n = a n 2 n The indices differ by two, so for k = 2, 3,, n = 3, 4, a 2k = a 2k 2 (2k) = a 2k 4 (2k 2)(2k) = = ( )k a (2k) = ( )k (a ) (2k), and for k =, 2, a 2k+ = a 2k (2k + ) = a 2k 3 (2k )(2k + ) = = ( ) k 3 5 (2k + ) Hence the general solution is y(x) = a + x + a x 2 x a 2 2 2! + x5 3 5 a 2 3 3! Collecting the terms containing a, [ ] y(x) = a x2 2 + x ! x ! + ] + [x + x2 2 x3 3 x ! + x x ! x Upon inspection, we find that ] y(x) = a e x2 /2 + [x + x2 2 x3 3 x ! + x x ! x x 4 x 6

152 52 Chapter 5 Series Solutions of Second Order Linear Equations Note that the given ODE is first order linear, with integrating factor µ(x) = e x2 /2 The general solution is given by y(x) = e x2 /2 x e u2 /2 du + (y() )e x2 / If α =, then y (x) = If α = 2n, then a 2m = for m n + As a result, y (x) = + n ( ) m 2m n(n ) (n m + )(2n + )(2n + 3) (2n + 2m ) x 2m (2m)! m = α = α = 2 3x 2 α = 4 x x4 If α = 2n +, then a 2m+ = for m n + As a result, y 2 (x) = x + n ( ) m 2m n(n ) (n m + )(2n + 3)(2n + 5) (2n + 2m + ) x 2m+ (2m + )! m = 24(a) Based on Problem 23, Normalizing the polynomials, we obtain α = x α = 3 α = 5 x 5 3 x3 x 4 3 x x5 α = y () = α = 2 3x 2 y () = 2 α = 4 x x4 y () = 8 3 P (x) = Similarly, P 2 (x) = x2 P 4 (x) = x x4 α = x y 2 () = α = 3 α = 5 x 5 3 x3 x 4 3 x x5 y 2 () = 2 3 y 2 () = 8 5 P (x) = x P 3 (x) = 3 2 x x3 P 5 (x) = 5 8 x 35 4 x x5

153 53 53 (b) (c) P (x) has no roots P (x) has one root at x = The zeros of P 2 (x) are at x = ± / 3 The zeros of P 3 (x) are x =, ± 3/5 The roots of P 4 (x) are given by x 2 = ( )/35, (5 2 3)/35 The roots of P 5 (x) are given by x = and x 2 = ( )/63, (35 2 7)/63 25 Observe that P n ( ) = ( )n 2 n [n/2] k = ( ) k (2n 2k)! k!(n k)!(n 2k)! = ( )n P n () But P n () = for all nonnegative integers n 27 We have (x 2 ) n = n k = ( ) n k n! k!(n k)! x2k, which is a polynomial of degree 2n Differentiating n times, d n n dx n (x2 ) n ( ) n k n! = k!(n k)! (2k)(2k ) (2k n + )x2k n, k = µ in which the lower index is µ = [n/2] + Note that if n = 2m +, then µ = m + Now shift the index, by setting k = n j Hence d n [n/2] dx n (x2 ) n ( ) j n! = (2n 2j)(2n 2j ) (n 2j + )xn 2j (n j)!j! j = = n! [n/2] j = ( ) j (2n 2j)! (n j)!j!(n 2j)! xn 2j Based on Problem 25, d n dx n (x2 ) n = n! 2 n P n (x)

154 54 Chapter 5 Series Solutions of Second Order Linear Equations 29 Since the n + polynomials P, P,, P n are linearly independent, and the degree of P k is k, any polynomial f of degree n can be expressed as a linear combination n f(x) = a k P k (x) k = Multiplying both sides by P m and integrating, n f(x)p m (x)dx = a k P k (x)p m (x)dx Based on Problem 28, k = P k (x)p m (x)dx = 2 2m + δ km Hence 2 f(x)p m (x)dx = 2m + a m 54 Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = r(r ) + 4r + 2 = r 2 + 3r + 2 The roots are r = 2, Hence the general solution, for x, is y = c x 2 + c 2 x 3 Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = r(r ) 3r + 4 = r 2 4r + 4 The root is r = 2, with multiplicity two Hence the general solution, for x, is y = (c + c 2 ln x ) x 2 5 Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = r(r ) r + = r 2 2r + The root is r =, with multiplicity two Hence the general solution, for x, is y = (c + c 2 ln x ) x 6 Substitution of y = (x ) r results in the quadratic equation F (r) =, where F (r) = r 2 + 7r + 2 The roots are r = 3, 4 Hence the general solution, for x, is y = c (x ) 3 + c 2 (x ) 4 7 Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = r 2 + 5r The roots are r = (5 ± 29)/2 Hence the general solution, for x, is y = c x (5 + 29)/2 + c 2 x (5 29)/2 8 Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = r 2 3r + 3 The roots are complex, with r = (3 ± i 3 )/2 Hence the general solution, for x, is y = c x 3/2 cos( 3 2 ln x ) + c 2 x 3/2 sin( 3 2 ln x )

155 54 55 Substitution of y = (x 2) r results in the quadratic equation F (r) =, where F (r) = r 2 + 4r + 8 The roots are complex, with r = 2 ± 2i Hence the general solution, for x 2, is y = c (x 2) 2 cos(2 ln x 2 ) + c 2 (x 2) 2 sin(2 ln x 2 ) Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = r 2 + r + 4 The roots are complex, with r = ( ± i 5 )/2 Hence the general solution, for x, is y = c x /2 5 cos( 2 ln x ) + c 2 x /2 5 sin( 2 ln x ) 2 Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = r 2 5r + 4 The roots are r =, 4 Hence the general solution is y = c x + c 2 x 4 4 Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = 4r 2 + 4r + 7 The roots are complex, with r = /2 ± 2i Hence the general solution, for x >, is y = c x /2 cos(2 ln x) + c 2 x /2 sin(2 ln x) Invoking the initial conditions, we obtain the system of equations c = 2, 2 c + 2c 2 = 3 Hence the solution of the initial value problem is y(x) = 2 x /2 cos(2 ln x) x /2 sin(2 ln x) As x +, the solution decreases without bound 5 Substitution of y = x r results in the quadratic equation F (r) =, where F (r) = r 2 4r + 4 The root is r = 2, with multiplicity two Hence the general solution, for x <, is y = (c + c 2 ln x ) x 2 Invoking the initial conditions, we obtain the system of equations c = 2, 2c c 2 = 3 Hence the solution of the initial value problem is y(x) = (2 7 ln x ) x 2

156 56 Chapter 5 Series Solutions of Second Order Linear Equations We find that y(x) as x 8 We see that P (x) = when x = and Since the three coefficients have no factors in common, both of these points are singular points Near x =, lim x p(x) = lim x x x 2x x 2 ( x) 2 = 2 lim x x2 q(x) = lim x 2 4 x x 2 ( x) 2 = 4 The singular point x = is regular Considering x =, lim x 2x (x )p(x) = lim (x ) x x 2 ( x) 2 The latter limit does not exist Hence x = is an irregular singular point 9 P (x) = when x = and Since the three coefficients have no common factors, both of these points are singular points Near x =, lim x p(x) = lim x x 2 x x x 2 ( x) The limit does not exist, and so x = is an irregular singular point Considering x =, lim (x )p(x) = lim (x ) x 2 x x x 2 ( x) = lim (x x )2 q(x) = lim (x ) 2 3x x x 2 ( x) = Hence x = is a regular singular point 2 P (x) = when x = and ± Since the three coefficients have no common factors, both of these points are singular points Near x =, lim x p(x) = lim x 2 x x x 3 ( x 2 ) The limit does not exist, and so x = is an irregular singular point Near x =, lim (x + )p(x) = lim (x + ) 2 x x x 3 ( x 2 ) =

157 54 57 lim (x + x )2 q(x) = lim (x + 2 x )2 x 3 ( x 2 ) = Hence x = is a regular singular point At x =, lim (x )p(x) = lim (x ) 2 x x x 3 ( x 2 ) = lim (x x )2 q(x) = lim (x ) 2 2 x x 3 ( x 2 ) = Hence x = is a regular singular point 22 The only singular point is at x = We find that Hence x = is a regular singular point lim x p(x) = lim x x x x x 2 = lim x x2 q(x) = lim x 2 x2 ν 2 x x 2 = ν 2 23 The only singular point is at x = 3 We find that 2x lim (x + 3)p(x) = lim (x + 3) x 3 x 3 x + 3 = 6 lim (x + x 3 3)2 x2 q(x) = lim (x + 3)2 x 3 x + 3 = Hence x = 3 is a regular singular point 24 Dividing the ODE by x( x 2 ) 3, we find that p(x) = x( x 2 ) and q(x) = The singular points are at x = and ± For x =, 2 x( + x) 2 ( x) 3 lim x p(x) = lim x x x x( x 2 ) = lim x x2 q(x) = lim x 2 2 x x( + x) 2 ( x) 3 = Hence x = is a regular singular point For x =, lim (x + )p(x) = lim (x + ) x x x( x 2 ) = 2 lim (x + x )2 q(x) = lim (x + 2 x )2 x( + x) 2 ( x) 3 = 4 Hence x = is a regular singular point For x =, lim (x )p(x) = lim (x ) x x x( x 2 ) = 2

158 58 Chapter 5 Series Solutions of Second Order Linear Equations lim (x x )2 q(x) = lim (x ) 2 2 x x( + x) 2 ( x) 3 The latter limit does not exist Hence x = is an irregular singular point 25 Dividing the ODE by (x + 2) 2 (x ), we find that p(x) = 3 2 (x + 2) 2 and q(x) = (x + 2)(x ) The singular points are at x = 2 and For x = 2, lim (x + 2)p(x) = lim (x + 2) 3 x 2 x 2 (x + 2) 2 The limit does not exist Hence x = 2 is an irregular singular point For x =, lim (x )p(x) = lim (x ) 3 x x (x + 2) 2 = lim (x x )2 q(x) = lim (x ) 2 2 x (x + 2)(x ) = Hence x = is a regular singular point 26 P (x) = when x = and 3 Since the three coefficients have no common factors, both of these points are singular points Near x =, lim x p(x) = lim x x + x x x(3 x) = 3 lim x x2 q(x) = lim x x 2 Hence x = is a regular singular point For x = 3, 2 x(3 x) = lim (x 3)p(x) = lim (x 3) x + x 3 x 3 x(3 x) = 4 3 lim (x x 3 3)2 q(x) = lim (x 3) 2 2 x 3 x(3 x) = Hence x = 3 is a regular singular point 27 Dividing the ODE by (x 2 + x 2), we find that p(x) = x + (x + 2)(x ) and q(x) = The singular points are at x = 2 and For x = 2, 2 (x + 2)(x ) x + lim (x + 2)p(x) = lim x 2 x 2 x = 3 lim (x + 2(x + 2) x 2 2)2 q(x) = lim = x 2 x

159 54 59 Hence x = 2 is a regular singular point For x =, x + lim (x )p(x) = lim x x x + 2 = 2 3 lim (x 2(x ) x )2 q(x) = lim x (x + 2) Hence x = is a regular singular point = 29 Note that p(x) = ln x and q(x) = 3x Evidently, p(x) is not analytic at x = Furthermore, the function x p(x) = x ln x does not have a Taylor series about x = Hence x = is an irregular singular point 3 P (x) = when x = Since the three coefficients have no common factors, x = is a singular point The Taylor series of e x, about x =, is e x = x + x 2 /2 + x 3 /6 + Hence the function x p(x) = 2(e x )/x is analytic at x = Similarly, the Taylor series of e x cos x, about x =, is e x cos x = x + x 3 /3 x 4 /6 + The function x 2 q(x) = e x cos x is also analytic at x = Hence x = is a regular singular point 3 P (x) = when x = Since the three coefficients have no common factors, x = is a singular point The Taylor series of sin x, about x =, is sin x = x x 3 /3! + x 5 /5! Hence the function x p(x) = 3 sin x/x is analytic at x = q(x) is a rational function, with On the other hand, Hence x = is a regular singular point lim x x2 q(x) = lim x 2 + x2 x x 2 = 32 P (x) = when x = Since the three coefficients have no common factors, x = is a singular point We find that lim x p(x) = lim x x x x = Although the function R(x) = cot x does not have a Taylor series about x =, note that x 2 q(x) = x cot x = x 2 /3 x 4 /45 2x 6 /945 Hence x = is a regular singular point Furthermore, q(x) = cot x/x 2 is undefined at x = ± nπ Therefore the points x = ± nπ are also singular points First note that lim (x nπ)p(x) = lim (x nπ) x ±nπ x ±nπ x =

160 6 Chapter 5 Series Solutions of Second Order Linear Equations Furthermore, since cot x has period π, Therefore From above, q(x) = cot x/x = cot(x nπ)/x = cot(x nπ) (x nπ) ± nπ [ ] (x nπ) 2 (x nπ) q(x) = (x nπ) cot(x nπ) (x nπ) ± nπ (x nπ) cot(x nπ) = (x nπ) 2 /3 (x nπ) 4 /45 Note that the function in brackets is analytic near x = ± nπ It follows that the function (x nπ) 2 q(x) is also analytic near x = ± nπ Hence all the singular points are regular 34 The singular points are located at x = ± nπ, n =,, Dividing the ODE by x sin x, we find that x p(x) = 3 csc x and x 2 q(x) = x 2 csc x Evidently, x p(x) is not even defined at x = Hence x = is an irregular singular point On the other hand, the Taylor series of x csc x, about x =, is Noting that csc(x nπ) = ( ) n csc x, x csc x = + x 2 /6 + 7x (x nπ)p(x) = 3( ) n (x nπ) csc(x nπ)/x [ = 3( ) n (x nπ) csc(x nπ) (x nπ) ± nπ It is apparent that (x nπ)p(x) is analytic at x = ± nπ Similarly, (x nπ) 2 q(x) = (x nπ) 2 csc x = ( ) n (x nπ) 2 csc(x nπ), which is also analytic at x = ± nπ Hence all other singular points are regular 36 Substitution of y = x r results in the quadratic equation r 2 r + β = The roots are r = ± 4β 2 If β > /4, the roots are complex, with r,2 = ( ± i 4β )/2 Hence the general solution, for x, is y = c x /2 cos( 4β ln x ) + c2 x /2 sin( 4β ln x ) 2 2 Since the trigonometric factors are bounded, y(x) as x If β = /4, the roots are equal, and y = c x /2 + c 2 x /2 ln x Since lim x x ln x =, y(x) as x If β < /4, the roots are real, with r,2 = ( ± 4β )/2 Hence the general solution, for x, is y = c x /2+ 4β/2 + c 2 x /2 4β/2 ]

161 54 6 Evidently, solutions approach zero as long as /2 4β/2 > That is, < β < /4 Hence all solutions approach zero for β > 37 Substitution of y = x r results in the quadratic equation r 2 r 2 = The roots are r =, 2 Hence the general solution, for x, is y = c x + c 2 x 2 Invoking the initial conditions, we obtain the system of equations c + c 2 =, c + 2c 2 = γ Hence the solution of the initial value problem is y(x) = 2 γ x + + γ x The solution is bounded, as x, if γ = 2 38 Substitution of y = x r results in the quadratic equation r 2 + (α )r + 5/2 = Formally, the roots are given by r = α ± α 2 2α 9 = α ± (α )(α + ) 2 2 (i) The roots will be complex if α < For solutions to approach zero, as x, we need < α < (ii) The roots will be equal if α = In this case, all solutions approach zero as long as α = (iii) The roots will be real and distinct if α > It follows that r max = α + α 2 2α 9 2 For solutions to approach zero, we need α + α 2 2α 9 < That is, α < Hence all solutions approach zero, as x, as long as α > 42 x = is the only singular point Dividing the ODE by 2x 2, we have p(x) = 3/(2x) and q(x) = x 2 ( + x)/2 It follows that lim x p(x) = lim x 3 x x 2x = 3 2, lim x x2 2 ( + x) q(x) = lim x x 2x 2 = 2, so x = is a regular singular point Let y = a + a x + a 2 x a n x n + Substitution into the ODE results in 2x 2 (n + 2)(n + )a n+2 x n + 3x (n + )a n+ x n ( + x) a n x n = n = n = n =

162 62 Chapter 5 Series Solutions of Second Order Linear Equations That is, 2 n(n )a n x n + 3 n = 2 It follows that a + (2a a )x + n a n x n n = a n x n n = a n x n = n = [2n(n )a n + 3n a n a n a n ] x n = n = 2 Equating the coefficients to zero, we find that a =, 2a a =, and (2n )(n + )a n = a n, n = 2, 3, We conclude that all the a n are equal to zero Hence y(x) = is the only solution that can be obtained 44 Based on Problem 43, the change of variable, x = /ξ, transforms the ODE into the form ξ 4 d2 y dy + 2ξ3 dξ2 dξ + y = Evidently, ξ = is a singular point Now p(ξ) = 2/ξ and q(ξ) = /ξ 4 Since the value of lim ξ ξ 2 q(ξ) does not exist, ξ = (x = ) is an irregular singular point 46 Under the transformation x = /ξ, the ODE becomes ξ 4 ( [ y ξ 2 )d2 dξ 2 + 2ξ 3 ( ξ 2 ) + ] dy 2ξ2 + α(α + )y =, ξ dξ that is, (ξ 4 ξ 2 ) d2 y dy + 2ξ3 + α(α + )y = dξ2 dξ Therefore ξ = is a singular point Note that It follows that p(ξ) = 2ξ ξ 2 lim ξ p(ξ) = lim ξ and q(ξ) = α(α + ) ξ 2 (ξ 2 ) ξ ξ 2ξ ξ 2 =, lim ξ ξ2 2 α(α + ) q(ξ) = lim ξ ξ ξ 2 (ξ 2 = α(α + ) ) Hence ξ = (x = ) is a regular singular point 48 Under the transformation x = /ξ, the ODE becomes [ ξ 4 d2 y dξ 2 + 2ξ 3 + 2ξ 2 ] dy ξ dξ + λ y =, that is, ξ 4 d2 y dξ 2 + 2(ξ3 + ξ) dy dξ + λ y =

163 55 63 Therefore ξ = is a singular point Note that p(ξ) = 2(ξ2 + ) ξ 3 and q(ξ) = λ ξ 4 It immediately follows that the limit lim ξ ξ p(ξ) does not exist Hence ξ = (x = ) is an irregular singular point 49 Under the transformation x = /ξ, the ODE becomes ξ 4 d2 y dy + 2ξ3 dξ2 dξ ξ y = Therefore ξ = is a singular point Note that We find that but p(ξ) = 2 ξ and q(ξ) = ξ 5 lim ξ p(ξ) = lim ξ 2 ξ ξ ξ = 2, lim ξ ξ2 q(ξ) = lim ξ 2 ( ) ξ ξ 5 The latter limit does not exist Hence ξ = (x = ) is an irregular singular point 55 (a) P (x) = when x = Since the three coefficients have no common factors, x = is a singular point Near x =, Hence x = is a regular singular point lim x p(x) = lim x x x 2x = 2 lim x x2 q(x) = lim x 2 x 2 = (b) Let Then and y = x r (a + a x + a 2 x a n x n + ) = a n x r+n y = (r + n)a n x r+n n = n = y = (r + n)(r + n )a n x r+n 2 n =

164 64 Chapter 5 Series Solutions of Second Order Linear Equations Substitution into the ODE results in 2 (r + n)(r + n )a n x r+n + (r + n)a n x r+n + n = That is, 2 (r + n)(r + n )a n x r+n + n = It follows that n = (r + n)a n x r+n + n = a [2r(r ) + r] x r + a [2(r + )r + r + ] x r+ + a n x r+n+ = n = a n 2 x r+n = n = 2 [2(r + n)(r + n )a n + (r + n)a n + a n 2 ] x r+n = n = 2 Assuming that a, we obtain the indicial equation 2r 2 r =, with roots r = /2 and r 2 = It immediately follows that a = Setting the remaining coefficients equal to zero, we have a n = a n 2, n = 2, 3, (r + n) [2(r + n) ] (c) For r = /2, the recurrence relation becomes a n = a n 2, n = 2, 3, n( + 2n) Since a =, the odd coefficients are zero Furthermore, for k =, 2,, a 2k = a 2k 2 2k( + 4k) = a 2k 4 (2k 2)(2k)(4k 3)(4k + ) = ( ) k a 2 k k! (4k + ) (d) For r =, the recurrence relation becomes a n = a n 2, n = 2, 3, n(2n ) Since a =, the odd coefficients are zero, and for k =, 2,, a 2k = a 2k 2 2k(4k ) = a 2k 4 (2k 2)(2k)(4k 5)(4k ) = ( ) k a 2 k k! 3 7 (4k ) The two linearly independent solutions are y (x) = [ ] ( ) k x 2k x + 2 k k! (4k + ) y 2 (x) = + k = k = ( ) k x 2k 2 k k! 3 7 (4k ) 3(a) Note that x p(x) = and x 2 q(x) = x, which are both analytic at x =

165 55 65 (b) Set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in (r + n)(r + n )a n x r+n + a n x r+n =, n = n = and after multiplying both sides of the equation by x, (r + n)(r + n )a n x r+n + a n x r+n = It follows that n = a [r(r )] x r + n = [(r + n)(r + n )a n + a n ] x r+n = n = Setting the coefficients equal to zero, the indicial equation is r(r ) = The roots are r = and r 2 = Here r r 2 = The recurrence relation is a n = a n, n =, 2, (r + n)(r + n ) (c) For r =, Hence for n, a n = a n, n =, 2, n(n + ) a n = Therefore one solution is a n n(n + ) = a n 2 (n )n 2 (n + ) = = ( )n a n!(n + )! y (x) = x n = ( ) n x n n!(n + )! 5(a) Here x p(x) = 2/3 and x 2 q(x) = x 2 /3, which are both analytic at x = (b) Set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in 3 (r + n)(r + n )a n x r+n + 2 (r + n)a n x r+n + a n x r+n+2 = n = It follows that n = a [3r(r ) + 2r] x r + a [3(r + )r + 2(r + )] x r+ + n = [3(r + n)(r + n )a n + 2(r + n)a n + a n 2 ] x r+n = n = 2 Assuming a, the indicial equation is 3r 2 r =, with roots r = /3, r 2 = Setting the remaining coefficients equal to zero, we have a =, and a n = a n 2, n = 2, 3, (r + n) [3(r + n) ]

166 66 Chapter 5 Series Solutions of Second Order Linear Equations It immediately follows that the odd coefficients are equal to zero (c) For r = /3, So for k =, 2,, a n = a n 2, n = 2, 3, n( + 3n) a 2k = a 2k 2 2k(6k + ) = a 2k 4 (2k 2)(2k)(6k 5)(6k + ) = ( ) k a 2 k k! 7 3 (6k + ) (d) For r =, So for k =, 2,, a n = a n 2, n = 2, 3, n(3n ) a 2k = a 2k 2 2k(6k ) = a 2k 4 (2k 2)(2k)(6k 7)(6k ) = ( ) k a 2 k k! 5 (6k ) The two linearly independent solutions are ] y (x) = x [ /3 ( ) k + k! 7 3 (6k + ) (x2 2 )k y 2 (x) = + k = k = ( ) k k! 5 (6k ) (x2 2 )k 6(a) Note that x p(x) = and x 2 q(x) = x 2, which are both analytic at x = (b) Set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in (r + n)(r + n )a n x r+n + (r + n)a n x r+n n = n = + a n x r+n+ 2 n = a n x r+n = n = After adjusting the indices in the second-to-last series, we obtain a [r(r ) + r 2] x r + [(r + n)(r + n )a n + (r + n)a n 2 a n + a n ] x r+n = n = Assuming a, the indicial equation is r 2 2 =, with roots r = ± 2 Setting the remaining coefficients equal to zero, the recurrence relation is a n = a n (r + n) 2, n =, 2, 2 Note that (r + n) 2 2 = (r + n + 2 )(r + n 2 )

167 55 67 (c) For r = 2, It follows that a n = a n n(n + 2, n =, 2, 2 ) a n = ( ) n a n!( )( ) (n + 2, n =, 2, 2 ) (d) For r = 2, and therefore a n = a n n(n 2, n =, 2,, 2 ) a n = ( ) n a n!( 2 2 )(2 2 2 ) (n 2, n =, 2, 2 ) The two linearly independent solutions are [ ] y (x) = x 2 ( ) n x n + n!( )( ) (n ) y 2 (x) = x 2 [ + n = n = ( ) n x n n!( 2 2 )(2 2 2 ) (n 2 2 ) 7(a) Here x p(x) = x and x 2 q(x) = x, which are both analytic at x = (b) Set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in (r + n)(r + n )a n x r+n + (r + n)a n x r+n n = After multiplying both sides by x, (r + n)(r + n )a n x r+n + n = n = (r + n)a n x r+n n = (r + n)a n x r+n n = (r + n)a n x r+n+ n = After adjusting the indices in the last two series, we obtain a [r(r ) + r] x r + ] a n x r+n = n = a n x r+n+ = n = [(r + n)(r + n )a n + (r + n)a n (r + n)a n ] x r+n = n =

168 68 Chapter 5 Series Solutions of Second Order Linear Equations Assuming a, the indicial equation is r 2 =, with roots r = r 2 = Setting the remaining coefficients equal to zero, the recurrence relation is (c) With r =, Hence one solution is a n = a n, n =, 2, r + n a n = a n n, n =, 2, y (x) = + x! + x2 2! + + xn n! + = ex 8(a) Note that xp(x) = 3/2 and x 2 q(x) = x 2 /2, which are both analytic at x = (b) Set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in 2 (r + n)(r + n )a n x r+n + 3 (r + n)a n x r+n n = n = + 2 a n x r+n+2 n = a n x r+n = n = After adjusting the indices in the second-to-last series, we obtain a [2r(r ) + 3r ] x r + a [2(r + )r + 3(r + ) ] + [2(r + n)(r + n )a n + 3(r + n)a n a n + 2 a n 2 ] x r+n = n = 2 Assuming a, the indicial equation is 2r 2 + r =, with roots r = /2 and r 2 = Setting the remaining coefficients equal to zero, the recurrence relation is 2a n 2 a n =, n = 2, 3, (r + n + ) [2(r + n) ] Setting the remaining coefficients equal to zero, we have a =, which implies that all of the odd coefficients are zero (c) With r = /2, So for k =, 2,, a n = 2a n 2, n = 2, 3, n(2n + 3) a 2k = a 2k 2 k(4k + 3) = a 2k 4 (k )k(4k 5)(4k + 3) = ( ) k a k! 7 (4k + 3) (d) With r =, a n = 2a n 2, n = 2, 3, n(2n 3)

169 55 69 So for k =, 2,, a 2k = a 2k 2 k(4k 3) = a 2k 4 (k )k(4k )(4k 3) = ( ) k a k! 5 9 (4k 3) The two linearly independent solutions are ] y (x) = x [ /2 ( ) n x 2n + n! 7 (4n + 3) n = ] y 2 (x) = x [ ( ) n x 2n + n! 5 9 (4n 3) n = 9(a) Note that x p(x) = x 3 and x 2 q(x) = x + 3, which are both analytic at x = (b) Set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in (r + n)(r + n )a n x r+n (r + n)a n x r+n+ 3 (r + n)a n x r+n n = n = + a n x r+n+ + 3 n = n = a n x r+n = n = After adjusting the indices in the second-to-last series, we obtain a [r(r ) 3r + 3] x r + [(r + n)(r + n )a n (r + n 2)a n 3(r + n )a n ] x r+n = n = Assuming a, the indicial equation is r 2 4r + 3 =, with roots r = 3 and r 2 = Setting the remaining coefficients equal to zero, the recurrence relation is (c) With r = 3, a n = It follows that for n, a n = (n + )a n n(n + 2) Therefore one solution is (r + n 2)a n, n =, 2, (r + n )(r + n 3) a n = (n + )a n n(n + 2) = y (x) = x 3 [ +, n =, 2, a n 2 (n )(n + 2) = = 2 a n! (n + 2) n = ] 2 x n n! (n + 2) (a) Here x p(x) = and x 2 q(x) = x 2 + /4, which are both analytic at x =

170 7 Chapter 5 Series Solutions of Second Order Linear Equations (b) Set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in (r + n)(r + n )a n x r+n + a n x r+n+2 + a n x r+n = 4 n = n = n = After adjusting the indices in the second series, we obtain [ a r(r ) + ] [ x r + a (r + )r + ] x r [(r + n)(r + n )a n + 4 ] a n + a n 2 x r+n = n = 2 Assuming a, the indicial equation is r 2 r + 4 =, with roots r = r 2 = /2 Setting the remaining coefficients equal to zero, we find that a = The recurrence relation is 4a n 2 a n =, n = 2, 3, (2r + 2n ) 2 (c) With r = /2, a n = a n 2 n 2, n = 2, 3, Since a =, the odd coefficients are zero So for k, Therefore one solution is a 2k = a 2k 2 a 2k 4 4k 2 = 4 2 (k ) 2 k 2 = = ( )k a 4 k (k!) 2 y (x) = x [ + n = ] ( ) n x 2n 2 2n (n!) 2 2(a) Dividing through by the leading coefficient, the ODE can be written as For x =, For x =, y x x 2 y + α2 x 2 y = x p = lim (x )p(x) = lim x x x + = 2 q = lim (x ) 2 α 2 ( x) q(x) = lim = x x x + q = p = x lim (x + )p(x) = lim x x x = 2 lim (x + x )2 q(x) = lim x α 2 (x + ) ( x) = Hence x = and x = are regular singular points As shown in Example, the indicial equation is given by r(r ) + p r + q =

171 55 7 In this case, both sets of roots are r = /2 and r 2 = (b) Let t = x, and u(t) = y(t + ) Under this change of variable, the differential equation becomes (t 2 + 2t)u + (t + )u α 2 u = Based on part (a), t = is a regular singular point Set u = n = a n t r+n Substitution into the ODE results in (r + n)(r + n )a n t r+n + 2 (r + n)(r + n )a n t r+n n = + (r + n)a n t r+n + n = n = (r + n)a n t r+n α 2 a n t r+n = n = n = Upon inspection, we can also write (r + n) 2 a n t r+n + 2 (r + n)(r + n 2 )a n t r+n α 2 a n t r+n = n = n = After adjusting the indices in the second series, it follows that [ a 2r(r ] 2 ) t r + n = n = [(r + n) 2 a n + 2(r + n + )(r + n + 2 )a n+ α 2 a n ] t r+n = Assuming that a, the indicial equation is 2r 2 r =, with roots r =, /2 The recurrence relation is (r + n) 2 a n + 2(r + n + )(r + n + 2 )a n+ α 2 a n =, n =,, 2, With r = /2, we find that for n, a n = 4α2 (2n ) 2 4n(2n + ) With r 2 =, we find that for n, a n = α2 (n ) 2 n(2n ) a n = ( ) n [ 4α 2 ] [ 9 4α 2] [ (2n ) 2 4α 2] 2 n (2n + )! a n = ( ) n α( α) [ α 2] [ 4 α 2] [ (n ) 2 α 2] n! 3 5 (2n ) The two linearly independent solutions of the Chebyshev equation are ) y (x) = x ( /2 + ( ) n ( 4α2 )(9 4α 2 ) ((2n ) 2 4α 2 ) 2 n (x ) n (2n + )! y 2 (x) = + n = ( ) n α( α)( α2 )(4 α 2 ) ((n ) 2 α 2 ) (x ) n n! 3 5 (2n ) n = a a

172 72 Chapter 5 Series Solutions of Second Order Linear Equations 3(a) Here xp(x) = x and x 2 q(x) = λx, which are both analytic at x = In fact, p = lim x p(x) = and q = lim x 2 q(x) = x x (b) The indicial equation is r(r ) + r =, with roots r,2 = (c) Set y = a + a x + a 2 x a n x n + Substitution into the ODE results in n(n )a n x n + na n x n n = 2 That is, n(n + )a n+ x n + n = It follows that a + λ a + n = na n x n + λ n = (n + )a n+ x n n = a n x n = n = na n x n + λ n = a n x n = n = [ (n + ) 2 ] a n+ (n λ)a n x n = n = Setting the coefficients equal to zero, we find that a = λa, and That is, for n 2, a n = a n = (n λ) n 2 a n, n = 2, 3, (n λ) ( λ)( λ) (n λ) n 2 a n = = (n!) 2 a Therefore one solution of the Laguerre equation is ( λ)( λ) (n λ) y (x) = + (n!) 2 x n n = Note that if λ = m, a positive integer, then a n = for n m + In that case, the solution is a polynomial m ( λ)( λ) (n λ) y (x) = + (n!) 2 x n n =

173 (a) P (x) = only for x = Furthermore, xp(x) = 2 x and x 2 q(x) = 2 + x 2 It follows that p = lim x ( 2 x) = 2 q = lim x (2 + x 2 ) = 2 and therefore x = is a regular singular point (b) The indicial equation is given by r(r ) 2r + 2 =, that is, r 2 3r + 2 =, with roots r = 2 and r 2 = 4 The coefficients P (x), Q(x), and R(x) are analytic for all x R Hence there are no singular points 5(a) P (x) = only for x = Furthermore, xp(x) = 3 sin x/ x and x 2 q(x) = 2 It follows that p = lim 3 sin x = 3 x x q = lim 2 = 2 x and therefore x = is a regular singular point (b) The indicial equation is given by r(r ) + 3r 2 =, that is, r 2 + 2r 2 =, with roots r = + 3 and r 2 = 3 6(a) P (x) = for x = and x = 2 We note that p(x) = x (x + 2) /2, and q(x) = (x + 2) /2 For the singularity at x =, p = lim x 2(x + 2) = 4 x 2 q = lim x 2(x + 2) = and therefore x = is a regular singular point For the singularity at x = 2, p = lim (x + 2)p(x) = lim x 2 x 2 q = lim x 2 (x + 2)2 q(x) = lim x 2 and therefore x = 2 is a regular singular point 2x = 4 (x + 2) 2 (b) For x = : the indicial equation is given by r(r ) + r/4 =, that is, r 2 3r/4 =, with roots r = 3/4 and r 2 = For x = 2: the indicial equation is given by r(r ) r/4 =, that is, r 2 5r/4 =, with roots r = 5/4 and r 2 = =

174 74 Chapter 5 Series Solutions of Second Order Linear Equations 7(a) P (x) = only for x = Furthermore, xp(x) = /2 + sin x/ 2x and x 2 q(x) = It follows that p = lim xp(x) = x q = lim x 2 q(x) = x and therefore x = is a regular singular point (b) The indicial equation is given by r(r ) + r + =, that is, r 2 + =, with complex conjugate roots r = ± i 8(a) Note that P (x) = only for x = We find that p(x) = 3(x )/(x + ), and q(x) = 3/(x + ) 2 It follows that p = lim (x + )p(x) = lim 3(x ) = 6 x x q = lim x (x + )2 q(x) = lim x 3 = 3 and therefore x = is a regular singular point (b) The indicial equation is given by r(r ) 6r + 3 =, that is, r 2 7r + 3 =, with roots r = ( )/2 and r 2 = (7 37 )/2 (a) P (x) = for x = 2 and x = 2 We note that p(x) = 2x(x 2) 2 (x + 2), and q(x) = 3(x 2) (x + 2) For the singularity at x = 2, lim (x 2)p(x) = lim x 2 x 2 2x x 2 4, which is undefined Therefore x = 2 is an irregular singular point For the singularity at x = 2, p = lim (x + 2)p(x) = lim x 2 x 2 q = lim x 2 (x + 2)2 q(x) = lim x 2 and therefore x = 2 is a regular singular point 2x (x 2) 2 = 4 3(x + 2) = x 2 (b) The indicial equation is given by r(r ) r/4 =, that is, r 2 5r/4 =, with roots r = 5/4 and r 2 =

175 56 75 (a) P (x) = for x = 2 and x = 2 We note that p(x) = 2x/(4 x 2 ), and q(x) = 3/(4 x 2 ) For the singularity at x = 2, p = lim x 2 (x 2)p(x) = lim x 2 2x x + 2 = q = lim (x 2) 2 3(2 x) q(x) = lim = x 2 x 2 x + 2 and therefore x = 2 is a regular singular point For the singularity at x = 2, p = lim (x + 2)p(x) = lim x 2 x 2 q = lim x 2 (x + 2)2 q(x) = lim x 2 and therefore x = 2 is a regular singular point 2x 2 x = 3(x + 2) 2 x = (b) For x = 2: the indicial equation is given by r(r ) r =, that is, r 2 2r =, with roots r = 2 and r 2 = For x = 2: the indicial equation is given by r(r ) r =, that is, r 2 2r =, with roots r = 2 and r 2 = 2(a) P (x) = for x = and x = 3 We note that p(x) = 2x (x + 3), and q(x) = /(x + 3) 2 For the singularity at x =, p = lim x x p(x) = lim x 2 x + 3 = 2 3 q = lim x 2 x 2 q(x) = lim x x (x + 3) 2 = and therefore x = is a regular singular point For the singularity at x = 3, p = lim (x + 3)p(x) = lim x 3 x 3 q = lim x 3 (x + 3)2 q(x) = lim x 3 and therefore x = 3 is a regular singular point 2 x = 2 3 ( ) = (b) For x = : the indicial equation is given by r(r ) 2r/3 =, that is, r 2 5r/3 =, with roots r = 5/3 and r 2 = For x = 3: the indicial equation is given by r(r ) + 2r/3 =, that is, r 2 r/3 =, with roots r = ( + 37 )/6 and r 2 = ( 37 )/6 4(a) Here x p(x) = 2x and x 2 q(x) = 6 xe x Both of these functions are analytic at x =, therefore x = is a regular singular point Note that p = q = (b) The indicial equation is given by r(r ) =, that is, r 2 r =, with roots r = and r 2 =

176 76 Chapter 5 Series Solutions of Second Order Linear Equations (c) In order to find the solution corresponding to r =, set y = x n = a nx n Upon substitution into the ODE, we have (n + 2)(n + )a n+ x n+ + 2 n = (n + )a n x n+ + 6 e x a n x n+ = n = After adjusting the indices in the first two series, and expanding the exponential function, n(n + )a n x n + 2 n = n a n x n + 6 a x + (6a + 6a )x 2 n = n = +(6a 2 + 6a + 3a )x 3 + (6a 3 + 6a 2 + 3a + a )x 4 + = Equating the coefficients, we obtain the system of equations 2a + 2a + 6a = 6a 2 + 4a + 6a + 6a = 2a 3 + 6a 2 + 6a 2 + 6a + 3a = 2a 4 + 8a 3 + 6a 3 + 6a 2 + 3a + a = Setting a =, solution of the system results in a = 4, a 2 = 7/3, a 3 = 47/2, a 4 = 9/2, Therefore one solution is y (x) = x 4x x x4 + The exponents differ by an integer So for a second solution, set y 2 (x) = a y (x) ln x + + c x + c 2 x c n x n + Substituting into the ODE, we obtain a L [y (x)] ln x + 2a y (x) + 2a y (x) a y (x) x + L [ + ] c n x n = n = Since L [y (x)] =, it follows that [ ] L + c n x n = 2a y (x) 2a y (x) + a y (x) x More specifically, n = n(n + )c n+ x n + 2 n = n c n x n (6 + 6c )x n = +(6c 2 + 6c + 3)x 2 + = a + ax 6 3 ax ax3 +

177 56 77 Equating the coefficients, we obtain the system of equations 6 = a 2c 2 + 8c + 6 = a 6c 3 + c 2 + 6c + 3 = 6 3 a 2c 4 + 2c 3 + 6c 2 + 3c + = 93 2 a Solving these equations for the coefficients, a = 6 In order to solve the remaining equations, set c = Then c 2 = 33, c 3 = 449/6, c 4 = 595/24, Therefore a second solution is [ y 2 (x) = 6 y (x) ln x + 33x x3 595 ] 24 x4 + 5(a) Note the p(x) = 6x/(x ) and q(x) = 3x (x ) Furthermore, x p(x) = 6x 2 /(x ) and x 2 q(x) = 3x/(x ) It follows that 6x 2 p = lim x x = q = lim x 3x x = and therefore x = is a regular singular point (b) The indicial equation is given by r(r ) =, that is, r 2 r =, with roots r = and r 2 = (c) In order to find the solution corresponding to r =, set y = x n = a nx n Upon substitution into the ODE, we have n(n + )a n x n+ n = n(n + )a n x n + 6 n = After adjusting the indices, it follows that n(n )a n x n n = 2 That is, 2a + 3a + n(n + )a n x n + 6 n = (n + )a n x n n = (n )a n 2 x n + 3 n = 2 a n x n+ = n = a n x n = n = [ n(n + )an + (n 2 ] n + 3)a n + 6(n )a n 2 x n = n = 2 Setting the coefficients equal to zero, we have a = 3a /2, and for n 2, n(n + )a n = (n 2 n + 3)a n + 6(n )a n 2

178 78 Chapter 5 Series Solutions of Second Order Linear Equations If we assign a =, then we obtain a = 3/2, a 2 = 9/4, a 3 = 5/6, Hence one solution is y (x) = x x x x4 + 4 x5 + The exponents differ by an integer So for a second solution, set y 2 (x) = a y (x) ln x + + c x + c 2 x c n x n + Substituting into the ODE, we obtain 2ax y (x) 2a y (x) + 6ax y (x) a y (x) + a y (x) x + L [ + ] c n x n =, n = since L [y (x)] = It follows that [ ] L + c n x n = 2a y (x) 2ax y (x) + a y (x) 6ax y (x) a y (x) x Now [ L + n = n = ] c n x n = 3 + ( 2c 2 + 3c )x + ( 6c 3 + 5c 2 + 6c )x ( 2c 4 + 9c 3 + 2c 2 )x 3 + ( 2c 5 + 5c 4 + 8c 3 )x 4 + Substituting for y (x), the right hand side of the ODE is a ax ax ax ax4 44 ax5 + Equating the coefficients, we obtain the system of equations 3 = a 2c 2 + 3c = 7 2 a 6c 3 + 5c 2 + 6c = 3 4 a 2c 4 + 9c 3 + 2c 2 = 33 6 a We find that a = 3 In order to solve the second equation, set c = Solution of the remaining equations results in c 2 = 2/4, c 3 = 9/4, c 4 = 597/64, Hence a second solution is [ y 2 (x) = 3 y (x) ln x x2 9 4 x3 597 ] 64 x4 + 6(a) After multiplying both sides of the ODE by x, we find that x p(x) = and x 2 q(x) = x Both of these functions are analytic at x =, hence x = is a regular singular point (b) Furthermore, p = q = So the indicial equation is r(r ) =, with roots r = and r 2 =

179 56 79 (c) In order to find the solution corresponding to r =, set y = x n = a nx n Upon substitution into the ODE, we have n(n + )a n x n + n = a n x n+ = n = That is, [n(n + )a n + a n ] x n = n = Setting the coefficients equal to zero, we find that for n, It follows that a n = Hence one solution is a n = a n n(n + ) a n n(n + ) = a n 2 (n )n 2 (n + ) = = ( )n a (n!) 2 (n + ) y (x) = x 2 x2 + 2 x3 44 x x5 + The exponents differ by an integer So for a second solution, set y 2 (x) = a y (x) ln x + + c x + c 2 x c n x n + Substituting into the ODE, we obtain a L [y (x)] ln x + 2a y (x) a y (x) x Since L [y (x)] =, it follows that [ ] L + c n x n Now [ L + n = + L [ + ] c n x n = n = = 2a y (x) + a y (x) x ] c n x n = + (2c 2 + c )x + (6c 3 + c 2 )x 2 + (2c 4 + c 3 )x 3 n = +(2c 5 + c 4 )x 4 + (3c 6 + c 5 )x 5 + Substituting for y (x), the right hand side of the ODE is a ax 5 2 ax ax3 32 ax4 +

180 8 Chapter 5 Series Solutions of Second Order Linear Equations Equating the coefficients, we obtain the system of equations = a 2c 2 + c = 3 2 a 6c 3 + c 2 = 5 2 a 2c 4 + c 3 = 7 44 a Evidently, a = In order to solve the second equation, set c = We then find that c 2 = 3/4, c 3 = 7/36, c 4 = 35/728, Therefore a second solution is [ y 2 (x) = y (x) ln x x x3 35 ] 728 x4 + 9(a) After dividing by the leading coefficient, we find that γ ( + α + β)x p = lim x p(x) = lim = γ x x x q = lim x x 2 q(x) = lim x αβ x x = Hence x = is a regular singular point The indicial equation is r(r ) + γ r =, with roots r = γ and r 2 = (b) For x =, γ + ( + α + β)x p = lim (x )p(x) = lim = γ + α + β x x x q = lim (x ) 2 αβ(x ) q(x) = lim = x x x Hence x = is a regular singular point The indicial equation is with roots r = γ α β and r 2 = r 2 (γ α β) r =, (c) Given that r r 2 is not a positive integer, we can set y = n = a nx n Substitution into the ODE results in x( x) n(n )a n x n 2 + [γ ( + α + β)x] n a n x n αβ a n x n = That is, n = 2 n(n + )a n+ x n n = n(n )a n x n + γ n = 2 n = (n + )a n+ x n n = n =

181 56 8 ( + α + β) Combining the series, we obtain in which n a n x n αβ n = a n x n = n = γ a αβ a + [(2 + 2γ)a 2 ( + α + β + αβ)a ] x + A n x n =, n = 2 A n = (n + )(n + γ)a n+ [n(n ) + ( + α + β)n + αβ] a n Note that n(n ) + ( + α + β)n + αβ = (n + α)(n + β) Setting the coefficients equal to zero, we have γ a αβ a =, and a n+ = for n Hence one solution is y (x) = + αβ γ! α(α + )β(β + ) x + x 2 + γ(γ + ) 2! (n + α)(n + β) (n + )(n + γ) a n α(α + )(α + 2)β(β + )(β + 2) x 3 + γ(γ + )(γ + 2) 3! Since the nearest other singularity is at x =, the radius of convergence of y (x) will be at least ρ = (d) Given that r r 2 is not a positive integer, we can set y = x γ n = b nx n Then substitution into the ODE results in x( x) (n + γ)(n γ)a n x n γ n = + [γ ( + α + β)x] (n + γ)a n x n γ αβ n = That is, (n + γ)(n γ)a n x n γ n = +γ a n x n+ γ = n = (n + γ)(n γ)a n x n+ γ n = (n + γ)a n x n γ ( + α + β) n = αβ a n x n+ γ = n = After adjusting the indices, (n + γ)(n γ)a n x n γ n = +γ (n + γ)a n x n+ γ n = (n γ)(n γ)a n x n γ n = (n + γ)a n x n γ ( + α + β) n = (n γ)a n x n γ αβ n = a n x n γ = n =

182 82 Chapter 5 Series Solutions of Second Order Linear Equations Combining the series, we obtain in which B n x n γ =, n = B n = n(n + γ)b n [(n γ)(n γ + α + β) + αβ] b n Note that (n γ)(n γ + α + β) + αβ = (n + α γ)(n + β γ) Setting B n =, it follows that for n, b n = (n + α γ)(n + β γ) n(n + γ) Therefore a second solution is [ y 2 (x) = x γ ( + α γ)( + β γ) + x (2 γ)! + b n ( + α γ)(2 + α γ)( + β γ)(2 + β γ) x 2 + (2 γ)(3 γ)2! (e) Under the transformation x = /ξ, the ODE becomes ξ 4 ξ ( y {2ξ ξ )d2 dξ ξ ( ξ [ ) ξ2 γ ( + α + β) ]} dy ξ dξ αβ y = That is, (ξ 3 ξ 2 ) d2 y dξ 2 + [ 2ξ 2 γ ξ 2 + ( + α + β)ξ ] dy dξ αβ y = Therefore ξ = is a singular point Note that It follows that p(ξ) = (2 γ) ξ + ( + α + β) ξ 2 ξ and q(ξ) = αβ ξ 3 ξ 2 (2 γ) ξ + ( + α + β) p = lim ξ p(ξ) = lim = α β, ξ ξ ξ q = lim ξ 2 αβ q(ξ) = lim ξ ξ ξ = αβ Hence ξ = (x = ) is a regular singular point The indicial equation is r(r ) + ( α β)r + αβ =, or r 2 (α + β)r + αβ = Evidently, the roots are r = α and r = β ]

183 Here x p(x) = and x 2 q(x) = 2x, which are both analytic everywhere We set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in (r + n)(r + n )a n x r+n + n = (r + n)a n x r+n + 2 n = a n x r+n+ = After adjusting the indices in the last series, we obtain a [r(r ) + r] x r + [(r + n)(r + n )a n + (r + n)a n + 2 a n ] x r+n = n = Assuming a, the indicial equation is r 2 =, with double root r = Setting the remaining coefficients equal to zero, we have for n, It follows that a n (r) = Since r =, one solution is given by y (x) = 2 a n (r) = (n + r) 2 a n (r) n = ( ) n 2 n [(n + r)(n + r ) ( + r)] 2 a, n n = ( ) n 2 n (n!) 2 x n For a second linearly independent solution, we follow the discussion in Section 56 First note that a n(r) [ a n (r) = 2 n + r + n + r + + ] + r Setting r =, Therefore, a n() = 2 H n a n () = 2 H n ( ) n 2 n (n!) 2 y 2 (x) = y (x) ln x 2 n = ( ) n 2 n H n (n!) 2 x n 4 Here x p(x) = 4 and x 2 q(x) = 2 + x, which are both analytic everywhere We set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in (r + n)(r + n )a n x r+n + 4 (r + n)a n x r+n n = + n = a n x r+n+ + 2 n = a n x r+n = n =

184 84 Chapter 5 Series Solutions of Second Order Linear Equations After adjusting the indices in the second-to-last series, we obtain a [r(r ) + 4r + 2] x r + [(r + n)(r + n )a n + 4(r + n)a n + 2 a n + a n ] x r+n = n = Assuming a, the indicial equation is r 2 + 3r + 2 =, with roots r = and r 2 = 2 Setting the remaining coefficients equal to zero, we have for n, It follows that a n (r) = a n (r) = (n + r + )(n + r + 2) a n (r) ( ) n [(n + r + )(n + r) (2 + r)] [(n + r + 2)(n + r) (3 + r)] a, n Since r =, one solution is given by y (x) = x n = ( ) n (n)!(n + )! xn For a second linearly independent solution, we follow the discussion in Section 56 Since r r 2 = N =, we find that a (r) = (r + 2)(r + 3), with a = Hence the leading coefficient in the solution is Further, (r + 2) a n (r) = a = lim r 2 (r + 2) a (r) = Let A n (r) = (r + 2) a n (r) It follows that A n(r) [ A n (r) = n + r ( ) n (n + r + 2) [(n + r + )(n + r) (3 + r)] 2 n + r + + n + r r Setting r = r 2 = 2, A n( 2) A n ( 2) = [ n 2 n + ] n = H n H n Hence Therefore, ( ) n c n ( 2) = (H n + H n ) A n ( 2) = (H n + H n ) n!(n )! y 2 (x) = y (x) ln x + x 2 [ n = ( ) n (H n + H n ) n!(n )! ] x n ]

185 Let y(x) = v(x)/ x Then y = x /2 v x 3/2 v/2 and y = x /2 v x 3/2 v + 3 x 5/2 v/4 Substitution into the ODE results in [ ] [ ] x 3/2 v x /2 v + 3 x /2 v/4 + x /2 v x /2 v/2 + (x 2 4 )x /2 v = Simplifying, we find that v + v =, with general solution v(x) = c cos x + c 2 sin x Hence y(x) = c x /2 cos x + c 2 x /2 sin x 8 The absolute value of the ratio of consecutive terms is a 2m+2 x 2m+2 a 2m x 2m = x 2m+2 2 2m (m + )! m! x 2 x 2m = 2 2m+2 (m + 2)!(m + )! 4(m + 2)(m + ) Applying the ratio test, lim a 2m+2 x 2m+2 m a 2m x 2m = lim m x 2 4(m + 2)(m + ) = Hence the series for J (x) converges absolutely for all values of x Furthermore, since the series for J (x) also converges absolutely for all x, term-by-term differentiation results in J (x) = m = Therefore, J (x) = J (x) ( ) m x 2m 2 2m m!(m )! = = x 2 m = m = ( ) m x 2m 2 2m (m + )! m! ( ) m+ x 2m+ 2 2m+ (m + )! m! 9(a) Note that x p(x) = and x 2 q(x) = x 2 ν 2, which are both analytic at x = Thus x = is a regular singular point Furthermore, p = and q = ν 2 Hence the indicial equation is r 2 ν 2 =, with roots r = ν and r 2 = ν (b) Set y = x r (a + a x + a 2 x a n x n + ) results in (r + n)(r + n )a n x r+n + (r + n)a n x r+n n = + n = a n x r+n+2 ν 2 n = n = a n x r+n = After adjusting the indices in the second-to-last series, we obtain a [ r(r ) + r ν 2 ] x r + a [ (r + )r + (r + ) ν 2 ] + = Substitution into the ODE [ (r + n)(r + n )an + (r + n)a n ν 2 ] a n + a n 2 x r+n = n = 2

186 86 Chapter 5 Series Solutions of Second Order Linear Equations Setting the coefficients equal to zero, we find that a =, and a n = (r + n) 2 ν 2 a n 2, for n 2 It follows that a 3 = a 5 = = a 2m+ = = Furthermore, with r = ν, a n = n(n + 2ν) a n 2 So for m =, 2,, a 2m = 2m(2m + 2ν) a ( ) m 2m 2 = 2 2m m!( + ν)(2 + ν) (m + ν)(m + ν) a Hence one solution is y (x) = x ν [ + m = ] ( ) m m!( + ν)(2 + ν) (m + ν)(m + ν) (x 2 )2m (c) Assuming that r r 2 = 2ν is not an integer, simply setting r = ν in the above results in a second linearly independent solution ] y 2 (x) = x [ ν ( ) m + m!( ν)(2 ν) (m ν)(m ν) (x 2 )2m m = (d) The absolute value of the ratio of consecutive terms in y (x) is a 2m+2 x 2m+2 a 2m x 2m = x 2m+2 2 2m m!( + ν) (m + ν) x 2m 2 2m+2 (m + )!( + ν) (m + + ν) = x 2 4(m + )(m + + ν) Applying the ratio test, lim a 2m+2 x 2m+2 m a 2m x 2m = lim m x 2 4(m + )(m + + ν) = Hence the series for y (x) converges absolutely for all values of x The same can be shown for y 2 (x) Note also, that if ν is a positive integer, then the coefficients in the series for y 2 (x) are undefined (a) It suffices to calculate L [J (x) ln x] Indeed, and Hence [J (x) ln x] = J (x) ln x + J (x) x [J (x) ln x] = J (x) ln x + 2 J (x) x J (x) x 2 L [J (x) ln x] = x 2 J (x) ln x + 2x J (x) J (x) + x J (x) ln x + J (x) + x 2 J (x) ln x

187 57 87 Since x 2 J (x) + x J (x) + x 2 J (x) =, L [J (x) ln x] = 2x J (x) (b) Given that L [y 2 (x)] =, after adjusting the indices in part (a), we have b x b 2 x 2 + (n 2 b n + b n 2 )x n = 2x J (x) n = 3 Using the series representation of J (x) in Problem 8, b x b 2 x 2 + (n 2 b n + b n 2 )x n = 2 n = 3 n = ( ) n (2n)x 2n 2 2n (n!) 2 (c) Equating the coefficients on both sides of the equation, we find that b = b 3 = = b 2m+ = = Also, with n =, 2 2 b 2 = /(!) 2, that is, b 2 = / [ 2 2 (!) 2] Furthermore, for m 2, More explicitly, It can be shown, in general, that Bessel s equation of order one is (2m) 2 b 2m + b 2m 2 = 2 ( )m (2m) 2 2m (m!) 2 b 4 = ( + 2 ) b 6 = ( ) H m b 2m = ( ) m+ 2 2m (m!) 2 x 2 y + x y + (x 2 )y = Based on Problem 9, the roots of the indicial equation are r = and r 2 = Set y = x r (a + a x + a 2 x a n x n + ) Substitution into the ODE results in (r + n)(r + n )a n x r+n + (r + n)a n x r+n n = + n = a n x r+n+2 n = a n x r+n = n = After adjusting the indices in the second-to-last series, we obtain a [r(r ) + r ] x r + a [(r + )r + (r + ) ]

188 88 Chapter 5 Series Solutions of Second Order Linear Equations + [(r + n)(r + n )a n + (r + n)a n a n + a n 2 ] x r+n = n = 2 Setting the coefficients equal to zero, we find that a =, and a n (r) = (r + n) 2 a n 2(r) = (n + r + )(n + r ) a n 2(r), for n 2 It follows that a 3 = a 5 = = a 2m+ = = Solving the recurrence relation, a 2m (r) = With r = r =, ( ) m (2m + r + )(2m + r ) 2 (r + 3) 2 (r + ) a a 2m () = ( ) m 2 2m (m + )! m! a For a second linearly independent solution, we follow the discussion in Section 56 Since r r 2 = N = 2, we find that a 2 (r) = (r + 3)(r + ), with a = Hence the leading coefficient in the solution is Further, (r + ) a 2m (r) = a = lim r (r + ) a 2(r) = 2 Let A n (r) = (r + ) a n (r) It follows that A 2m(r) [ A 2m (r) = 2m + r + 2 Setting r = r 2 =, we calculate c 2m ( ) = 2 (H m + H m )A 2m ( ) ( ) m (2m + r + ) [(2m + r ) (3 + r)] 2 2m + r r = 2 (H ( ) m m + H m ) 2m [(2m 2) 2] 2 = 2 (H ( ) m m + H m ) 2 2m m!(m )! Note that a 2m+ (r) = implies that A 2m+ (r) =, so [ ] d c 2m+ ( ) = dr A 2m+(r) = r=r 2 Therefore, y 2 (x) = 2 [ x m = ( ) m (m + )! m! (x 2 )2m ] ln x ]

189 x [ m = Based on the definition of J (x), [ y 2 (x) = J (x) ln x + x ] ( ) m (H m + H m ) ( x m!(m )! 2 )2m m = ] ( ) m (H m + H m ) ( x m!(m )! 2 )2m

190 9 Chapter 5 Series Solutions of Second Order Linear Equations

191 C H A P T E R 6 The Laplace Transform 6 3 The function f(t) is continuous 9

192 92 Chapter 6 The Laplace Transform 4 The function f(t) has a jump discontinuity at t =, and is thus piecewise continuous 7 Integration is a linear operation It follows that Hence A A cosh bt e st dt = 2 = 2 A cosh bt e st dt = 2 Taking a limit, as A, A cosh bt e st dt = 2 e (b s)t dt + 2 Note that the above is valid for s > b 8 Proceeding as in Problem 7, A sinh bt e st dt = 2 Taking a limit, as A, sinh bt e st dt = 2 The limit exists as long as s > b e bt e st dt + 2 A [ e (b s)a s b A e (b+s)t dt ] + 2 e bt e st dt = [ e (b+s)a s + b ] [ ] + [ ] s = s b 2 s + b s 2 b 2 [ e (b s)a s b ] 2 [ e (b+s)a s + b ] [ ] [ ] b = s b 2 s + b s 2 b 2 Observe that e at sinh bt = (e (a+b)t e (a b)t )/2 It follows that A e at sinh bt e st dt = [ ] e (a+b s)a [ ] e (b a+s)a 2 s a + b 2 s + b a

193 6 93 Taking a limit, as A, e at sinh bt e st dt = 2 [ s a + b The limit exists as long as s a > b ] [ 2 Using the linearity of the Laplace transform, Since we have Therefore s + b a L [ sin bt] = 2i L [ e ibt] 2i L [ e ibt] The formula holds for s > e (a+ib)t e st dt = s a ib, e ± ibt e st dt = s ib ] = L [sin bt] = [ 2i s ib ] b = s + ib s 2 + b 2 2 Using the linearity of the Laplace transform, From Problem, we have Therefore The formula holds for s > L [ cos bt] = 2 L [ e ibt] + 2 L [ e ibt] e ± ibt e st dt = s ib L [cos bt] = [ 2 s ib + ] s = s + ib s 2 + b 2 4 Using the linearity of the Laplace transform, L [ e at cos bt ] = 2 L [ e (a+ib)t] + 2 L [e (a ib)t] Based on the integration in Problem, e (a ± ib)t e st dt = s a ib b (s a) 2 b 2 Therefore [ L e at cos bt ] = [ ] 2 s a ib + s a = s a + ib (s a) 2 + b 2 The above is valid for s > a

194 94 Chapter 6 The Laplace Transform 5 Integrating by parts, A Taking a limit, as A, te at e st dt = t e(a s)t s a A A + s a e(a s)t dt = = ea(a s) + A(a s)e A(a s) (s a) 2 te at e st dt = Note that the limit exists as long as s > a (s a) 2 7 Observe that t cosh at = (t e at + t e at )/2 For any value of c, A t e ct e st dt = t e(c s)t A A s c + s c e(c s)t dt = Taking a limit, as A, = ea(c s) + A(c s)e A(c s) (s c) 2 te ct e st dt = (s c) 2 Note that the limit exists as long as s > c Therefore, t cosh at e st dt = [ ] 2 (s a) 2 + s 2 + a 2 (s + a) 2 = (s a) 2 (s + a) 2 8 Integrating by parts, A t n e at e st dt = tn e (a s)t s a = An e (s a)a s a + A Continuing to integrate by parts, it follows that That is, A A t n e at e st dt = An e (a s)a A s a A n + s a tn e (a s)t dt = n s a tn e (a s)t dt nan e (a s)a (s a) 2 n!ae(a s)a (n 2)!(s a) 3 n!(e(a s)a ) (s a) n+ t n e at e st dt = p n (A) e (a s)a + n! (s a) n+, in which p n (ξ) is a polynomial of degree n For any given polynomial, lim p n(a) e (s a)a =, A

195 6 95 as long as s > a Therefore, t n e at e st dt = n! (s a) n+ 9 First observe that sin at e st sin at e st dt = + a cos at e st dt = s s sin at e st = + a cos at e st [ a ] sin at e st dt s s s s This implies that sin at e st (s sin at + a cos at)e st dt = s 2 + a 2 Integrating by parts we obtain that A A t 2 sin at e st 2 (s sin at + a cos at)e st dt = t s 2 + a A 2t (s sin at + a cos at)e st s 2 + a 2 dt Taking the limit A and using the results of Problem 6 (from the Student Solutions Manual), we obtain that t 2 sin at e st dt = This is valid for s > 2s 2as s 2 + a 2 (s 2 + a 2 ) 2 + 2a s 2 a 2 s 2 + a 2 (s 2 + a 2 ) 2 = 2a(3s2 a 2 ) (s 2 + a 2 ) 3 2 Observe that t 2 sinh at = (t 2 e at t 2 e at )/2 Using the result in Problem 8, t 2 sinh at e st dt = [ ] 2! 2 (s a) 3 2! (s + a) 3 = 2a(3s2 + a 2 ) (s 2 a 2 ) 3 The above is valid for s > a 22 Using the fact that f(t) = when t and integration by parts, we obtain that ] L [f(t)] = e st f(t) dt = e st t dt = [ e st s t e st + dt s = e s s + ] [ e st s 2 = e s s e s s 2 + s 2 23 Using the definition of the Laplace transform and Problem 22, we get that L [f(t)] = e st f(t) dt = e st t dt + e st dt =

196 96 Chapter 6 The Laplace Transform = e s s e s s 2 + s 2 + e s s = e s s 2 + s 2 26 Integrating by parts, A Taking a limit, as A, A t e t dt = t e t + Hence the integral converges A e t dt = e A Ae A t e t dt = 27 Based on a series expansion, note that for t >, e t > + t + t 2 /2 > t 2 /2 It follows that for t >, t 2 e t > /2 Hence for any finite A >, A t 2 e t dt > A 2 It is evident that the limit as A does not exist 28 Using the fact that cos t, and the fact that it follows that the given integral converges 3(a) Let p > Integrating by parts, A A e x x p dx = e x x p + p Taking a limit, as A, That is, Γ(p + ) = p Γ(p) A e x x p dx = p e t dt =, e x x p dx = A p e A + p e x x p dx A e x x p dx (b) Setting p =, Γ() = e x dx = (c) Let p = n Using the result in part (a), Γ(n + ) =n Γ(n) =n(n )Γ(n ) =n(n )(n 2) 2 Γ() Since Γ() =, Γ(n + ) = n!

197 62 97 (d) Using the result in part (a), Hence Γ(p + n) =(p + n ) Γ(p + n ) =(p + n )(p + n 2)Γ(p + n 2) Γ(p + n) Γ(p) Given that Γ(/2) = π, it follows that and =(p + n )(p + n 2) (p + )p Γ(p) = p(p + )(p + 2) (p + n ) Γ( 3 2 ) = 2 Γ( 2 ) = π 2 Γ( 2 ) = Γ(3 2 ) = 945 π Write the function as Hence L [Y (s)] = 3 sin 2t /2 3 Using partial fractions, 3 s = s 2 + 3s 4 = 2 5 Hence L [Y (s)] = 2(e t e 4t )/5 2 s [ s ] s Note that the denominator s 2 + 2s + 5 is irreducible over the reals Completing the square, s 2 + 2s + 5 = (s + ) Now convert the function to a rational function of the variable ξ = s + That is, We know that 2s + 2 2(s + ) s 2 = + 2s + 5 (s + ) [ ] 2 ξ L ξ 2 = 2 cos 2t + 4 Using the fact that L [e at f(t)] = L [f(t)] s s a, [ ] L 2s + 2 s 2 = 2e t cos 2t + 2s + 5

198 98 Chapter 6 The Laplace Transform 6 Using partial fractions, 2s 3 s 2 4 = [ 4 s ] s + 2 Hence L [Y (s)] = (e 2t + 7e 2t )/4 Note that we can also write 8 Using partial fractions, 2s 3 s 2 4 = 2 8s 2 4s + 2 s(s 2 + 4) s s Hence L [Y (s)] = cos 2t 2 sin 2t 2 s 2 4 = 3 s + 5 s s s The denominator s 2 + 4s + 5 is irreducible over the reals Completing the square, s 2 + 4s + 5 = (s + 2) 2 + Now convert the function to a rational function of the variable ξ = s + 2 That is, We find that 2s 5 2(s + 2) s 2 = + 4s + 5 (s + 2) 2 + [ L 5 ξ ξ ] ξ 2 = 5 sin t 2 cos t + Using the fact that L [e at f(t)] = L [f(t)] s s a, [ ] L 2s s 2 = e 2t (5 sin t 2 cos t) + 4s + 5 Note that the denominator s 2 + 2s + is irreducible over the reals Completing the square, s 2 + 2s + = (s + ) Now convert the function to a rational function of the variable ξ = s + That is, We find that 2s 3 2(s + ) 5 s 2 = + 2s + (s + ) [ 2 ξ L ξ ] ξ 2 = 2 cos 3t 5 sin 3t Using the fact that L [e at f(t)] = L [f(t)] s s a, [ ] L 2s 3 s 2 = e t (2 cos 3t 5 sin 3t) + 2s Taking the Laplace transform of the ODE, we obtain s 2 Y (s) s y() y () + 3 [s Y (s) y()] + 2 Y (s) =

199 62 99 Applying the initial conditions, s 2 Y (s) + 3s Y (s) + 2 Y (s) s 3 = Solving for Y (s), the transform of the solution is Using partial fractions, Y (s) = Hence y(t) = L [Y (s)] = 2 e t e 2t s + 3 s 2 + 3s + 2 s + 3 s 2 + 3s + 2 = 2 s + s Taking the Laplace transform of the ODE, we obtain s 2 Y (s) s y() y () 2 [s Y (s) y()] + 2 Y (s) = Applying the initial conditions, s 2 Y (s) 2s Y (s) + 2 Y (s) = Solving for Y (s), the transform of the solution is Y (s) = s 2 2s + 2 Since the denominator is irreducible, write the transform as a function of ξ = s That is, s 2 2s + 2 = (s ) 2 + First note that [ ] L ξ 2 = sin t + Using the fact that L [e at f(t)] = L [f(t)] s s a, [ ] L s 2 = e t sin t 2s + 2 Hence y(t) = e t sin t 6 Taking the Laplace transform of the ODE, we obtain s 2 Y (s) s y() y () + 2 [s Y (s) y()] + 5 Y (s) = Applying the initial conditions, s 2 Y (s) + 2s Y (s) + 5 Y (s) 2s 3 = Solving for Y (s), the transform of the solution is Y (s) = 2s + 3 s 2 + 2s + 5

200 2 Chapter 6 The Laplace Transform Since the denominator is irreducible, write the transform as a function of ξ = s + That is, 2s + 3 2(s + ) + s 2 = + 2s + 5 (s + ) We know that [ 2 ξ L ξ ] ξ 2 = 2 cos 2t + sin 2t Using the fact that L [e at f(t)] = L [f(t)] s s a, the solution of the IVP is [ ] y(t) = L 2s + 3 s 2 = e t (2 cos 2t + sin 2t ) + 2s Taking the Laplace transform of the ODE, we obtain s 4 Y (s) s 3 y() s 2 y () s y () y () Y (s) = Applying the initial conditions, s 4 Y (s) Y (s) s 3 s = Solving for the transform of the solution, Y (s) = s s 2 By inspection, it follows that y(t) = L [Y (s)] = cosh t 9 Taking the Laplace transform of the ODE, we obtain s 4 Y (s) s 3 y() s 2 y () s y () y () 4 Y (s) = Applying the initial conditions, s 4 Y (s) 4Y (s) s 3 + 2s = Solving for the transform of the solution, Y (s) = s s It follows that y(t) = L [Y (s)] = cos 2 t 2 Taking the Laplace transform of both sides of the ODE, we obtain s 2 Y (s) s y() y s () 2 [s Y (s) y()] + 2 Y (s) = s 2 + Applying the initial conditions, s 2 Y (s) 2s Y (s) + 2 Y (s) s + 2 = Solving for Y (s), the transform of the solution is Y (s) = s s 2 + s (s 2 2s + 2)(s 2 + ) + s 2 s 2 2s + 2

201 62 2 Using partial fractions on the first term, Thus we can write s (s 2 2s + 2)(s 2 + ) = 5 Y (s) = 5 s s [ ] s 2 s 2 + s 4 s 2 2s + 2 s s 3 s 2 2s + 2 For the last term, we note that s 2 2s + 2 = (s ) 2 + So that We know that 2s 3 2(s ) s 2 = 2s + 2 (s ) 2 + [ 2 ξ L ξ 2 + ] ξ 2 = 2 cos t sin t + Based on the translation property of the Laplace transform, [ ] L 2s 3 s 2 = e t (2 cos t sin t) 2s + 2 Combining the above, the solution of the IVP is y(t) = 5 cos t 2 5 sin t et (2 cos t sin t) 23 Taking the Laplace transform of both sides of the ODE, we obtain s 2 Y (s) s y() y () + 2 [s Y (s) y()] + Y (s) = 4 s + Applying the initial conditions, s 2 Y (s) + 2s Y (s) + Y (s) 2s 3 = 4 s + Solving for Y (s), the transform of the solution is First write We note that Y (s) = 4 (s + ) 3 + 2s + 3 (s + ) 2 2s + 3 2(s + ) + = (s + ) 2 (s + ) 2 = 2 s + + (s + ) 2 L [ 4 ξ ξ + ξ 2 ] = 2 t t So based on the translation property of the Laplace transform, the solution of the IVP is y(t) = 2 t 2 e t + t e t + 2 e t

202 22 Chapter 6 The Laplace Transform 25 Let f(t) be the forcing function on the right-hand-side Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, s 2 Y (s) s y() y () + Y (s) = L [f(t)] s 2 Y (s) + Y (s) = L [f(t)] Based on the definition of the Laplace transform, L [f(t)] = Solving for the transform, f(t) e st dt = Y (s) = t e st dt = s 2 e s s s 2 (s 2 + ) e s s + s 2 (s 2 + ) e s s 2 26 Let f(t) be the forcing function on the right-hand-side Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, s 2 Y (s) s y() y () + 4 Y (s) = L [f(t)] s 2 Y (s) + 4 Y (s) = L [f(t)] Based on the definition of the Laplace transform, L [f(t)] = Solving for the transform, f(t) e st dt = Y (s) = t e st dt + s 2 (s 2 + 4) e s s 2 (s 2 + 4) e st dt = s 2 e s s 2 29(a) Assuming that the conditions of Theorem 62 are satisfied, F (s) = d ds = e st f(t)dt = [ e st f(t) ] dt = s [ t e st f(t) ] dt = e st [ tf(t)] dt (b) Using mathematical induction, suppose that for some k, Differentiating both sides, F (k+) (s) = d ds = F (k) (s) = e st [ ( t) k f(t) ] dt = e st [ ( t) k f(t) ] dt [ e st ( t) k f(t) ] dt = s [ t e st ( t) k f(t) ] dt = e st [ ( t) k+ f(t) ] dt

203 We know that Based on Problem 29, Therefore, L [ e at] = s a L [ t e at] = d ds L [ t e at] = [ ] s a (s a) 2 32 Based on Problem 29, Therefore, L [( t) n ] = dn ds n L [ ] = dn ds n [ ] s L [ t n ] = ( ) n ( )n n! s n+ = n! s n+ 34 Using the translation property of the Laplace transform, Therefore, L [ e at sin bt ] = b (s a) 2 + b 2 [ L t e at sin bt ] = d [ ] b 2b(s a) ds (s a) 2 + b 2 = ((s a) 2 + b 2 ) 2 35 Using the translation property of the Laplace transform, Therefore, L [ e at cos bt ] = L [ t e at cos bt ] = d ds s a (s a) 2 + b 2 [ ] s a (s a) 2 + b 2 = (s a)2 b 2 ((s a) 2 + b 2 ) 2 36(a) Taking the Laplace transform of the given Bessel equation, L [ t y ] + L [ y ] + L [ t y ] = Using the differentiation property of the transform, That is, d ds d ds L [ y ] + L [ y ] d ds L [ y ] = [ s 2 Y (s) s y() y () ] + s Y (s) y() d ds Y (s) = It follows that ( + s 2 )Y (s) + s Y (s) =

204 24 Chapter 6 The Laplace Transform (b) We obtain a first-order linear ODE in Y (s): with integrating factor Y (s) + µ(s) = e s s 2 + Y (s) =, s s 2 + ds = s 2 + The first-order ODE can be written as d [ s2 + Y (s)] =, ds with solution Y (s) = c s2 + (c) In order to obtain negative powers of s, first write s2 + = [ + ] /2 s s 2 Expanding ( + /s 2 ) /2 in a binomial series, + (/s2 ) = 2 s s s 6 +, valid for s 2 < Hence, we can formally express Y (s) as [ Y (s) = c s 2 s s Assuming that term-by-term inversion is valid, [ y(t) = c t 2 2 2! s 7 + t 4 4! 3 5 t ! + [ = c 2! t ! + 4! t ! 6! t ! + It follows that [ y(t) = c 2 2 t2 + ] t t6 + = c The series is evidently the expansion, about x =, of J (t) ] n = ] ] ( ) n 2 2n (n!) 2 t2n 38 By definition of the Laplace transform, given the appropriate conditions, [ t ] t L [ g(t)] = e st f(τ)dτ dt = e st f(τ)dτdt Assuming that the order of integration can be exchanged, [ ] [ e L [ g(t)] = f(τ) e st sτ dt dτ = f(τ) s τ ] dτ

205 63 25 (Note the region of integration is the area between the lines τ(t) = t and τ(t) = ) Hence L [ g(t)] = s f(τ) e sτ dτ = s L [ f(t)]

206 26 Chapter 6 The Laplace Transform 6 8(a) (b) f(t) = 2u (t) + 2u 2 (t) 2u 3 (t) + u 4 (t) 9(a) (b) f(t) = + (e (t 2) )u 2 (t)

207 63 27 (a) (b) f(t) = t u (t) u 2 (t) + (2 t)u 3 (t) 2(a) (b) f(t) = t + (2 t)u 2 (t) + (5 t)u 5 (t) + (t 7)u 7 (t) 3 Using the Heaviside function, we can write f(t) = (t 2) 2 u 2 (t) The Laplace transform has the property that L [ u c (t)f(t c)] = e cs L [f(t)] Hence L [ (t 2) 2 u 2 (t) ] = 2 e 2s s 3 5 The function can be expressed as f(t) = (t π) [u π (t) u 2π (t)] Before invoking the translation property of the transform, write the function as f(t) = (t π) u π (t) (t 2π) u 2π (t) π u 2π (t) It follows that L [ f(t)] = e πs s 2 e 2πs s 2 πe 2πs s 6 It follows directly from the translation property of the transform that L [ f(t)] = e s s + 2e 3s s 6 e 4s s

208 28 Chapter 6 The Laplace Transform 7 Before invoking the translation property of the transform, write the function as f(t) = (t 2) u 2 (t) u 2 (t) (t 3) u 3 (t) u 3 (t) It follows that L [ f(t)] = e 2s s 2 e 2s s e 3s s 2 e 3s s 8 It follows directly from the translation property of the transform that L [ f(t)] = s 2 e s s 2 9 Using the fact that L [e at f(t)] = L [f(t)] s s a, [ ] L 3! (s 2) 4 = t 3 e 2t 22 The inverse transform of the function 2/(s 2 4) is f(t) = sinh 2t Using the translation property of the transform, [ ] 2 e L 2s s 2 = sinh (2(t 2)) u 2 (t) 4 23 First consider the function G(s) = Completing the square in the denominator, G(s) = (s 2) s 2 4s + 3 (s 2) (s 2) 2 It follows that L [G(s)] = e 2t cosh t Hence [ ] (s 2)e L s s 2 = e 2(t ) cosh (t ) u (t) 4s Write the function as F (s) = e s s + e 2s s e 3s s e 4s s It follows from the translation property of the transform, that [ e L s + e 2s e 3s e 4s ] = u (t) + u 2 (t) u 3 (t) u 4 (t) s 25(a) By definition of the Laplace transform, L [ f(ct)] = e st f(ct)dt

209 63 29 Making a change of variable, τ = ct, we have L [ f(ct)] = c e s(τ/c) f(τ)dτ = c Hence L [ f(ct)] = (/c) F (s/c), where s/c > a (b) Using the result in part (a), Hence L [ f ( )] t = k F (ks) k L [F (ks)] = k f ( t k ) e (s/c)τ f(τ)dτ (c) From part (b), L [F (as)] = (/a)f(t/a) Note that as + b = a(s + b/a) Using the fact that L [e ct f(t)] = L [f(t)] s s c, L [F (as + b)] = e bt/a ( ) t a f a 26 First write F (s) = n! ( s 2 )n+ Let G(s) = n!/s n+ Based on the results in Problem 25, [ ( s )] 2 L G = g(2t), 2 in which g(t) = t n Hence L [F (s)] = 2 (2t) n = 2 n+ t n 29 First write Now consider Using the result in Problem 25(b), F (s) = e 4(s /2) 2(s /2) G(s) = e 2s s L [G(2s)] = 2 g ( t 2 in which g(t) = u 2 (t) Hence L [G(2s)] = u 2 (t/2)/2 = u 4 (t)/2 It follows that ), L [F (s)] = 2 et/2 u 4 (t) 3 By definition of the Laplace transform, L [ f(t)] = e st u (t)dt

210 2 Chapter 6 The Laplace Transform That is, L [ f(t)] = e st dt = e s s 3 First write the function as f(t) = u (t) u (t) + u 2 (t) u 3 (t) It follows that That is, L [ f(t)] = L [ f(t)] = e s s e st dt + + e 2s e 3s s 3 2 e st dt = e s + e 2s e 3s s 32 The transform may be computed directly On the other hand, using the translation property of the transform, [ L [f(t)] = 2n+ s + ( ) k e ks = 2n+ ] ( e s ) k = ( e s ) 2n+2 s s s + e s That is, k = k = L [f(t)] = (e 2s ) n+ s( + e s ) 35 The given function is periodic, with T = 2 Using the result of Problem 34, That is, L [f(t)] = e 2s L [f(t)] = 2 e st f(t)dt = e 2s e s s( e 2s ) = s( + e s ) e st dt 37 The function is periodic, with T = Using the result of Problem 34, It follows that L [f(t)] = e s t e st dt L [f(t)] = e s ( + s) s 2 ( e s ) 38 The function is periodic, with T = π Using the result of Problem 34, We first calculate Hence L [f(t)] = π e πs π sin t e st dt sin t e st dt = + e πs + s 2 L [f(t)] = + e πs ( e πs )( + s 2 )

211 (a) We get that L [f(t)] = L [] L [u (t)] = s e s s (b) Let F (s) = L [ u (t)] Then [ t ] L [ u (τ)] dτ = (c) s F (s) = e s s 2

212 22 Chapter 6 The Laplace Transform Let G(s) = L [g(t)] Then 4(a) L [h(t)] = G(s) e s G(s) = e s s 2 s e s e s 2 = ( e s ) 2 s 2 (b) The given function is periodic, with T = 2 Using the result of Problem 34, L [f(t)] = e 2s Based on the piecewise definition of p(t), Hence 2 e st p(t)dt = t e st dt e st p(t)dt L [p(t)] = ( e s ) s 2 ( + e s ) (2 t)e st dt = s 2 ( e s ) 2 (c) Since p(t) satisfies the hypotheses of Theorem 62, L [p (t)] = s L [p(t)] p() Using the result of Problem 36 from the Student Solutions Manual, We note the p() =, hence L [p (t)] = ( e s ) s( + e s ) L [p(t)] = s [ ( e s ] ) s( + e s ) 64 2(a) Let h(t) be the forcing function on the right-hand-side Taking the Laplace transform of both sides of the ODE, we obtain s 2 Y (s) s y() y () + 2 [s Y (s) y()] + 2 Y (s) = L [h(t)]

213 64 23 Applying the initial conditions, s 2 Y (s) + 2s Y (s) + 2 Y (s) = L [h(t)] The forcing function can be written as h(t) = u π (t) u 2π (t) Its transform is L [h(t)] = e πs e 2πs s Solving for Y (s), the transform of the solution is First note that Using partial fractions, Y (s) = s 2 + 2s e πs e 2πs s(s 2 + 2s + 2) s 2 + 2s + 2 = (s + ) 2 + s(s 2 + 2s + 2) = 2 s 2 Taking the inverse transform, term-by-term, [ ] [ ] L s 2 = L + 2s + 2 (s + ) 2 + Now let Then Using Theorem 63, G(s) = s(s 2 + 2s + 2) (s + ) + (s + ) 2 + = e t sin t L [G(s)] = 2 2 e t cos t 2 e t sin t L [ e cs G(s) ] = 2 u c(t) 2 e (t c) [cos(t c) + sin(t c)] u c (t) Hence the solution of the IVP is That is, y(t) = e t sin t + 2 u π(t) 2 e (t π) [cos(t π) + sin(t π)] u π (t) 2 u 2π(t) + 2 e (t 2π) [cos(t 2π) + sin(t 2π)] u 2π (t) y(t) = e t sin t + 2 [u π(t) u 2π (t)] + 2 e (t π) [cos t + sin t] u π (t)+ + 2 e (t 2π) [cos t + sin t] u 2π (t)

214 24 Chapter 6 The Laplace Transform (b) The solution starts out as free oscillation, due to the initial conditions The amplitude increases, as long as the forcing is present Thereafter, the solution rapidly decays 4(a) Let h(t) be the forcing function on the right-hand-side Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, s 2 Y (s) s y() y () + 4 Y (s) = L [h(t)] The transform of the forcing function is s 2 Y (s) + 4 Y (s) = L [h(t)] L [h(t)] = s e πs s 2 + Solving for Y (s), the transform of the solution is Y (s) = Using partial fractions, It follows that (s 2 + 4)(s 2 + ) + e πs (s 2 + 4)(s 2 + ) (s 2 + 4)(s 2 + ) = 3 [ ] L (s 2 + 4)(s 2 = + ) 3 [ s 2 + ] s [sin t 2 sin 2t ] Based on Theorem 63, [ L e πs ] (s 2 + 4)(s 2 = [sin(t π) 2 ] + ) 3 sin(2t 2π) u π (t) Hence the solution of the IVP is y(t) = 3 [sin t 2 sin 2t ] 3 [sin t + 2 sin 2t ] u π (t)

215 64 25 (b) Since there is no damping term, the solution follows the forcing function, after which the response is a steady oscillation about y = 5(a) Let f(t) be the forcing function on the right-hand-side Taking the Laplace transform of both sides of the ODE, we obtain s 2 Y (s) s y() y () + 3 [s Y (s) y()] + 2 Y (s) = L [f(t)] Applying the initial conditions, The transform of the forcing function is Solving for the transform, Using partial fractions, Hence s 2 Y (s) + 3s Y (s) + 2 Y (s) = L [f(t)] Y (s) = L [f(t)] = s e s s s(s 2 + 3s + 2) e s s(s 2 + 3s + 2) s(s 2 + 3s + 2) = 2 [ ] L s(s 2 = + 3s + 2) [ s + s ] s e 2t 2 e t Based on Theorem 63, [ L e s ] s(s 2 = [ + e 2(t ) 2e (t )] u (t) + 3s + 2) 2 Hence the solution of the IVP is y(t) = 2 [ u (t)] + e 2t 2 e t 2 [ e (2t 2) 2e (t )] u (t)

216 26 Chapter 6 The Laplace Transform (b) The solution increases to a temporary steady value of y = /2 After the forcing ceases, the response decays exponentially to y = 6(a) Taking the Laplace transform of both sides of the ODE, we obtain s 2 Y (s) s y() y () + 3 [s Y (s) y()] + 2 Y (s) = e 2s s Applying the initial conditions, s 2 Y (s) + 3s Y (s) + 2 Y (s) = e 2s s Solving for the transform, Using partial fractions, and Y (s) = s 2 + 3s e 2s s(s 2 + 3s + 2) s 2 + 3s + 2 = s + s + 2 s(s 2 + 3s + 2) = [ 2 s + s ] s + Taking the inverse transform term-by-term, the solution of the IVP is [ y(t) = e t e 2t + 2 e (t 2) + ] 2 e 2(t 2) u 2 (t) (b)

217 64 27 Due to the initial conditions, the response has a transient overshoot, followed by an exponential convergence to a steady value of y s = /2 7(a) Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, Solving for the transform, Using partial fractions, Hence s 2 Y (s) s y() y () + Y (s) = e 3πs s Y (s) = s 2 Y (s) + Y (s) s = e 3πs s Y (s) = s s e 3πs s(s 2 + ) s(s 2 + ) = s s s 2 + s [ s e 3πs s s ] s 2 + Taking the inverse transform, the solution of the IVP is y(t) = cos t + [ cos(t 3π)] u 3π (t) = cos t + [ + cos t ] u 3π (t) (b) Due to initial conditions, the solution temporarily oscillates about y = After the forcing is applied, the response is a steady oscillation about y m = 9(a) Let g(t) be the forcing function on the right-hand-side Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, s 2 Y (s) s y() y () + Y (s) = L [g(t)] s 2 Y (s) + Y (s) = L [g(t)]

218 28 Chapter 6 The Laplace Transform The forcing function can be written as with Laplace transform Solving for the transform, Using partial fractions, g(t) = t 2 [ u 6(t)] + 3 u 6 (t) = t 2 2 (t 6)u 6(t) L [g(t)] = 2s 2 e 6s 2s 2 Y (s) = s s 2 (s 2 + ) e 6s 2s 2 (s 2 + ) 2s 2 (s 2 + ) = [ 2 s 2 ] s 2 + Taking the inverse transform, and using Theorem 63, the solution of the IVP is y(t) = sin t + 2 [t sin t] 2 [(t 6) sin(t 6)] u 6(t) = 2 [t + sin t] 2 [(t 6) sin(t 6)] u 6(t) (b) The solution increases, in response to the ramp input, and thereafter oscillates about a mean value of y m = 3 (a) Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, Solving for the transform, s 2 Y (s) s y() y () + 4 Y (s) = e πs s s 2 Y (s) + 4 Y (s) = e πs s Y (s) = e 3πs s e πs s(s 2 + 4) e 3πs s(s 2 + 4) e 3πs s

219 64 29 Using partial fractions, s(s 2 + 4) = [ 4 s s ] s Taking the inverse transform, and applying Theorem 63, (b) y(t) = 4 [ cos(2t 2π)] u π(t) 4 [ cos(2t 6π)] u 3π(t) = 4 [u π(t) u 3π (t)] 4 cos 2t [u π(t) u 3π (t)] Since there is no damping term, the solution responds immediately to the forcing input There is a temporary oscillation about y = /4 2(a) Taking the Laplace transform of the ODE, we obtain s 4 Y (s) s 3 y() s 2 y () s y () y () Y (s) = e s s Applying the initial conditions, s 4 Y (s) Y (s) = e s s Solving for the transform of the solution, Using partial fractions, It follows that Y (s) = s(s 4 ) = 4 e 2s s e s s(s 4 ) e 2s s(s 4 ) [ 4 s + s + + s + 2s ] s 2 + [ ] L s(s 4 = [ 4 + e t + e t + 2 cos t ] ) 4 e 2s s Based on Theorem 63, the solution of the IVP is y(t) = [u (t) u 2 (t)] + [ ] e (t ) + e (t ) + 2 cos(t ) u (t) 4 [ ] e (t 2) + e (t 2) + 2 cos(t 2) u 2 (t) 4

220 22 Chapter 6 The Laplace Transform (b) The solution increases without bound, exponentially 3(a) Taking the Laplace transform of the ODE, we obtain s 4 Y (s) s 3 y() s 2 y () s y () y ()+ + 5 [ s 2 Y (s) s y() y () ] + 4 Y (s) = s e πs s Applying the initial conditions, s 4 Y (s) + 5s 2 Y (s) + 4Y (s) = s e πs s Solving for the transform of the solution, Using partial fractions, It follows that Y (s) = s(s 4 + 5s 2 + 4) e πs s(s 4 + 5s 2 + 4) s(s 4 + 5s 2 + 4) = 2 [ ] L s(s 4 + 5s 2 + 4) [ 3 s + s s Based on Theorem 63, the solution of the IVP is y(t) = 4 [ u π(t)] + 2 That is, y(t) = 4 [ u π(t)] + 2 4s ] s 2 + = [3 + cos 2t 4 cos t] 2 [cos 2t 4 cos t] 2 [cos 2(t π) 4 cos(t π)] u π(t) [cos 2t 4 cos t] 2 [cos 2t + 4 cos t] u π(t)

221 64 22 (b) After an initial transient, the solution oscillates about y m = 4 The specified function is defined by, t < t h f(t) = k (t t ), t t < t + k h, t t + k which can conveniently be expressed as f(t) = h k (t t ) u t (t) h k (t t k) u t+k(t) 5 The function is defined by, t < t h g(t) = k (t t ), t t < t + k h k (t t 2k), t + k t < t + 2k, t t + 2k which can also be written as g(t) = h k (t t ) u t (t) 2h k (t t k) u t+k(t) + h k (t t 2k) u t+2k(t) 7 We consider the initial value problem y + 4y = k [(t 5) u 5(t) (t 5 k) u 5+k (t)], with y() = y () = (a) The specified function is defined by, t < 5 f(t) = k (t 5), 5 t < 5 + k, t 5 + k so k controls the point at which f(t) reaches When k = 5, f(t) = g(t) in Ex2

222 222 Chapter 6 The Laplace Transform (b) Taking the Laplace transform of both sides of the ODE, we obtain s 2 Y (s) s y() y () + 4 Y (s) = e 5s ks 2 Applying the initial conditions, s 2 Y (s) + 4 Y (s) = e 5s ks 2 Solving for the transform, Using partial fractions, It follows that Y (s) = e (5+k)s ks 2 e 5s ks 2 (s 2 + 4) e (5+k)s ks 2 (s 2 + 4) s 2 (s 2 + 4) = [ 4 s 2 ] s [ ] L s 2 (s 2 = + 4) 4 t sin 2t 8 Using Theorem 63, the solution of the IVP is in which h(t) = t/4 sin 2t /8 e (5+k)s ks 2 y(t) = k [h(t 5) u 5(t) h(t 5 k) u 5+k (t)], (c) Note that for t > 5 + k, the solution is given by y(t) = 4 sin(2t ) + 8k 8k sin(2t 2k) = 4 sin k cos(2t k) 4k So for t > 5 + k, the solution oscillates about y m = /4, with an amplitude of A = sin(k) 4k

223 (a) k = 3/2 (b) k = (c) k = /2 8(a) The graph shows f k for k = 2, k = and k = /2 (b) The forcing function can be expressed as f k (t) = 2k [u 4 k(t) u 4+k (t)] Taking the Laplace transform of both sides of the ODE, we obtain s 2 Y (s) s y() y () + 3 Applying the initial conditions, Solving for the transform, [s Y (s) y()] + 4 Y (s) = e (4 k)s 2ks s 2 Y (s) + e (4 k)s s Y (s) + 4 Y (s) = 3 2ks Y (s) = Using partial fractions, Let s(3s 2 + s + 2) = 2 e (4+k)s 2ks 3 e (4 k)s 2ks(3s 2 + s + 2) 3 e (4+k)s 2ks(3s 2 + s + 2) [ s ] + 3s 3s 2 = [ + s H(s) = [ 8k s 6 (s + 6 ) s 6 s + ] 6 (s + 6 ) e (4+k)s 2ks + 6(s + 6 ) (s + 6 ) ]

224 224 Chapter 6 The Laplace Transform It follows that h(t) = L [H(s)] = 8k e t/6 8k Based on Theorem 63, the solution of the IVP is [ ( ) ( )] 43 t 43 t sin + cos y(t) = h(t 4 + k) u 4 k (t) h(t 4 k) u 4+k (t) (c) (a) k = 2 (b) k = (c) k = /2 As the parameter k decreases, the solution remains null for a longer period of time Since the magnitude of the impulsive force increases, the initial overshoot of the response also increases The duration of the impulse decreases All solutions eventually decay to y = 2(a) (a) n = 3 (b) n = 4 (c) n = 5 (b) Taking the Laplace transform of both sides of the ODE, we obtain s 2 Y (s) s y() y () + Y (s) = s + Applying the initial conditions, s 2 Y (s) + Y (s) = s + n k = n k = ( ) k e kπs s ( ) k e kπs s

225 Solving for the transform, Using partial fractions, Let Y (s) = s(s 2 + ) + n k = ( ) k e kπs s(s 2 + ) s(s 2 + ) = s s s 2 + [ ] h(t) = L s(s 2 = cos t + ) Applying Theorem 63, term-by-term, the solution of the IVP is Note that y(t) = h(t) + n ( ) k h(t kπ) u kπ (t) k = h(t kπ) = u (t kπ) cos(t kπ) = u kπ (t) ( ) k cos t Hence n n y(t) = cos t + ( ) k u kπ (t) (cos t) u kπ (t) k = k = (c) The ODE has no damping term Each interval of forcing adds to the energy of the system, so the amplitude will increase For n = 5, g(t) = when t > 5π Therefore the oscillation will eventually become steady, with an amplitude depending on the values of y(5π) and y (5π) (d) As n increases, the interval of forcing also increases Hence the amplitude of the transient will increase with n Eventually, the forcing function will be constant In fact, for large values of t, {, n even g(t) =, n odd

226 226 Chapter 6 The Laplace Transform Further, for t > nπ, y(t) = cos t n cos t ( )n 2 Hence the steady state solution will oscillate about or, depending on n, with an amplitude of A = n + In the limit, as n, the forcing function will be a periodic function, with period 2π From Problem 33, in Section 63, L [g(t)] = s( + e s ) As n increases, the duration and magnitude of the transient will increase without bound 22(a) Taking the initial conditions into consideration, the transform of the ODE is Solving for the transform, s 2 Y (s) + s Y (s) + Y (s) = s + Y (s) = Using partial fractions, s(s 2 + s + ) + n k = n k = ( ) k e kπs s ( ) k e kπs s(s 2 + s + ) s(s 2 + s + ) = s s + s 2 + s + Since the denominator in the second term is irreducible, write s + s 2 + s + = (s + 5) + 5 (s + 5) 2 + (399/4) Let [ ] h(t) = L s (s + 5) (s + 5) 2 + (399/4) 5 (s + 5) 2 + (399/4) 399 = e [cos( t/2 t) + ] 399 sin( t) Applying Theorem 63, term-by-term, the solution of the IVP is n y(t) = h(t) + ( ) k h(t kπ) u kπ (t) k = For odd values of n, the solution approaches y = (On the next figure, n = 5)

227 For even values of n, the solution approaches y = (On the next figure, n = 6) (b) The solution is a sum of damped sinusoids, each of frequency ω = 399 /2 Each term has an initial amplitude of approximately For any given n, the solution contains n + such terms Although the amplitude will increase with n, the amplitude will also be bounded by n + (c) Suppose that the forcing function is replaced by g(t) = sin t Based on the methods in Chapter 3, the general solution of the differential equation is 399 y(t) = e [c t/2 cos( t) + c 2 sin( 2 t) ] + y p (t) Note that y p (t) = A cos t + B sin t Using the method of undetermined coefficients, A = and B = Based on the initial conditions, the solution of the IVP is 399 y(t) = e [cos( t/2 2 t) sin( t) ] cos t Observe that both solutions have the same frequency, ω = 399/2

228 228 Chapter 6 The Laplace Transform 23(a) Taking the initial conditions into consideration, the transform of the ODE is Solving for the transform, Using partial fractions, s 2 Y (s) + Y (s) = s + 2 Y (s) = s(s 2 + ) + 2 n k = n k = ( ) k e (k/4)s s ( ) k e (k/4)s s(s 2 + ) s(s 2 + ) = s s s 2 + Let [ ] h(t) = L s(s 2 = cos t + ) Applying Theorem 63, term-by-term, the solution of the IVP is That is, y(t) = h(t) + 2 y(t) = cos t + 2 k = n ( ) k h(t k 4 ) u k/4(t) k = n [ ( ) k cos(t k ] 4 ) u k/4 (t) (b) On the figure we see the solution for n = 35

229 (c) Based on the plot, the slow period appears to be 88 The fast period appears to be about 6 These values correspond to a slow frequency of ω s = 74 and a fast frequency ω f = 472 (d) The natural frequency of the system is ω = The forcing function is initially periodic, with period T = /2 = 55 Hence the corresponding forcing frequency is w = 424 Using the results in Section 38, the slow frequency is given by ω s = ω ω = 72 2 and the fast frequency is given by ω f = ω + ω = 72 2 Based on theses values, the slow period is predicted as and the fast period is given as (a) Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, Solving for the transform, s 2 Y (s) s y() y () + 4Y (s) = e πs e 2πs s 2 Y (s) + 4Y (s) = e πs e 2πs Y (s) = e πs e 2πs s Applying Theorem 63, the solution of the IVP is = e πs s e 2πs s y(t) = 2 sin(2t 2π)u π(t) 2 sin(2t 4π)u 2π(t) = 2 sin(2t) [ u π(t) u 2π (t) ]

230 23 Chapter 6 The Laplace Transform (b) 4(a) Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, Solving for the transform, s 2 Y (s) s y() y () Y (s) = 2 e 3s s 2 Y (s) Y (s) s = 2 e 3s Y (s) = s 2 e 3s s 2 s 2 Using a table of transforms, and Theorem 63, the solution of the IVP is y(t) = cosh t 2 sinh(t 3)u 3 (t) (b) 6(a) Taking the initial conditions into consideration, the transform of the ODE is s 2 Y (s) + 4Y (s) s/2 = e 4πs Solving for the transform, Y (s) = s/2 s e 4πs s 2 + 4

231 65 23 Using a table of transforms, and Theorem 63, the solution of the IVP is y(t) = 2 cos 2t + 2 sin(2t 8π)u 4π(t) = 2 cos 2t + 2 sin(2t) u 4π(t) (b) 8(a) Taking the Laplace transform of both sides of the ODE, we obtain Applying the initial conditions, Solving for the transform, s 2 Y (s) s y() y () + 4Y (s) = 2 e (π/4)s s 2 Y (s) + 4Y (s) = 2 e (π/4)s Y (s) = 2 e (π/4)s s Applying Theorem 63, the solution of the IVP is y(t) = sin(2t π 2 )u π/4(t) = cos(2t) u π/4 (t) (b)

232 232 Chapter 6 The Laplace Transform 9(a) Taking the initial conditions into consideration, the transform of the ODE is s 2 Y (s) + Y (s) = e (π/2)s s Solving for the transform, Using partial fractions, Hence Y (s) = e (π/2)s s + 3 e (3π/2)s e 2πs s Y (s) = e (π/2)s s(s 2 + ) + 3 e (3π/2)s s 2 e 2πs + s(s 2 + ) s(s 2 + ) = s s s 2 + s e (π/2)s s e (3π/2)s s 2 + Based on Theorem 63, the solution of the IVP is e 2πs s + s e 2πs s 2 + y(t) = u π/2 (t) cos(t π 2 )u π/2(t) + 3 sin(t 3π 2 )u 3π/2(t) u 2π (t) + cos(t 2π)u 2π (t) That is, y(t) = [ sin(t)] u π/2 (t) + 3 cos(t) u 3π/2 (t) [ cos(t)] u 2π (t) (b) (a) Taking the transform of both sides of the ODE, 2s 2 Y (s) + sy (s) + 4Y (s) = Solving for the transform, First write Y (s) = e st δ(t π 6 ) sin t dt = 2 e (π/6)s e (π/6)s 2(2s 2 + s + 4) 2(2s 2 + s + 4) = 4 (s + 4 )

233 It follows that (b) y(t) = L [Y (s)] = 3 e (t π/6)/4 sin 3 4 (t π 6 )u π/6(t) (a) Taking the initial conditions into consideration, the transform of the ODE is s 2 Y (s) + 2s Y (s) + 2 Y (s) = s s e (π/2)s Solving for the transform, Y (s) = Using partial fractions, We can also write Let Then s (s 2 + )(s 2 + 2s + 2) = 5 s (s 2 + )(s 2 + 2s + 2) + e (π/2)s s 2 + 2s + 2 [ s s ] s 2 + s + 4 s 2 + 2s + 2 s + 4 (s + ) + 3 s 2 = + 2s + 2 (s + ) 2 + Y (s) = s (s 2 + )(s 2 + 2s + 2) L [Y (s)] = 5 cos t sin t 5 e t [cos t + 3 sin t] Applying Theorem 63, [ ] e L (π/2)s ( s 2 = e (t π 2 ) sin t π ) u π/2 (t) + 2s Hence the solution of the IVP is y(t) = 5 cos t sin t 5 e t [cos t + 3 sin t] e (t π 2 ) cos(t) u π/2 (t)

234 234 Chapter 6 The Laplace Transform (b) 2(a) Taking the initial conditions into consideration, the transform of the ODE is Solving for the transform, s 4 Y (s) Y (s) = e s Using partial fractions, It follows that Y (s) = s 4 = [ 2 e s s 4 s 2 s 2 + ] [ ] L s 4 = 2 sinh t 2 sin t Applying Theorem 63, the solution of the IVP is y(t) = 2 [sinh(t ) sin(t )] u (t) (b)

235 (a) The Laplace transform of the ODE is Solving for the transform of the solution, First write s 2 Y (s) + 2 s Y (s) + Y (s) = e s Y (s) = e s s 2 + s/2 + s 2 + s/2 + = (s + 4 ) Taking the inverse transform and applying both shifting theorems, y(t) = 4 5 e (t )/4 sin 5 4 (t ) u (t) (b) As shown on the graph, the maximum is attained at some t > 2 Note that for t > 2, y(t) = 4 5 e (t )/4 sin (t ) 5 4 Setting y (t) =, we find that t 2363 The maximum value is calculated as y(2363) 753 (c) Setting γ = /4, the transform of the solution is Y (s) = e s s 2 + s/4 + Following the same steps, it follows that y(t) = e (t )/8 sin (t ) u (t)

236 236 Chapter 6 The Laplace Transform Once again, the maximum is attained at some t > 2 Setting y (t) =, we find that t 24569, with y(t ) 8335 (d) Now suppose that < γ < Then the transform of the solution is First write Y (s) = e s s 2 + γs + s 2 + γs + = (s + γ/2) 2 + ( γ 2 /4) It follows that [ ] h(t) = L 2 s 2 = + γs + 4 γ 2 e γt/2 sin( γ 2 /4 t) Hence the solution is y(t) = h(t ) u (t) The solution is nonzero only if t >, in which case y(t) = h(t ) Setting y (t) =, we obtain [ ] tan γ2 /4 (t ) = 4 γ2, γ that is, [ ] tan γ2 /4 (t ) = 2 γ2 /4 γ As γ, we obtain the formal equation tan(t ) = Hence t + π 2 Setting t = π/2 in h(t), and letting γ, we find that y These conclusions agree with the case γ =, for which it is easy to show that the solution is y(t) = sin(t ) u (t) 5(a) See Problem 4 It follows that the solution of the IVP is y(t) = 4k 5 e (t )/4 sin 5 4 (t ) u (t)

237 This function is a multiple of the answer in Problem 4(a) Hence the peak value occurs at t 2363 The maximum value is calculated as y(2363) 753 k We find that the appropriate value of k is k = 2/ (b) Based on Problem 4(c), the solution is y(t) = 8 k 3 7 e (t )/8 sin (t ) u (t) Since this function is a multiple of the solution in Problem 4(c), we have that t 24569, with y(t ) 8335 k The solution attains a value of y = 2 when k = 2/8335, that is, k (c) Similar to Problem 4(d), for < γ <, the solution is y(t) = h(t ) u (t), in which h(t) = 2 k 4 γ 2 e γt/2 sin( γ 2 /4 t) It follows that t π/2 Setting t = π/2 in h(t), and letting γ, we find that y k Requiring that the peak value remains at y = 2, the limiting value of k is k = 2 These conclusions agree with the case γ =, for which it is easy to show that the solution is y(t) = k sin(t ) u (t) 6(a) Taking the initial conditions into consideration, the transformation of the ODE is s 2 Y (s) + Y (s) = [ ] e (4 k)s e (4+k)s 2k s s Solving for the transform of the solution, Y (s) = [ ] e (4 k)s 2k s(s 2 + ) e (4+k)s s(s 2 + ) Using partial fractions, Now let s(s 2 + ) = s s s 2 + [ ] h(t) = L s(s 2 = cos t + ) Applying Theorem 63, the solution is φ(t, k) = 2k [h(t 4 + k) u 4 k(t) h(t 4 k) u 4+k (t)]

238 238 Chapter 6 The Laplace Transform That is, φ(t, k) = 2k [ u 4 k(t) u 4+k (t)] 2k [cos(t 4 + k) u 4 k(t) cos(t 4 k) u 4+k (t)] (b) Consider various values of t For any fixed t < 4, φ(t, k) =, as long as 4 k > t If t 4, then for 4 + k < t, It follows that Hence φ(t, k) = [cos(t 4 + k) cos(t 4 k) ] 2k 4 + k) cos(t 4 k) lim φ(t, k) = lim cos(t = sin(t 4) k k 2k lim φ(t, k) = sin(t 4) u 4(t) k (c) The Laplace transform of the differential equation with y() = y () =, is Solving for the transform of the solution, y + y = δ(t 4), s 2 Y (s) + Y (s) = e 4s Y (s) = e 4s s 2 + It follows that the solution is φ (t) = sin(t 4) u 4 (t), and this means that lim φ(t, k) = φ (t) k (d) We can see the convergence on the graphs

239 (b) The transform of the ODE (given the specified initial conditions) is Solving for the transform of the solution, 2 s 2 Y (s) + Y (s) = ( ) k+ e kπs Y (s) = s 2 + Applying Theorem 63, term-by-term, k = 2 k = ( ) k+ e kπs 2 y(t) = ( ) k+ sin(t kπ) u kπ (t) = sin(t) k = 2 k = u kπ (t) (c) For t > 2π, y(t) = 2 sin t 9(b) Taking the initial conditions into consideration, the transform of the ODE is s 2 Y (s) + Y (s) = Solving for the transform of the solution, Y (s) = s 2 + Applying Theorem 63, term-by-term, y(t) = 2 k = 2 k = 2 k = e (kπ/2)s e (kπ/2)s sin(t kπ 2 ) u kπ/2(t)

240 24 Chapter 6 The Laplace Transform (c) For t > π, y(t) = 2(b) The transform of the ODE (given the specified initial conditions) is 2 s 2 Y (s) + Y (s) = ( ) k+ e (kπ/2)s Solving for the transform of the solution, k = Y (s) = 2 k+ e (kπ/2)s ( ) k = s 2 + Applying Theorem 63, term-by-term, 2 y(t) = ( ) k+ sin(t kπ 2 ) u kπ/2(t) k = (c) For t > π, y(t) = 22(b) Taking the initial conditions into consideration, the transform of the ODE is 4 s 2 Y (s) + Y (s) = ( ) k+ e (k/4)s k =

241 65 24 Solving for the transform of the solution, Y (s) = 4 k+ e (k/4)s ( ) k = Applying Theorem 63, term-by-term, s y(t) = ( ) k+ sin(t k 4 ) u k/4(t) k = (c) For t >, 4 ( y(t) = ( ) k+ sin t k ) cos(56375 t) 4 k= 23(b) The transform of the ODE (given the specified initial conditions) is 2 s 2 Y (s) + s Y (s) + Y (s) = ( ) k+ e kπs Solving for the transform of the solution, First write It follows that Y (s) = 2 k = k = e kπs s 2 + s + s 2 + s + = (s + 2 ) [ ] L s 2 + s + Applying Theorem 63, term-by-term, = e t/2 sin( t) 2 y(t) = ( ) k+ h(t kπ) u kπ (t), k =

242 242 Chapter 6 The Laplace Transform in which h(t) = e t/2 sin( t) (c) For t > 2π, each term in the sum for y(t) will contain a decaying exponential term multiplied by a bounded term Thus y(t) 24(b) Taking the initial conditions into consideration, the transform of the ODE is s 2 Y (s) + s Y (s) + Y (s) = Solving for the transform of the solution, Y (s) = As shown in Problem 23, [ ] L s 2 + s + 5 k = Applying Theorem 63, term-by-term, in which y(t) = 5 k = 5 k = e (2k )πs s 2 + s + e (2k )πs = e t/2 sin( t) h [t (2k )π] u (2k )π (t), h(t) = e t/2 sin( t)

243 (c) For t > 29π, each term in the sum for y(t) will contain a decaying exponential term multiplied by a bounded term Thus y(t) 66 2 Let f(t) = e t Then (f ) (t) = 3 It follows directly that (f f) (t) = t t sin(t τ) sin(τ) dτ = 2 The range of the resulting function is R t e t τ dτ = e t e τ dτ = e t t [cos(t 2τ) cos(t)]dτ = (sin t t cos t) 2 5 We have L [e t ] = /(s + ) and L [sin t] = /(s 2 + ) Based on Theorem 66, [ t ] L e (t τ) sin(τ) dτ = s + s 2 + = (s + )(s 2 + ) 6 Let g(t) = t and h(t) = e t Then f(t) = (g h) (t) Applying Theorem 66, [ t ] L g(t τ)h(τ) dτ = s 2 s = s 2 (s ) 7 We have f(t) = (g h) (t), in which g(t) = sin t and h(t) = cos t The transform of the convolution integral is [ t ] L g(t τ)h(τ) dτ = s 2 + s s 2 + = s (s 2 + ) 2 9 It is easy to see that [ ] [ ] L = e t and L s s + s 2 = cos 2t + 4

244 244 Chapter 6 The Laplace Transform Applying Theorem 66, [ ] L s (s + )(s 2 = + 4) t e (t τ) cos 2τ dτ We first note that [ ] [ ] L (s + ) 2 = t e t and L s Based on the convolution theorem, [ ] L (s + ) 2 (s 2 + 4) = 2 = 2 t t = sin 2t 2 (t τ)e (t τ) sin 2τ dτ τ e τ sin(2t 2τ) dτ Let g(t) = L [G(s)] Since [ L /(s 2 + ) ] = sin t, the inverse transform of the product is [ ] G(s) t t L s 2 = g(t τ) sin τ dτ = sin(t τ) g(τ) dτ + 2(a) By definition, f g = Set τ = t t u, dτ = t du, so that t (t τ) m τ n dτ = t (t τ) m τ n dτ (tu) m (t tu) n t du = t m+n+ u m ( u) n du (b) The Convolution Theorem states that L [f g] = L [f] L [g] Noting that it follows that Therefore L [ t k] = k! s k+, m! n! (m + n + )! s m+ = sm+ s m+n+2 u m ( u) n du = (c) If k is not an integer, we can write and L[t k ] = u m ( u) n du m! n! (m + n + )! Γ(k + ) s k+ Γ(m + ) Γ(n + ) Γ(m + ) s m+ s n+ = s m+n+2 u m ( u) n du,

245 so u m ( u) n du = Γ(m + )Γ(n + ) Γ(m + n + 2) 3 Taking the initial conditions into consideration, the transform of the ODE is Solving for the transform of the solution, s 2 Y (s) + ω 2 Y (s) = G(s) Y (s) = s 2 + ω 2 + G(s) s 2 + ω 2 As shown in a related situation, Problem, [ ] G(s) L s 2 + ω 2 = t sin(ω(t τ)) g(τ) dτ ω Hence the solution of the IVP is y(t) = ω sin(ωt) + ω t sin(ω(t τ)) g(τ) dτ 5 The transform of the ODE (given the specified initial conditions) is Solving for the transform of the solution, First write 4s 2 Y (s) + 4s Y (s) + 7 Y (s) = G(s) Y (s) = G(s) 4s 2 + 4s + 7 4s 2 + 4s + 7 = 4 (s + 2 )2 + 4 Based on the elementary properties of the Laplace transform, [ ] L 4s 2 = + 4s e t/2 sin 2t Applying the convolution theorem, the solution of the IVP is y(t) = 8 t e (t τ)/2 sin 2(t τ) g(τ) dτ 7 Taking the initial conditions into consideration, the transform of the ODE is s 2 Y (s) 2s [s Y (s) 2] + 4 Y (s) = G(s) Solving for the transform of the solution, We can write Y (s) = 2s + 5 (s + 2) 2 + G(s) (s + 2) 2 2s + 5 (s + 2) 2 = 2 s (s + 2) 2

246 246 Chapter 6 The Laplace Transform It follows that L [ 2 s + 2 ] = 2e 2t and L [ ] (s + 2) 2 Based on the convolution theorem, the solution of the IVP is y(t) = 2e 2t + t e 2t + t = t e 2t (t τ)e 2(t τ) g(τ) dτ 9 The transform of the ODE (given the specified initial conditions) is Solving for the transform of the solution, First write It follows that s 4 Y (s) Y (s) = G(s) Y (s) = s 4 = [ 2 G(s) s 4 s 2 s 2 + ] [ ] L s 4 = [sinh t sin t] 2 Based on the convolution theorem, the solution of the IVP is y(t) = 2 t [sinh(t τ) sin(t τ)] g(τ) dτ 2 Taking the initial conditions into consideration, the transform of the ODE is s 4 Y (s) s 3 + 5s 2 Y (s) 5s + 4 Y (s) = G(s) Solving for the transform of the solution, Y (s) = Using partial fractions, we find that and It follows that and s 3 + 5s (s 2 + )(s 2 + 4) + G(s) (s 2 + )(s 2 + 4) s 3 + 5s (s 2 + )(s 2 + 4) = 3 (s 2 + )(s 2 + 4) = [ 3 [ 4s s 2 + s ] s 2, + 4 s 2 + s ] [ L s(s 2 ] + 5) (s 2 + )(s 2 = 4 + 4) 3 cos t cos 2t, 3 [ ] L (s 2 + )(s 2 = + 4) 3 sin t sin 2t 6

247 Based on the convolution theorem, the solution of the IVP is y(t) = 4 3 cos t 3 cos 2t + 6 t [2 sin(t τ) sin 2(t τ)] g(τ) dτ 22(a) Taking the Laplace transform of the integral equation, with Φ(s) = L [φ(t)], Φ(s) + s 2 Φ(s) = 2 s Note that the convolution theorem was applied Solving for the transform Φ(s), Φ(s) = Using partial fractions, we can write 2s 2 (s 2 + )(s 2 + 4) = 2 3 2s 2 (s 2 + )(s 2 + 4) Therefore the solution of the integral equation is [ 4 s ] s 2 + φ(t) = 4 3 sin 2t 2 3 sin t (b) Differentiate both sides of the equation, we get φ (t) + (t t)φ(t) + t φ(ξ)dξ = 2 cos 2t Clearly, t t =, so differentiating this equation again we obtain φ (t) + φ(t) = 4 sin 2t Plugging t = into the original equation gives us φ() = Also, t = in the first equation here in part (b) gives φ () = 2 (c) Taking the Laplace transform of the ODE, with Φ(s) = L [φ(t)], Solving for the transform of the solution, s 2 Φ(s) 2 + Φ(s) = 8 s Φ(s) = 2s 2 (s 2 + )(s 2 + 4) This is identical to the Laplace transform we obtained in part (a), so the solution will be the same 23(a) Taking the Laplace transform of both sides of the integral equation (using the Convolution Theorem) Φ(s) + s 2 Φ(s) = s It follows that Φ(s) = s s 2 +

248 248 Chapter 6 The Laplace Transform and [ ] φ(t) = L s s 2 = cos t + (b) Differentiating both sides of the equation twice, we get φ (t) + t φ(ξ) dξ =, and then φ (t) + φ(t) = Plugging t = into the original equation and the first equation above gives φ() = and φ () = (c) The function φ(t) = cos t clearly solves the initial value problem in part (b) 25(a) The Laplace transform of both sides of the integral equation (using the Convolution Theorem) is Solving for Φ(s): Rewriting, Φ(s) + 2 s s 2 + Φ(s) = s + Φ(s) = s2 + (s + ) 3 Φ(s) = (s + )2 2(s + ) + 2 (s + ) 3 = s + 2 (s + ) (s + ) 3 The solution of the integral equation is [ ] φ(t) = L s + 2 (s + ) (s + ) 3 (b) Differentiating both sides of the equation twice, we get and then φ (t) + 2φ(t) 2 φ (t) + 2φ (t) 2 t t = e t 2 t e t + t 2 e t sin(t ξ)φ(ξ) dξ = e t, cos(t ξ)φ(ξ) dξ = e t Using the original equation, we can convert the second equation to φ (t) + 2φ (t) + φ(t) = 2e t Plugging t = into the original equation and the first equation above gives φ() = and φ () = 3 (c) It is easily confirmed that the function φ(t) = e t 2te t + t 2 e t solves the initial value problem in part (b)

249 (a) Taking the Laplace transform of both sides of the integro-differential equation Solving for Φ(s): s Φ(s) s 3 Φ(s) = s 2 Φ(s) = Taking the inverse Laplace transform, s s 2 + φ(t) = cos t (b) Differentiating both sides of the equation three times, we get then φ (t) t φ (t) (t ξ)φ(ξ) dξ =, t φ(ξ) dξ =, and then φ (4) (t) φ(t) = Plugging t = into the original equation and the first two equations above gives φ () =, φ () = and φ () = (c) It is easily confirmed that the function φ(t) = cos t solves the initial value problem in part (b) 28(a) The Laplace transform of both sides of the integro-differential equation is Solving for Φ(s): Note further that Φ(s) = s Φ(s) + Φ(s) = s 2 + Φ(s) s 2 + s(s 2 + s + ) = s (s 2 + s + ) (s 2 + s + ) = (s + /2) 2 + 3/4 = 2 3 /2 3 (s + /2) 2 + 3/4 Taking the inverse Laplace transform, φ(t) = 2 3 e t/2 sin( 3 2 t) (b) Differentiating both sides of the equation twice, we get and then φ (t) + φ (t) t φ (t) + φ (t) φ(t) + cos(t ξ)φ(ξ) dξ =, t sin(t ξ)φ(ξ) dξ =

250 25 Chapter 6 The Laplace Transform Using the original equation, we can convert the second equation to φ (t) + φ (t) + φ (t) = Plugging t = into the original equation and the first equation above gives φ () = and φ () = (c) It is easily confirmed that the function φ(t) = (2/ 3)e t/2 sin( 3 t/2) solves the initial value problem in part (b) 29(a) First note that b ( f(y) dy = b y y ) f (b) Take the Laplace transformation of both sides of the equation Using the convolution theorem, with F (s) = L [f(y)], T s = [ ] F (s) L 2g y It was shown in Problem 3(c), Section 6, that [ ] π L = y s Hence and T s = π F (s) 2g s, F (s) = Taking the inverse transform, we obtain (b) Combining equations (i) and (iv), Solving for the derivative dx/dy, in which α = gt 2 /π 2 f(y) = T π 2g T 2 π 2 y = + dx dy = 2g π T s 2g y ( ) 2 dx dy 2α y (c) Consider the change of variable y = 2α sin 2 (θ/2) Using the chain rule, dy dθ = 2α sin(θ/2) cos(θ/2) dx dx y,

251 66 25 and dx dy = 2α sin(θ/2) cos(θ/2) dx dθ It follows that dx cos = 2α sin(θ/2) cos(θ/2) 2 (θ/2) dθ sin 2 (θ/2) = 2α cos2 (θ/2) = α + α cos θ Direct integration results in x(θ) = α θ + α sin θ + C Since the curve passes through the origin, we require y() = x() = C =, and x(θ) = α θ + α sin θ We also have y(θ) = 2α sin 2 (θ/2) = α α cos θ Hence

252 252 Chapter 6 The Laplace Transform

253 C H A P T E R 7 Systems of First Order Linear Equations 7 Introduce the variables x = u and x 2 = u It follows that x = x 2 and x 2 = u = 2u 5 u In terms of the new variables, we obtain the system of two first order ODEs x = x 2 x 2 = 2x 5 x 2 3 First divide both sides of the equation by t 2, and write u = t u ( 4t 2 )u Set x = u and x 2 = u It follows that x = x 2 We obtain the system of equations and x 2 = u = t u ( 4t 2 )u x = x 2 x 2 = ( 4t 2 )x t x 2 253

254 254 Chapter 7 Systems of First Order Linear Equations 5 Let x = u and x 2 = u ; then u = x 2 In terms of the new variables, we have x x x = 2 cos 3t with the initial conditions x () = and x 2 () = 2 The equivalent first order system is x = x 2 x 2 = 4 x 25 x cos 3t with the above initial conditions 7(a) Solving the first equation for x 2, we have x 2 = x + 2x Substitution into the second equation results in (x + 2x ) = x 2(x + 2x ) That is, x + 4 x + 3 x = The resulting equation is a second order differential equation with constant coefficients The general solution is x (t) = c e t + c 2 e 3t With x 2 given in terms of x, it follows that x 2 (t) = c e t c 2 e 3t (b) Imposing the specified initial conditions, we obtain c + c 2 = 2, c c 2 = 3, with solution c = 5/2 and c 2 = /2 Hence x (t) = 5 2 e t 2 e 3t and x 2 (t) = 5 2 e t + 2 e 3t (c) (a) Solving the first equation for x 2, we obtain x 2 = (x x )/2 Substitution into the second equation results in (x x ) /2 = 3 x 2(x x ) Rearranging the terms, the single differential equation for x is x + 3 x + 2 x = (b) The general solution is x (t) = c e t + c 2 e 2t With x 2 given in terms of x, it follows that x 2 (t) = c e t + 3c 2 e 2t /2 Invoking the specified initial conditions, c = 7 and c 2 = 6 Hence x (t) = 7e t + 6 e 2t and x 2 (t) = 7e t + 9 e 2t

255 7 255 (c) (a) Solving the first equation for x 2, we have x 2 = x /2 Substitution into the second equation results in x /2 = 2 x The resulting equation is x + 4 x = (b) The general solution is x (t) = c cos 2t + c 2 sin 2t With x 2 given in terms of x, it follows that x 2 (t) = c sin 2t + c 2 cos 2t Imposing the specified initial conditions, we obtain c = 3 and c 2 = 4 Hence x (t) = 3 cos 2t + 4 sin 2t and x 2 (t) = 3 sin 2t + 4 cos 2t (c) 3 Solving the first equation for V, we obtain V = L I Substitution into the second equation results in L I = I C L RC I Rearranging the terms, the single differential equation for I is LRC I + L I + R I = 5 Let x = c x (t) + c 2 x 2 (t) and y = c y (t) + c 2 y 2 (t) Then x = c x (t) + c 2 x 2(t) y = c y (t) + c 2 y 2(t)

256 256 Chapter 7 Systems of First Order Linear Equations Since x (t), y (t) and x 2 (t), y 2 (t) are solutions for the original system, Rearranging terms gives x = c (p x (t) + p 2 y (t)) + c 2 (p x 2 (t) + p 2 y 2 (t)) y = c (p 2 x (t) + p 22 y (t)) + c 2 (p 2 x 2 (t) + p 22 y 2 (t)) x = p (c x (t) + c 2 x 2 (t)) + p 2 (c y (t) + c 2 y 2 (t)) y = p 2 (c x (t) + c 2 x 2 (t)) + p 22 (c y (t) + c 2 y 2 (t)), and so x and y solve the original system 6 Based on the hypothesis, Subtracting the two equations, Similarly, x (t) = p (t)x (t) + p 2 (t)y (t) + g (t) x 2(t) = p (t)x 2 (t) + p 2 (t)y 2 (t) + g (t) x (t) x 2(t) = p (t) [x (t) x 2(t)] + p 2 (t) [y (t) y 2(t)] y (t) y 2(t) = p 2 (t) [x (t) x 2(t)] + p 22 (t) [y (t) y 2(t)] Hence the difference of the two solutions satisfies the homogeneous ODE 7 For rectilinear motion in one dimension, Newton s second law can be stated as F = m x The resisting force exerted by a linear spring is given by F s = k δ, in which δ is the displacement of the end of a spring from its equilibrium configuration Hence, with < x < x 2, the first two springs are in tension, and the last spring is in compression The sum of the spring forces on m is F s = k x k 2 (x 2 x ) The total force on m is F = k x + k 2 (x 2 x ) + F (t) Similarly, the total force on m 2 is F 2 = k 2 (x 2 x ) k 3 x 2 + F 2 (t) 8 One of the ways to transform the system is to assign the variables Before proceeding, note that y = x, y 2 = x 2, y 3 = x, y 4 = x 2 x = m [ (k + k 2 )x + k 2 x 2 + F (t)] x 2 = m 2 [k 2 x (k 2 + k 3 )x 2 + F 2 (t)]

257 7 257 Differentiating the new variables, we obtain the system of four first order equations y = y 3 y 2 = y 4 y 3 = m ( (k + k 2 )y + k 2 y 2 + F (t)) y 4 = m 2 (k 2 y (k 2 + k 3 )y 2 + F 2 (t)) 9(a) Taking a clockwise loop around each of the paths, it is easy to see that voltage drops are given by V V 2 =, and V 2 V 3 = (b) Consider the right node The current in is given by I + I 2 The current leaving the node is I 3 Hence the current passing through the node is (I + I 2 ) ( I 3 ) Based on Kirchhoff s first law, I + I 2 + I 3 = (c) In the capacitor, In the resistor, In the inductor, C V = I V 2 = R I 2 L I 3 = V 3 (d) Based on part (a), V 3 = V 2 = V Based on part (b), C V + R V 2 + I 3 = It follows that C V = R V I 3 and L I 3 = V 2 Let I, I 2, I 3,and I 4 be the current through the resistors, inductor, and capacitor, respectively Assign V, V 2, V 3,and V 4 as the respective voltage drops Based on Kirchhoff s second law, the net voltage drops, around each loop, satisfy V + V 3 + V 4 =, V + V 3 + V 2 = and V 4 V 2 = Applying Kirchhoff s first law to the upper-right node, Likewise, in the remaining nodes, That is, I 3 (I 2 + I 4 ) = I I 3 = and I 2 + I 4 I = V 4 V 2 =, V + V 3 + V 4 = and I 2 + I 4 I 3 = Using the current-voltage relations, V = R I, V 2 = R 2 I 2, L I 3 = V 3, C V 4 = I 4

258 258 Chapter 7 Systems of First Order Linear Equations Combining these equations, R I 3 + L I 3 + V 4 = and C V 4 = I 3 V 4 R 2 Now set I 3 = I and V 4 = V, to obtain the system of equations L I = R I V and C V = I V R 2 23(a) Let Q (t) and Q 2 (t) be the amount of salt in the respective tanks at time t Note that the volume of each tank remains constant Based on conservation of mass, the rate of increase of salt, in any given tank, is given by The rate of salt flowing into Tank is [ ] [ oz r in = q 3 gal ] [ Q2 + gal min rate of increase = rate in rate out oz gal The rate at which salt flows out of Tank is [ ] [ Q oz r out = 4 gal ] 6 gal min Hence Similarly, for Tank 2, dq dt dq 2 dt ] [ gal ] = 3 q + Q 2 min = Q 5 = 3 q + Q 2 Q 5 = q 2 + Q 3 3Q 2 The process is modeled by the system of equations Q = Q 5 + Q q Q 2 = Q 3 3Q 2 + q 2 oz min The initial conditions are Q () = Q and Q 2 () = Q 2 oz min

259 7 259 (b) The equilibrium values are obtained by solving the system Q 5 + Q q = Q 3 3Q 2 + q 2 = Its solution leads to Q E = 54 q + 6 q 2 and Q E 2 = 6 q + 4 q 2 (c) The question refers to a possible solution of the system 54 q + 6 q 2 = 6 6 q + 4 q 2 = 5 It is possible to formally solve the system of equations, but the unique solution gives q = 7 oz and q 2 = oz 6 gal 2 gal, which is not physically possible (d) We can write q 2 = 9 q + QE 6 q 2 = 3 2 q + QE 2 4, which are the equations of two lines in the q -q 2 -plane: The intercepts of the first line are Q E /54 and Q E /6 The intercepts of the second line are Q E 2 /6 and Q E 2 /4 Therefore the system will have a unique solution, in the first quadrant, as long as Q E /54 Q E 2 /6 or Q E 2 /4 Q E /6 That is, 9 QE 2 Q E 2 3

260 26 Chapter 7 Systems of First Order Linear Equations 72 2(a) (b) (c) (d) 3(c,d) 4(b) A 2 B = 3 A + B = ( ) + i 2i + 2i 6 = 3 + 2i 4 2 i + 4i ( i 7 + 2i + 2i 2 + 3i ( ) ( ) 3 + 3i + i 3 + 6i i 6i = 9 + 6i i 2i + 6i 6 5i ( ) ( + i)i + 2( + 2i) 3( + i) + ( + 2i)( 2i) AB = (3 + 2i)i + 2(2 i) 3(3 + 2i) + (2 i)( 2i) ( ) 3 + 5i 7 + 5i = 2 + i 7 + 2i ( ) ( + i)i + 3(3 + 2i) ( + 2i)i + 3(2 i) BA = 2( + i) + ( 2i)(3 + 2i) 2( + 2i) + ( 2i)(2 i) ( ) 8 + 7i 4 4i = 6 4i A T + B T = = 3 = (A + B) T 5 4 A = ( ) 3 + 2i i 2 + i 2 3i ) (c) By definition, 5 A = A T = (A) T = ( ) 3 + 2i 2 + i i 2 3i 2(A + B) = = Let A= (a ij ) and B= (b ij ) The given operations in (a)-(d) are performed elementwise That is, (a) a ij + b ij = b ij + a ij

261 72 26 (b) a ij + (b ij + c ij ) = (a ij + b ij ) + c ij (c) α(a ij + b ij ) = α a ij + α b ij (d) (α + β) a ij = α a ij + β a ij In the following, let A= (a ij ), B= (b ij ) and C= (c ij ) (e) Calculating the generic element, n (BC) ij = b ik c kj k= Therefore n n n n n n [A(BC)] ij = a ir ( b rk c kj ) = a ir b rk c kj = ( a ir b rk ) c kj r= k= r= k= k= r= The inner summation is recognized as n a ir b rk = (AB) ik, r= which is the ik-th element of the matrix AB Thus [A(BC)] ij = [(AB)C] ij (f) Likewise, n n n [A(B + C)] ij = a ik ( b kj + c kj ) = a ik b kj + a ik c kj = (AB) ij + (AC) ij k= k= k= 8(a) x T y= 2( + i) + 2(3i) + ( i)(3 i) = 4i (b) y T y= ( + i) (3 i) 2 = 2 8i (c) (x, y) = 2( i) + 2(3i) + ( i)(3 + i) = 2 + 2i (d) (y, y) = ( + i)( i) (3 i)(3 + i) = 6 9 Indeed, and n 5 + 3i = x T y = x j y j = y T x, j= n n n 3 5i = (x, y) = x j y j = y j x j = y j x j = (y, x) j= j= j=

262 262 Chapter 7 Systems of First Order Linear Equations First augment the given matrix by the identity matrix: ( ) 3 [A I ] = 6 2 Divide the first row by 3, to obtain ( ) /3 /3 6 2 Adding 6 times the first row to the second row results in ( ) /3 /3 4 2 Divide the second row by 4, to obtain ( ) /3 /3 /2 /4 Finally, adding /3 times the second row to the first row results in ( ) /6 /2 /2 /4 Hence ( ) 3 = ( ) The augmented matrix is 2 2 Combining the elements of the first row with the elements of the second and third rows results in Divide the elements of the second row by 3, and the elements of the third row by 3 Now subtracting the new second row from the first row yields /3 /3 2/3 /3 /3 /3 Finally, combine the third row with the second row to obtain /3 /3 /3 /3 /3 /3 /3

263 Hence 2 2 = 3 5 Elementary row operations yield 2 /2 /2 2 /2 /2 2 /2 /4 /2 /4 /4 /2 /4 /2 /4 /2 /4 /2 /2 Finally, combining the first and third rows results in /2 /4 /8 /2 /4 /8 /2 /4, so A = /2 /4 /2 /2 6 Elementary row operations yield /3 /3 /3 / 3/ / 2/3 2/3 /3 /5 2/5 /5 /3 7/3 /3 /3 7/3 /3 Finally, normalizing the last row results in / 3/ / / 3/ / /5 2/5 /5, so A = /5 2/5 /5 7/ / 3/ 7/ / 3/ 7 Elementary row operations on the augmented matrix yield the row-reduced form of the augmented matrix /7 /7 2/7 3/7 4/7 /7 2 The left submatrix cannot be converted to the identity matrix Hence the given matrix is singular

264 264 Chapter 7 Systems of First Order Linear Equations 8 Elementary row operations on the augmented matrix yield, so A = 9 Elementary row operations on the augmented matrix yield Normalizing the last row and combining it with the others results in /5 8/5 2/ /5 6/5 4/5 /5 /5 /5, 2 4/5 4/5 /5 2 4/5 4/5 /5 so 6 3/5 8/5 2/5 A = 5 /5 6/5 4/5 /5 /5 /5 2 4/5 4/5 /5 2 Suppose that there exist matrices B and C, such that AB = I and CA = I Then CAB = IB = B, also, CAB = CI = C This shows that B = C 23 First note that x = ( ( ( ) e ) t + 2 (e ) t + t e t 3e ) = t + 2t e t 2e t + 2t e t

265 We also have (2 ) ( ( ) ( ( ) 2 x = e ) t (t e 3 2) t ) 2 ( ) ( ) ( ) 2 = e t 2 + (t e t 2e ) = t + 2t e t 3 2 3e t + 2t e t It follows that ( ) ( ( ) 2 x + e 3 2 ) t 3e = t + 2t e t 2e t + 2t e t 24 It is easy to see that 6 6e t x = 8 e t + 4 e 2t = 8e t + 4e 2t 4 4 4e t 4e 2t On the other hand, 2 x = e t e 2t = 8 e t + 4 e 2t Differentiation, elementwise, results in Ψ = et 2e 2t 3e 3t 4e t 2e 2t 6e 3t e t 2e 2t 3e 3t On the other hand, 4 4 e t e 2t e 3t 3 2 Ψ = 3 2 4e t e 2t 2e 3t 2 2 e t e 2t e 3t = et 2e 2t 3e 3t 4e t 2e 2t 6e 3t e t 2e 2t 3e 3t 73 4 The augmented matrix is 2 2 2

266 266 Chapter 7 Systems of First Order Linear Equations Adding 2 times the first row to the second row and subtracting the first row from the third row results in Adding the negative of the second row to the third row results in We evidently end up with an equivalent system of equations x + 2x 2 x 3 = x 2 + x 3 = Since there is no unique solution, let x 3 = α, where α is arbitrary It follows that x 2 = α, and x = α Hence all solutions have the form x = α 5 The augmented matrix is 3 2 Adding 3 times the first row to the second row and adding the first row to the last row yields 3 Now add the negative of the second row to the third row to obtain 3 2 We end up with an equivalent linear system x x 3 = x x 3 = x 3 = Hence the unique solution of the given system of equations is x = x 2 = x 3 = 6 The augmented matrix is

267 Adding 2 times the first row to the second row and subtracting 2 times the first row from the third row results in 2 2 We evidently end up with an equivalent system of equations x + 2x 2 x 3 = 2 Since there is no unique solution, let x 2 = α, and x 3 = β, where α, β are arbitrary It follows that x = 2 2α + β Hence all solutions have the form 2 2α + β x = α β 8 Write the given vectors as columns of the matrix X = 2 2 It is evident that det(x) = Hence the vectors are linearly dependent In order to find a linear relationship between them, write c x () + c 2 x (2) + c 3 x (3) = The latter equation is equivalent to 2 2 c c 2 = Performing elementary row operations, 2 2 /2 5/2 We obtain the system of equations c 3 c c 3 /2 = c 2 + 5c 3 /2 = Setting c 3 = 2, it follows that c = and c 3 = 5 Hence x () 5x (2) + 2x (3) = The matrix containing the given vectors as columns is 2 3 X = We find that det(x) = 7 Hence the given vectors are linearly independent

268 268 Chapter 7 Systems of First Order Linear Equations Write the given vectors as columns of the matrix X = The four vectors are necessarily linearly dependent Hence there are nonzero scalars such that c x () + c 2 x (2) + c 3 x (3) + c 4 x (4) = The latter equation is equivalent to c 2 3 c 2 c = Performing elementary row operations, We end up with an equivalent linear system c + c 4 = c 2 + c 4 = c 4 c 3 = Let c 4 = Then c = and c 2 = Therefore we find that x () + x (2) x (4) = 2 The matrix containing the given vectors as columns, X, is of size n m Since n < m, we can augment the matrix with m n rows of zeros The resulting matrix, X, is of size m m Since X is a square matrix, with at least one row of zeros, it follows that det( X) = Hence the column vectors of X are linearly dependent That is, there is a nonzero vector, c, such that Xc= m If we write only the first n rows of the latter equation, we have Xc= n Therefore the column vectors of X are linearly dependent 3 By inspection, we find that ( ) x () (t) 2x (2) e t (t) = Hence 3 x () (t) 6 x (2) (t)+x (3) (t) =, and the vectors are linearly dependent 7 The eigenvalues λ and eigenvectors x satisfy the equation ( ) ( ) ( ) 3 λ 2 x = 4 λ For a nonzero solution, we must have (3 λ)( λ) + 8 =, that is, x 2 λ 2 2λ + 5 =

269 The eigenvalues are λ = 2 i and λ 2 = + 2 i The components of the eigenvector x () are solutions of the system ( ) ( ) ( ) i 2 x = i The two equations reduce to ( + i)x = x 2 Hence x () = (, + i) T Now setting λ = λ 2 = + 2 i, we have ( ) ( ) ( ) 2 2 i 2 x =, i with solution given by x (2) = (, i) T 8 The eigenvalues λ and eigenvectors x satisfy the equation ( ) ( ) ( ) 2 λ x = 2 λ For a nonzero solution, we must have ( 2 λ)( 2 λ) =, that is, x 2 x 2 x 2 λ 2 + 4λ + 3 = The eigenvalues are λ = 3 and λ 2 = For λ = 3, the system of equations becomes ( ) ( ) ( ) x =, x 2 which reduces to x + x 2 = A solution vector is given by x () = (, ) T Substituting λ = λ 2 =, we have ( ) ( ) ( ) x = The equations reduce to x = x 2 Hence a solution vector is given by x (2) = (, ) T 2 The eigensystem is obtained from analysis of the equation ( ) ( ) ( ) λ 3 x = 3 λ x 2 For a nonzero solution, the determinant of the coefficient matrix must be zero That is, λ 2 4 = Hence the eigenvalues are λ = 2 and λ 2 = 2 Substituting the first eigenvalue, λ = 2, yields ( ) ( ) ( ) 3 3 x = 3 x 2 x 2 The system is equivalent to the equation 3 x + x 2 = given by x () = (, 3 ) T Substitution of λ = 2 results in ( ) ( ) ( ) 3 x =, 3 3 x 2 A solution vector is

270 27 Chapter 7 Systems of First Order Linear Equations which reduces to x = 3 x 2 A corresponding solution vector is x (2) = ( 3, ) T 2 The eigenvalues λ and eigenvectors x satisfy the equation ( ) ( ) ( ) 3 λ 3/4 x = 5 λ For a nonzero solution, we must have ( 3 λ)( λ) + 5/4 =, that is, x 2 λ 2 + 2λ + 3/4 = Hence the eigenvalues are λ = 3/2 and λ 2 = /2 In order to determine the eigenvector corresponding to λ, set λ = 3/2 The system of equations becomes ( ) ( ) ( ) 3/2 3/4 x =, 5 5/2 which reduces to 2 x + x 2 = A solution vector is given by x () = (, 2) T Substitution of λ = λ 2 = /2 results in ( ) ( ) ( ) 5/2 3/4 x =, 5 3/2 which reduces to x = 3 x 2 A corresponding solution vector is x (2) = (3, ) T 23 The eigensystem is obtained from analysis of the equation 3 λ 2 2 x 4 λ x 2 = 2 4 λ The characteristic equation of the coefficient matrix is λ 3 6λ 2 + λ 6 =, with roots λ =, λ 2 = 2 and λ 3 = 3 Setting λ = λ =, we have x 3 x 2 = This system is reduces to the equations x 2 x 2 x 3 x + x 3 = x 2 = A corresponding solution vector is given by x () = (,, ) T Setting λ = λ 2 = 2, the reduced system of equations is x + 2 x 2 = x 3 = x 3 A corresponding solution vector is given by x (2) = ( 2,, ) T λ = λ 3 = 3, the reduced system of equations is x = x 2 + x 3 = Finally, setting A corresponding solution vector is given by x (3) = (,, ) T

271 For computational purposes, note that if λ is an eigenvalue of B, then c λ is an eigenvalue of the matrix A= c B Eigenvectors are unaffected, since they are only determined up to a scalar multiple So with 2 8 B = 2 2, 8 5 the associated characteristic equation is µ 3 8µ 2 8µ =, with roots µ = 9, µ 2 = 9 and µ 3 = 8 Hence the eigenvalues of the given matrix, A, are λ =, λ 2 = and λ 3 = 2 Setting λ = λ =, (which corresponds to using µ = 9 in the modified problem) the reduced system of equations is 2 x + x 3 = x 2 + x 3 = A corresponding solution vector is given by x () = (, 2, 2) T Setting λ = λ 2 =, the reduced system of equations is x + 2 x 3 = x 2 2 x 3 = A corresponding solution vector is given by x (2) = (2, 2, ) T Finally, setting λ = λ 2 =, the reduced system of equations is x x 3 = 2 x 2 x 3 = A corresponding solution vector is given by x (3) = (2,, 2) T 26(b) By definition, (Ax, y) = n (Ax) i y i = i = n i = j = n a ij x j y i Let b ij = a ji, so that a ij = b ji Now interchanging the order or summation, Now note that Therefore (Ax, y) = n n x j j = i = n b ji y i = i = (Ax, y) = a ij y i = n n x j j = i = n b ji y i = (A y) j i = n x j (A y) j = (x, A y) j = (c) By definition of a Hermitian matrix, A=A b ji y i

272 272 Chapter 7 Systems of First Order Linear Equations 27 Suppose that Ax=, but that x Let A= (a ij ) Using elementary row operations, it is possible to transform the matrix into one that is not upper triangular If it were upper triangular, backsubstitution would imply that x= Hence a linear combination of all the rows results in a row containing only zeros That is, there are n scalars, β i, one for each row and not all zero, such that for each for column j, n β i a ij = i = Now consider A = (b ij ) By definition, b ij = a ji, or a ij = b ji It follows that for each j, n n n β i b ji = b jk β k = b jk β k = i = k = Let y= (β, β 2,, β n ) T Hence we have a nonzero vector, y, such that A y= 29 By linearity, k = A(x () + α ξ) = Ax () + α Aξ = b + = b 3 Let c ij = a ji By the hypothesis, there is a nonzero vector, y, such that n c ij y j = j = n a ji y j =, i =, 2,, n j = Taking the conjugate of both sides, and interchanging the indices, we have n a ij y i = i = This implies that a linear combination of each row of A is equal to zero consider the augmented matrix [A B] Replace the last row by [ ] n n y i [a i, a i2,, a in, b i ] =,,,, y i b i i = i = Now We find that if (B, y) =, then the last row of the augmented matrix contains only zeros Hence there are n remaining equations We can now set x n = α, some parameter, and solve for the other variables in terms of α Therefore the system of equations Ax=b has a solution 3 If λ = is an eigenvalue of A, then there is a nonzero vector, x, such that Ax = λ x = That is, Ax= has a nonzero solution This implies that the mapping defined by A is not -to-, and hence not invertible On the other hand, if A is singular, then det(a) = Thus, Ax= has a nonzero solution The latter equation can be written as Ax= x

273 (a) Based on Problem 26, (Ax, x) = (x, Ax) (b) Let x be an eigenvector corresponding to an eigenvalue λ It then follows that (Ax, x) = (λx, x) and (x, Ax) = (x, λx) Based on the properties of the inner product, (λx, x) = λ(x, x) and (x, λx) = λ(x, x) Then from part (a), λ(x, x) = λ(x, x) (c) From part (b), (λ λ)(x, x) = Based on the definition of an eigenvector, (x, x) = x 2 > Hence we must have λ λ =, which implies that λ is real 33 From Problem 26(c), (Ax (), x (2) ) = (x (), Ax (2) ) Hence λ (x (), x (2) ) = λ 2 (x (), x (2) ) = λ 2 (x (), x (2) ), since the eigenvalues are real Therefore (λ λ 2 )(x (), x (2) ) = Given that λ λ 2, we must have (x (), x (2) ) = 74 3 Equation (4) states that the Wronskian satisfies the first order linear ODE dw dt = (p + p p nn )W The general solution of this is given by Equation (5): W (t) = C e (p +p 22+ +p nn)dt, in which C is an arbitrary constant Let X and X 2 be matrices representing two sets of fundamental solutions It follows that det(x ) = W (t) = C e (p +p 22+ +p nn)dt det(x 2 ) = W 2 (t) = C 2 e (p +p 22+ +p nn)dt Hence det(x )/det(x 2 ) = C /C 2 Note that C 2 4 First note that p + p 22 = p(t) As shown in Problem 3, [ W x (), x (2)] = c e p(t)dt

274 274 Chapter 7 Systems of First Order Linear Equations For second order linear ODE, the Wronskian (as defined in Chapter 3) satisfies the first order differential equation W + p(t)w = It follows that W Alternatively, based on the hypothesis, [ y (), y (2)] = c e p(t)dt y () = α x + α 2 x 2 y (2) = α 2 x + α 22 x 2 Direct calculation shows that [ W y (), y (2)] = α x + α 2 x 2 α 2 x + α 22 x 2 α x + α 2 x 2 α 2 x + α 22 x 2 = (α α 22 α 2 α 2 )x x 2 (α α 22 α 2 α 2 )x 2 x = (α α 22 α 2 α 2 )x x 22 (α α 22 α 2 α 2 )x 2 x 2 Here we used the fact that x = x 2 Hence W [ y (), y (2)] [ = (α α 22 α 2 α 2 )W x (), x (2)] 5 The particular solution satisfies the ODE (x (p) ) =P(t)x (p) +g(t) Now let x be any solution of the homogeneous equation, x =P(t)x We know that x=x (c), in which x (c) is a linear combination of some fundamental solution By linearity of the differential equation, it follows that x =x (p) +x (c) is a solution of the ODE Based on the uniqueness theorem, all solutions must have this form 7(a) By definition, W [ x (), x (2)] = t2 2t e t e t = (t2 2t)e t (b) The Wronskian vanishes at t = and t = 2 Hence the vectors are linearly independent on D = (, ) (, 2) (2, ) (c) It follows from Theorem 743 that one or more of the coefficients of the ODE must be discontinuous at t = and t = 2 If not, the Wronskian would not vanish (d) Let Then ( ) ( ) t 2 e t x = c + c 2 2t e t ( ) ( ) 2t e x t = c + c 2 2 e t

275 On the other hand, ( ) ( ) ( ) ( ) ( ) p p 2 p p x = c 2 t 2 p p p 2 p + c 2 e t 22 p 2 p t p 2 p 22 e t ( [ c p t 2 + 2p 2 t ] + c 2 [p + p 2 ] e t ) = c [p 2 t 2 + 2p 22 t] + c 2 [p 2 + p 22 ] e t Comparing coefficients, we find that p t 2 + 2p 2 t = 2t p + p 2 = p 2 t 2 + 2p 22 t = 2 p 2 + p 22 = Solution of this system of equations results in p (t) =, p 2 (t) =, p 2 (t) = 2 2t t 2 2t, p 22(t) = t2 2 t 2 2t Hence the vectors are solutions of the ODE x = t 2 2t ( t 2 2t 2 2t t 2 2 ) x 8 Suppose that the solutions x (), x (2),, x (m) are linearly dependent at t = t Then there are constants c, c 2,, c m (not all zero) such that c x () (t ) + c 2 x (2) (t ) + + c m x (m) (t ) = Now let z(t) = c x () (t) + c 2 x (2) (t) + + c m x (m) (t) Then clearly, z(t) is a solution of x =P(t)x, with z(t ) = Furthermore, y(t) is also a solution, with y(t ) = By the uniqueness theorem, z(t) =y(t) = Hence c x () (t) + c 2 x (2) (t) + + c m x (m) (t) = on the entire interval α < t < β Going in the other direction is trivial 9(a) Let y(t) be any solution of x =P(t)x It follows that z(t) + y(t) = c x () (t) + c 2 x (2) (t) + + c n x (n) (t) + y(t) is also a solution Now let t (α, β) Then the collection of vectors x () (t ), x (2) (t ),, x (n) (t ), y(t ) constitutes n + vectors, each with n components Based on the assertion in Problem 2, Section 73, these vectors are necessarily linearly dependent That is, there are n + constants b, b 2,, b n, b n+ (not all zero) such that b x () (t ) + b 2 x (2) (t ) + + b n x (n) (t ) + b n+ y(t ) = From Problem 8, we have b x () (t) + b 2 x (2) (t) + + b n x (n) (t) + b n+ y(t) =

276 276 Chapter 7 Systems of First Order Linear Equations for all t (α, β) Now b n+, otherwise that would contradict the fact that the first n vectors are linearly independent Hence and the assertion is true y(t) = b n+ (b x () (t) + b 2 x (2) (t) + + b n x (n) (t)), (b) Consider z(t) = c x () (t) + c 2 x (2) (t) + + c n x (n) (t), and suppose that we also have z(t) = k x () (t) + k 2 x (2) (t) + + k n x (n) (t) Based on the assumption, (k c )x () (t) + (k 2 c 2 )x (2) (t) + + (k n c n )x (n) (t) = The collection of vectors x () (t), x (2) (t),, x (n) (t) is linearly independent on α < t < β It follows that k i c i =, for i =, 2,, n 75 2(a) Setting x= ξ e rt, and substituting into the ODE, we obtain the algebraic equations ( ) ( ) ( ) r 2 ξ = 3 4 r For a nonzero solution, we must have det(a ri) = r 2 + 3r + 2 = The roots of the characteristic equation are r = and r 2 = 2 For r =, the two equations reduce to ξ = ξ 2 The corresponding eigenvector is ξ () = (, ) T Substitution of r = 2 results in the single equation 3ξ = 2ξ 2 A corresponding eigenvector is ξ (2) = (2, 3) T Since the eigenvalues are distinct, the general solution is ( ) ( ) 2 x = c e t + c 2 e 2t 3 ξ 2 (b)

277 (a) Setting x= ξ e rt results in the algebraic equations ( ) ( ) ( ) 2 r ξ = 3 2 r For a nonzero solution, we must have det(a ri) = r 2 = The roots of the characteristic equation are r = and r 2 = For r =, the system of equations reduces to ξ = ξ 2 The corresponding eigenvector is ξ () = (, ) T Substitution of r = results in the single equation 3ξ = ξ 2 A corresponding eigenvector is ξ (2) = (, 3) T Since the eigenvalues are distinct, the general solution is ( ) ( ) x = c e t + c 2 e t 3 ξ 2 (b) The system has an unstable eigendirection along ξ () = (, ) T Unless c =, all solutions will diverge 4(a) Solution of the ODE requires analysis of the algebraic equations ( ) ( ) ( ) r ξ = 4 2 r For a nonzero solution, we must have det(a ri) = r 2 + r 6 = The roots of the characteristic equation are r = 2 and r 2 = 3 For r = 2, the system of equations reduces to ξ = ξ 2 The corresponding eigenvector is ξ () = (, ) T Substitution of r = 3 results in the single equation 4ξ + ξ 2 = A corresponding eigenvector is ξ (2) = (, 4) T Since the eigenvalues are distinct, the general solution is ( ) ( ) x = c e 2t + c 2 e 3t 4 ξ 2

278 278 Chapter 7 Systems of First Order Linear Equations (b) The system has an unstable eigendirection along ξ () = (, ) T Unless c =, all solutions will diverge 8(a) Setting x= ξ e rt results in the algebraic equations ( ) ( ) 3 r 6 ξ = 2 r ξ 2 ( ) For a nonzero solution, we must have det(a ri) = r 2 r = The roots of the characteristic equation are r = and r 2 = With r =, the system of equations reduces to ξ + 3ξ 2 = The corresponding eigenvector is ξ () = (3, ) T For the case r =, the system is equivalent to the equation ξ + 2ξ 2 = An eigenvector is ξ (2) = (2, ) T Since the eigenvalues are distinct, the general solution is ( ) ( ) 3 2 x = c e t + c 2 (b) The entire line along the eigendirection ξ (2) = (2, ) T consists of equilibrium points All other solutions diverge The direction field changes across the line x + 2x 2 = Eliminating the exponential terms in the solution, the trajectories are given by x + 3 x 2 = c 2

279 The characteristic equation is given by 2 r 2 + i i r = r2 ( i)r i = The equation has complex roots r = and r 2 = i For r =, the components of the solution vector must satisfy ξ + (2 + i)ξ 2 = Thus the corresponding eigenvector is ξ () = (2 + i, ) T Substitution of r = i results in the single equation ξ + ξ 2 = A corresponding eigenvector is ξ (2) = (, ) T Since the eigenvalues are distinct, the general solution is x = c ( 2 + i ) ( ) e t + c 2 e it Setting x= ξ e rt results in the algebraic equations r 2 ξ 2 r ξ 2 = 2 r For a nonzero solution, we must have det(a ri) = r 3 4r 2 r + 4 = The roots of the characteristic equation are r = 4, r 2 = and r 3 = Setting r = 4, we have 3 2 ξ 2 ξ 2 = 2 3 This system is reduces to the equations ξ 3 ξ ξ 3 = ξ 2 ξ 3 = A corresponding solution vector is given by ξ () = (,, ) T Setting λ =, the reduced system of equations is ξ ξ 3 = ξ ξ 3 = ξ 3 A corresponding solution vector is given by ξ (2) = (, 2, ) T λ =, the reduced system of equations is Finally, setting ξ + ξ 3 = ξ 2 = A corresponding solution vector is given by ξ (3) = (,, ) T Since the eigenvalues are distinct, the general solution is x = c e 4t + c 2 2 e t + c 3 e t

280 28 Chapter 7 Systems of First Order Linear Equations 2 The eigensystem is obtained from analysis of the equation 3 r r 2 ξ ξ 2 = r The characteristic equation of the coefficient matrix is r 3 6r 2 5r 8 =, with roots r = 8, r 2 = and r 3 = Setting r = r = 8, we have ξ ξ 2 = This system is reduced to the equations ξ 3 ξ ξ 3 = ξ 3 2 ξ 2 ξ 3 = A corresponding solution vector is given by ξ () = (2,, 2) T Setting r =, the system of equations is reduced to the single equation 2 ξ + ξ ξ 3 = Two independent solutions are obtained as ξ (2) = (, 2, ) T and ξ (3) = (, 2, ) T Hence the general solution is x = c 2 e 8t + c 2 2 e t + c 3 2 e t 2 3 Setting x= ξ e rt results in the algebraic equations r ξ 2 r ξ 2 = r For a nonzero solution, we must have det(a ri) = r 3 + r 2 4r 4 = The roots of the characteristic equation are r = 2, r 2 = 2 and r 3 = Setting r = 2, we have 2 ξ ξ 2 = This system is reduces to the equations ξ 3 ξ = ξ 2 + ξ 3 = A corresponding solution vector is given by ξ () = (,, ) T Setting λ =, the reduced system of equations is 2 ξ + 3 ξ 3 = ξ 2 2 ξ 3 = ξ 3

281 75 28 A corresponding solution vector is given by ξ (2) = (3, 4, 2) T Finally, setting λ = 2, the reduced system of equations is 7 ξ + 4 ξ 3 = 7ξ 2 5 ξ 3 = A corresponding solution vector is given by ξ (3) = (4, 5, 7) T Since the eigenvalues are distinct, the general solution is x = c e 2t + c e t + c e 2t Setting x= ξ e rt results in the algebraic equations ( ) ( ) ( ) 5 r ξ = 3 r For a nonzero solution, we must have det(a ri) = r 2 6r + 8 = The roots of the characteristic equation are r = 4 and r 2 = 2 With r = 4, the system of equations reduces to ξ ξ 2 = The corresponding eigenvector is ξ () = (, ) T For the case r = 2, the system is equivalent to the equation 3 ξ ξ 2 = An eigenvector is ξ (2) = (, 3) T Since the eigenvalues are distinct, the general solution is ( ) ( ) x = c e 4t + c 2 e 2t 3 Invoking the initial conditions, we obtain the system of equations c + c 2 = 2 ξ 2 c + 3 c 2 = Hence c = 7/2 and c 2 = 3/2, and the solution of the IVP is x = 7 ( ) e 4t 3 ( ) e 2t Setting x= ξ e rt results in the algebraic equations r 2 2 r 2 ξ ξ 2 = 3 r For a nonzero solution, we must have det(a ri) = r 3 6r 2 + r 6 = The roots of the characteristic equation are r =, r 2 = 2 and r 3 = 3 Setting r =, we have 2 2 ξ ξ 2 = 2 This system reduces to the equations ξ 3 ξ = ξ 3 ξ ξ 3 =

282 282 Chapter 7 Systems of First Order Linear Equations A corresponding solution vector is given by ξ () = (, 2, ) T Setting λ = 2, the reduced system of equations is ξ ξ 2 = ξ 3 = A corresponding solution vector is given by ξ (2) = (,, ) T Finally, upon setting λ = 3, the reduced system of equations is ξ 2 ξ 3 = ξ 2 2 ξ 3 = A corresponding solution vector is given by ξ (3) = (2, 2, ) T Since the eigenvalues are distinct, the general solution is 2 x = c 2 e t + c 2 e 2t + c 3 2 e 3t Invoking the initial conditions, the coefficients must satisfy the equations c c 3 = 2 2 c + c c 3 = c + c 3 = It follows that c =, c 2 = 2 and c 3 = Hence the solution of the IVP is x = 2 e t + 2 e 2t 8 The eigensystem is obtained from analysis of the equation r ξ 2 r ξ 2 = 2 4 r The characteristic equation of the coefficient matrix is r 3 4r 2 r + 4 =, with roots r =, r 2 = and r 3 = 4 Setting r = r =, we have 2 ξ ξ 2 = 2 3 This system is reduced to the equations ξ 3 ξ 3 ξ ξ 3 = ξ ξ 3 = A corresponding solution vector is given by ξ () = (, 2, ) T Setting r =, the system reduces to the equations ξ + ξ 3 = ξ ξ 3 =

283 The corresponding eigenvector is ξ (2) = (, 2, ) T Finally, upon setting r = 4, the system is equivalent to the equations 4 ξ + ξ 3 = 8 ξ 2 + ξ 3 = The corresponding eigenvector is ξ (3) = (2,, 8) T Hence the general solution is x = c 2 e t + c 2 2 e t + c 3 2 e 4t 8 Invoking the initial conditions, c + c c 3 = 7 2 c + 2 c 2 + c 3 = 5 c c 2 8 c 3 = 5 It follows that c = 3, c 2 = 6 and c 3 = Hence the solution of the IVP is 2 x = 3 2 e t e t e 4t 8 9 Set x= ξ t r Substitution into the system of differential equations results in t rt r ξ = A ξ t r, which upon simplification yields is, A ξ rξ = Hence the vector ξ and constant r must satisfy (A ri)ξ = 2 Setting x= ξ t r results in the algebraic equations ( ) ( ) ( ) 5 r ξ = 3 r For a nonzero solution, we must have det(a ri) = r 2 6r + 8 = The roots of the characteristic equation are r = 4 and r 2 = 2 With r = 4, the system of equations reduces to ξ ξ 2 = The corresponding eigenvector is ξ () = (, ) T For the case r = 2, the system is equivalent to the equation 3 ξ ξ 2 = An eigenvector is ξ (2) = (, 3) T It follows that x () = ( ξ 2 ) t 4 and x (2) = ( ) t 2 3 The Wronskian of this solution set is W [ x (), x (2)] = 2t 6 Thus the solutions are linearly independent for t > Hence the general solution is ( ) ( ) x = c t 4 + c 2 t As shown in Problem 9, solution of the ODE requires analysis of the equations ( ) ( ) ( ) 4 r 3 ξ = 8 6 r ξ 2

284 284 Chapter 7 Systems of First Order Linear Equations For a nonzero solution, we must have det(a ri) = r 2 + 2r = The roots of the characteristic equation are r = and r 2 = 2 For r =, the system of equations reduces to 4 ξ = 3 ξ 2 The corresponding eigenvector is ξ () = (3, 4) T Setting r = 2 results in the single equation 2 ξ ξ 2 = A corresponding eigenvector is ξ (2) = (, 2) T It follows that ( ) ( ) 3 x () = and x (2) = t The Wronskian of this solution set is W [ x (), x (2)] = 2 t 2 These solutions are linearly independent for t > Hence the general solution is ( ) ( ) 3 x = c + c 2 t Setting x= ξ t r results in the algebraic equations ( ) ( ) ( ) 3 r 2 ξ = 2 2 r For a nonzero solution, we must have det(a ri) = r 2 r 2 = The roots of the characteristic equation are r = 2 and r 2 = Setting r = 2, the system of equations reduces to ξ 2ξ 2 = The corresponding eigenvector is ξ () = (2, ) T With r =, the system is equivalent to the equation 2ξ ξ 2 = An eigenvector is ξ (2) = (, 2) T It follows that x () = ( 2 ξ 2 ) t 2 and x (2) = ( ) t 2 The Wronskian of this solution set is W [ x (), x (2)] = 3 t Thus the solutions are linearly independent for t > Hence the general solution is ( ) ( ) 2 x = c t 2 + c 2 t 2 24(a) The general solution is ( ) ( ) x = c e t + c 2 e 2t 2 2

285 (b) (c) 26(a) The general solution is ( ) ( ) x = c e t + c 2 e 2t 2 2

286 286 Chapter 7 Systems of First Order Linear Equations (b) (c) 28(a) We note that (A r i I)ξ (i) =, for i =, 2 (b) It follows that (A r 2 I)ξ () =A ξ () r 2 ξ () = r ξ () r 2 ξ () (c) Suppose that ξ () and ξ (2) are linearly dependent Then there exist constants c and c 2, not both zero, such that c ξ () + c 2 ξ (2) = Assume that c It is clear that (A r 2 I)(c ξ () + c 2 ξ (2) ) = On the other hand, (A r 2 I)(c ξ () + c 2 ξ (2) ) = c (r r 2 )ξ () + = c (r r 2 )ξ () Since r r 2, we must have c =, which leads to a contradiction (d) Note that (A r I)ξ (2) = (r 2 r )ξ (2) (e) Let n = 3, with r r 2 r 3 Suppose that ξ (), ξ (2) and ξ (3) are indeed linearly dependent Then there exist constants c, c 2 and c 3, not all zero, such that c ξ () + c 2 ξ (2) + c 3 ξ (3) = Assume that c It is clear that (A r 2 I)(c ξ () + c 2 ξ (2) + c 3 ξ (3) ) = On the other hand, (A r 2 I)(c ξ () + c 2 ξ (2) + c 3 ξ (3) ) = c (r r 2 )ξ () + c 3 (r 3 r 2 )ξ (3)

287 It follows that c (r r 2 )ξ () + c 3 (r 3 r 2 )ξ (3) = Based on the result of part (a), which is actually not dependent on the value of n, the vectors ξ () and ξ (3) are linearly independent Hence we must have c (r r 2 ) = c 3 (r 3 r 2 ) =, which leads to a contradiction 29(a) Let x = y and x 2 = y It follows that x = x 2 and x 2 = y = a (c y + b y ) In terms of the new variables, we obtain the system of two first order ODEs x = x 2 x 2 = a (c x + b x 2 ) (b) The coefficient matrix is given by ( A = c a b a ) Setting x= ξ e rt results in the algebraic equations ( ) ( ) r ξ b a r = c a For a nonzero solution, we must have ξ 2 ( ) det(a ri) = r 2 + b a r + c a = Multiplying both sides of the equation by a, we obtain a r 2 + b r + c = 3(a) Solution of the ODE requires analysis of the algebraic equations ( ) ( ) ( ) / r 3/4 ξ = / /5 r For a nonzero solution, we must have det(a ri) = The characteristic equation is 8 r r + =, with roots r = /4 and r 2 = /2 With r = /4, the system of equations reduces to 2 ξ + ξ 2 = The corresponding eigenvector is ξ () = (, 2) T Substitution of r = /2 results in the equation 2 ξ 3 ξ 2 = A corresponding eigenvector is ξ (2) = (3, 2) T Since the eigenvalues are distinct, the general solution is ( ) ( ) 3 x = c e t/4 + c 2 e t/2 2 2 Invoking the initial conditions, we obtain the system of equations ξ 2 c + 3 c 2 = 7 2 c + 2 c 2 = 2

288 288 Chapter 7 Systems of First Order Linear Equations Hence c = 29/8 and c 2 = 55/8, and the solution of the IVP is x = 29 ( ) e t/4 55 ( ) 3 e t/ (b) (c) Both functions are monotone increasing It is easy to show that 5 x (t) < and 5 x 2 (t) < provided that t > T (a) The system of differential equations is ( ) ( d I /2 /2 = dt V 3/2 5/2 ) ( I V Solution of the system requires analysis of the eigenvalue problem ( ) ( ) ( ) /2 r /2 ξ = 3/2 5/2 r The characteristic equation is r 2 + 3r + 2 =, with roots r = and r 2 = 2 With r =, the equations reduce to ξ ξ 2 = A corresponding eigenvector is given by ξ () = (, ) T Setting r = 2, the system reduces to the equation 3ξ ξ 2 = An eigenvector is ξ (2) = (, 3) T Hence the general solution is ( ) ( ) ( ) I = c e t + c 2 e 2t V 3 (b) The eigenvalues are distinct and both negative We find that the equilibrium point (, ) is a stable node Hence all solutions converge to (, ) 33(a) Solution of the ODE requires analysis of the algebraic equations ( R L r ) ( ) ( ) ξ L = r The characteristic equation is C CR 2 ξ 2 ξ 2 ) r 2 + ( L + CR R 2 LCR 2 )r + R + R 2 LCR 2 =

289 The eigenvectors are real and distinct, provided that the discriminant is positive That is, which simplifies to the condition ( L + CR R 2 LCR 2 ) 2 4( R + R 2 LCR 2 ) >, ( R CR 2 L )2 4 LC > (b) The parameters in the ODE are all positive Observe that the sum of the roots is L + CR R 2 LCR 2 < Also, the product of the roots is R + R 2 LCR 2 > It follows that both roots are negative Hence the equilibrium solution I =, V = represents a stable node, which attracts all solutions (c) If the condition in part (a) is not satisfied, that is, then the real part of the eigenvalues is ( R CR 2 L )2 4 LC, Re(r,2 ) = L + CR R 2 2 LCR 2 As long as the parameters are all positive, then the solutions will still converge to the equilibrium point (, ) 76 2(a) Setting x= ξ e rt results in the algebraic equations ( ) ( ) ( ) r 4 ξ = r For a nonzero solution, we require that det(a ri) = r 2 + 2r + 5 = The roots of the characteristic equation are r = ± 2i Substituting r = 2i, the two equations reduce to ξ + 2i ξ 2 = The two eigenvectors are ξ () = ( 2i, ) T and ξ (2) = (2i, ) T Hence one of the complex-valued solutions is given by ( 2i x () = ) e (+2i)t = = e t ( 2 sin 2t cos 2t ( 2i ) ξ 2 ) e t (cos 2t i sin 2t) = + i e t ( 2 cos 2t sin 2t )

290 29 Chapter 7 Systems of First Order Linear Equations Based on the real and imaginary parts of this solution, the general solution is ( ) ( ) 2 sin 2t 2 cos 2t x = c e t + c 2 e t cos 2t sin 2t (b) 3(a) Solution of the ODEs is based on the analysis of the algebraic equations ( ) ( ) ( ) 2 r 5 ξ = 2 r For a nonzero solution, we require that det(a ri) = r 2 + = The roots of the characteristic equation are r = ±i Setting r = i, the equations are equivalent to ξ (2 + i)ξ 2 = The eigenvectors are ξ () = (2 + i, ) T and ξ (2) = (2 i, ) T Hence one of the complex-valued solutions is given by Therefore the general solution is ( ) ( 2 + i 2 + i x () = e it = ( ) ( 2 cos t sin t cos t + 2 sin t = + i cos t sin t x = c ( 2 cos t sin t cos t ) ξ 2 ) (cos t + i sin t) = ) ( ) cos t + 2 sin t + c 2 sin t The solution may also be written as ( ) ( ) 5 cos t 5 sin t x = c + c 2 2 cos t + sin t cos t + 2 sin t

291 76 29 (b) 4(a) Setting x= ξ e rt results in the algebraic equations ( ) ( ) ( ) 2 r 5/2 ξ = 9/5 r For a nonzero solution, we require that det(a ri) = r 2 r = The roots of the characteristic equation are r = ( ± 3i)/2 With r = ( + 3 i)/2, the equations reduce to the single equation (3 3i)ξ 5 ξ 2 = The corresponding eigenvector is given by ξ () = (5, 3 3 i) T Hence one of the complex-valued solutions is ( ) ( ) i x () = e (+3i)t/2 = e t/2 (cos 3 3 3i 2 t + i sin 3 2 t) = ( 2 cos 3 = e t/2 2 t sin 3 2 t ) ( cos 3 cos 3 2 t + ie t/2 2 t + 2 sin 3 2 t ) sin 3 2 t The general solution is ( 2 cos 3 x = c e t/2 2 t sin 3 2 t ) ( cos 3 cos 3 2 t + c 2 e t/2 2 t + 2 sin 3 2 t ) sin 3 2 t The solution may also be written as ( x = c e t/2 5 cos 3 2 t ) ( 3 cos 3 2 t + 3 sin 3 2 t + c 2 e t/2 5 sin 3 2 t ) 3 cos 3 2 t + 3 sin 3 2 t ξ 2

292 292 Chapter 7 Systems of First Order Linear Equations (b) 5(a) Setting x= ξ t r results in the algebraic equations ( ) ( ) r ξ = 5 3 r ξ 2 ( ) The characteristic equation is r r + 2 =, with roots r = ± i Substituting r = i reduces the system of equations to (2 + i)ξ ξ 2 = The eigenvectors are ξ () = (, 2 + i) T and ξ (2) = (, 2 i) T Hence one of the complex-valued solutions is given by ( ) ( ) x () = e (+i)t = e t (cos t i sin t) = 2 + i 2 + i ( ) = e t cos t 2 cos t + sin t ( ) sin t + ie t cos t 2 sin t The general solution is ( ) ( ) x = c e t cos t + c 2 e t sin t 2 cos t + sin t cos t + 2 sin t (b)

293 (a) Solution of the ODEs is based on the analysis of the algebraic equations ( ) ( ) ( ) r 2 ξ = 5 r For a nonzero solution, we require that det(a ri) = r = The roots of the characteristic equation are r = ± 3 i Setting r = 3 i, the two equations reduce to ( 3 i)ξ + 2ξ 2 = The corresponding eigenvector is ξ () = ( 2, 3i) T Hence one of the complex-valued solutions is given by ( ) ( ) 2 2 x () = e 3it = (cos 3t + i sin 3t) = = 3i ( 2 cos 3t cos 3t + 3 sin 3t 3i ) ( + i ξ 2 2 sin 3t 3 cos 3t + sin 3t The general solution is ( ) ( ) 2 cos 3t 2 sin 3t x = c + c 2 cos 3t + 3 sin 3t 3 cos 3t sin 3t (b) ) 8 The eigensystem is obtained from analysis of the equation 3 r 2 r ξ ξ 2 = 2 r The characteristic equation of the coefficient matrix is r 3 + 4r 2 + 7r + 6 =, with roots r = 2, r 2 = 2 i and r 3 = + 2 i Setting r = 2, the equations reduce to ξ + 2 ξ 3 = ξ + ξ 2 = ξ 3 The corresponding eigenvector is ξ () = (2, 2, ) T system of equations is equivalent to (2 i 2 )ξ 2 ξ 3 = ξ + i 2 ξ 2 = With r = 2 i, the

294 294 Chapter 7 Systems of First Order Linear Equations An eigenvector is given by ξ (2) = ( i 2,, i 2 ) T Hence one of the complexvalued solutions is given by x (2) = i 2 i e (+i 2)t = i 2 2 i e t (cos 2 t i sin 2 t) = 2 = e t 2 sin 2 t cos 2 t cos 2 t 2 sin 2 t + ie t 2 cos 2 t sin 2 t 2 cos 2 t + sin 2 t The other complex-valued solution is x (3) = ξ (2) e r3t The general solution is 2 x = c 2 e 2t + 2 sin 2 t + c 2 e t cos 2 cos 2 t 2 t cos 2 t + 2 sin + c 3 e t sin 2 t 2 t 2 cos 2 t sin 2 t It is easy to see that all solutions converge to the equilibrium point (,, ) Solution of the system of ODEs requires that ( ) ( ) 3 r 2 ξ = r ξ 2 ( ) The characteristic equation is r r + 5 =, with roots r = 2 ± i Substituting r = 2 + i, the equations are equivalent to ξ ( i)ξ 2 = The corresponding eigenvector is ξ () = ( i, ) T One of the complex-valued solutions is given by ( ) ( ) i i x () = e ( 2+i)t = e 2t (cos t + i sin t) = ( ) ( ) cos t + sin t cos t + sin t = e 2t + ie 2t cos t sin t Hence the general solution is x = c e 2t ( cos t + sin t cos t ) ( ) cos t + sin t + c 2 e 2t sin t Invoking the initial conditions, we obtain the system of equations c c 2 = c = 2 Solving for the coefficients, the solution of the initial value problem is ( ) ( ) cos t + sin t cos t + sin t x = 2 e 2t 3 e 2t cos t sin t ( ) cos t 5 sin t = e 2t 2 cos t 3 sin t

295 The solution converges to (, ) as t 2 Solution of the ODEs is based on the analysis of the algebraic equations ( 4 5 r 2 ) ( ) ( ) ξ 6 5 r = ξ 2 The characteristic equation is 25 r 2 r + 26 =, with roots r = /5 ± i Setting r = /5 + i, the two equations reduce to ξ ( i)ξ 2 = The corresponding eigenvector is ξ () = ( i, ) T One of the complex-valued solutions is given by ( i x () = ) e ( 5 +i)t = = e t/5 ( cos t + sin t cos t Hence the general solution is ) x = c e t/5 ( cos t + sin t cos t ( i ) e t/5 (cos t + i sin t) = + ie t/5 ( cos t + sin t sin t ) ) ( ) cos t + sin t + c 2 e t/5 sin t (b) Let x() = (x, x 2) T The solution of the initial value problem is ( ) ( ) cos t + sin t cos t + sin t x = x 2 e t/5 + (x 2 x cos t )e t/5 sin t ( x = e t/5 cos t + (2 x 2 x ) ) sin t x 2 cos t + (x 2 x ) sin t With x() = (, 2) T, the solution is ( ) cos t + 3 sin t x = e t/5 2 cos t + sin t

296 296 Chapter 7 Systems of First Order Linear Equations (c) (d) 3(a) The characteristic equation is r 2 2αr + + α 2 =, with roots r = α ± i (b) When α < and α >, the equilibrium point (, ) is a stable spiral and an unstable spiral, respectively The equilibrium point is a center when α = (c) (a) α = /8 (b) α = /8

297 (a) The roots of the characteristic equation, r 2 α r + 5 =, are r,2 = α 2 ± 2 α2 2 (b) Note that the roots are complex when 2 < α < 2 For the case when α ( 2, ), the equilibrium point (, ) is a stable spiral On the other hand, when α (, 2 ), the equilibrium point is an unstable spiral For the case α =, the roots are purely imaginary, so the equilibrium point is a center When α 2 > 2, the roots are real and distinct The equilibrium point becomes a node, with its stability dependent on the sign of α Finally, the case α 2 = 2 marks the transition from spirals to nodes (c) (a) α = 5 (b) α = 3 (c) α = /2 (d) α = /2 7 The characteristic equation of the coefficient matrix is r 2 + 2r + + α =, with roots given formally as r,2 = ± α The roots are real provided that α First note that the sum of the roots is 2 and the product of the roots is + α For negative values of α, the roots are distinct, with one always negative When α <, the roots have opposite signs Hence the equilibrium point is a saddle For the case < α <, the roots are both negative, and the equilibrium point is a stable node α = represents a transition from saddle to node When α =, both roots are equal For the case α >, the roots are complex conjugates, with negative real part Hence the equilibrium point is a stable spiral (a) α = 3/2 (b) α = /2 (c) α = /2

298 298 Chapter 7 Systems of First Order Linear Equations 9 The characteristic equation for the system is given by The roots are r 2 + (4 α)r + 4α = r,2 = 2 + α 2 ± α 2 + 8α 24 First note that the roots are complex when 4 2 < α < We also find that when 4 2 < α < 4 + 2, the equilibrium point is a stable spiral For α > 4 + 2, the roots are real When α > 25, the roots have opposite signs, with the equilibrium point being a saddle For the case < α < 25, the roots are both negative, and the equilibrium point is a stable node Finally, when α < 4 2, both roots are negative, with the equilibrium point being a stable node (a) α = (b) α = 8 (c) α = 2 (d) α = 24 (e) α = 4 2 The characteristic equation is r r (24 + 8α) =, with roots r,2 = ± α The roots are complex when α < 25/8 Since the real part is negative, the origin is a stable spiral Otherwise the roots are real When 25/8 < α < 3, both roots are negative, and hence the equilibrium point is a stable node For α > 3, the roots are of opposite sign and the origin is a saddle

299 (a) α = 4 (b) α = 35 (c) α = 2 22 Based on the method in Problem 9 of Section 75, setting x= ξ t r results in the algebraic equations ( ) ( ) ( ) 2 r 5 ξ = 2 r The characteristic equation for the system is r 2 + =, with roots r,2 = ± i With r = i, the equations reduce to the single equation ξ (2 + i)ξ 2 = A corresponding eigenvector is ξ () = (2 + i, ) T One complex-valued solution is ( ) 2 + i x () = t i We can write t i = e i ln t Hence ( ) ( ) 2 + i 2 + i x () = e i ln t = [cos(ln t) + i sin(ln t)] = ( ) ( ) 2 cos(ln t) sin(ln t) cos(ln t) + 2 sin(ln t) = + i cos(ln t) sin(ln t) Therefore the general solution is x = c ( 2 cos(ln t) sin(ln t) cos(ln t) Other combinations are also possible 24(a) The characteristic equation of the system is ) ξ 2 r r r 7 6 =, ( ) cos(ln t) + 2 sin(ln t) + c 2 sin(ln t) with eigenvalues r = /, and r 2,3 = /4 ± i For r = /, simple calculations reveal that a corresponding eigenvector is ξ () = (,, ) T Setting r = /4 i, we obtain the system of equations ξ i ξ 2 = ξ 3 =

300 3 Chapter 7 Systems of First Order Linear Equations A corresponding eigenvector is ξ (2) = (i,, ) T Hence one solution is x () = e t/ Another solution, which is complex-valued, is given by x (2) = i e ( 4 +i)t = i e t/4 (cos t i sin t) = = e t/4 sin t cos t + ie t/4 cos t sin t Using the real and imaginary parts of x (2), the general solution is constructed as sin t cos t x = c e t/ + c 2 e t/4 cos t + c 3 e t/4 sin t (b) Let x() = (x, x 2, x 3) The solution can be written as x x = + e t/4 2 sin t + x cos t x x 3 e t/ 2 cos t x sin t With x() = (,, ), the solution of the initial value problem is sin t + cos t x = + e t/4 cos t sin t e t/ (a) x x 2 (b) x x 3 (c) x 2 x 3

301 76 3 (c) 25(a) Based on Problems 9-2 of Section 7, the system of differential equations is ( ) ( d I R = L ) ( ) L I, dt V V C CR 2 ) ( ) ( I = 2 8 V 2 2 since R = R 2 = 4 ohms, C = /2 farads and L = 8 henrys (b) The eigenvalue problem is ( 2 r ) ( ) ξ r = ξ 2 ( ) The characteristic equation of the system is r 2 + r + 2 =, with eigenvalues r,2 = 2 ± 2 i Setting r = /2 + i/2, the algebraic equations reduce to 4iξ + ξ 2 = It follows that ξ () = (, 4i) T Hence one complex-valued solution is ( ) () ( ) ( ) I = e ( +i)t/2 = e t/2 [cos(t/2) + i sin(t/2)] = V 4i 4i ( ) ( ) cos(t/2) sin(t/2) = e t/2 + ie t/2 4 sin(t/2) 4 cos(t/2) Therefore the general solution is ( ) ( I cos(t/2) = c e t/2 V 4 sin(t/2) ) ( ) sin(t/2) + c 2 e t/2 4 cos(t/2) (c) Imposing the initial conditions, we arrive at the equations c = 2 and c 2 = 3/4, and ( ) ( I 2 cos(t/2) = e t/2 3 4 sin(t/2) ) V 8 sin(t/2) + 3 cos(t/2) (d) Since the eigenvalues have negative real parts, all solutions converge to the origin

302 32 Chapter 7 Systems of First Order Linear Equations 26(a) The characteristic equation of the system is with eigenvalues r 2 + RC r + CL =, r,2 = 2RC ± 4R2 C 2RC L The eigenvalues are real and different provided that 4R2 C L > The eigenvalues are complex conjugates as long as 4R2 C L < (b) With the specified values, the eigenvalues are r,2 = ± i The eigenvector corresponding to r = + i is ξ () = (, 4i) T Hence one complex-valued solution is ( ) () ( ) ( ) I = e ( +i)t = e t (cos t + i sin t) = V + i + i ( ) ( ) = e t cos t + ie t sin t cos t sin t cos t sin t Therefore the general solution is ( ) ( ) ( ) I = c e t cos t + c 2 e t sin t V cos t sin t cos t sin t (c) Imposing the initial conditions, we arrive at the equations c = 2 c + c 2 =, with c = 2 and c 2 = 3 Therefore the solution of the IVP is ( ) ( ) I 2 cos t + 3 sin t = e t V cos t 5 sin t (d) Since Re(r,2 ) =, all solutions converge to the origin 27(a) Suppose that c a+c 2 b= Since a and b are the real and imaginary parts of the vector ξ (), respectively, a= (ξ () + ξ () )/2 and b= (ξ () ξ () )/2i Hence which leads to c (ξ () + ξ () ) ic 2 (ξ () ξ () ) =, (c ic 2 )ξ () + (c + ic 2 )ξ () =

303 76 33 (b) Now since ξ () and ξ () are linearly independent, we must have c ic 2 = c + ic 2 = It follows that c = c 2 = (c) Recall that u(t) = e λt (a cos µt b sin µt) v(t) = e λt (a cos µt + b sin µt) Consider the equation c u(t ) + c 2 v(t ) =, for some t We can then write c e λt (a cos µt b sin µt ) + c 2 e λt (a cos µt + b sin µt ) = ( ) Rearranging the terms, and dividing by the exponential, (c + c 2 ) cos µt a + (c 2 c ) sin µt b = From part (b), since a and b are linearly independent, it follows that (c + c 2 ) cos µt = (c 2 c ) sin µt = Without loss of generality, assume that the trigonometric factors are nonzero Otherwise proceed again from Equation ( ), above We then conclude that c + c 2 = and c 2 c =, which leads to c = c 2 = Thus u(t ) and v(t ) are linearly independent for some t, and hence the functions are linearly independent at every point 28(a) Let x = u and x 2 = u It follows that x = x 2 and x 2 = u = k m u In terms of the new variables, we obtain the system of two first order ODEs x = x 2 x 2 = k m x (b) The associated eigenvalue problem is ( ) ( ) r ξ = k/m r ξ 2 ( ) The characteristic equation is r 2 + k/m =, with roots r,2 = ± i k/m (c) Since the eigenvalues are purely imaginary, the origin is a center Hence the phase curves are ellipses, with a clockwise flow For computational purposes, let k = and m = 2

304 34 Chapter 7 Systems of First Order Linear Equations (a) k =, m = 2 (b) x x 2 (c) x, x 2 vs t (d) The general solution of the second order equation is k k u(t) = c cos m t + c 2 sin m t The general solution of the system of ODEs is given by ( m k sin k m t ) ( m k x = c cos + c 2 cos sin k m t k m t k m t It is evident that the natural frequency of the system is equal to r = r 2 29(a) Set x = (x, x 2 ) T We can rewrite Equation (22) in the form ( ) ( ) 2 d 2 x ( ) ( ) 4 3 x = 9/ x 2 dt 2 d 2 x 2 dt 2 Multiplying both sides of this equation by the inverse of the diagonal matrix, we obtain ( ) d 2 x ( ) ( ) dt 2 3/2 x 2 d 2 x 2 = 4/3 3 x dt 2 2 (b) Substituting x = ξ e rt, r 2 ( ξ ξ 2 ) e rt = ( ) ( ) 2 3/2 ξ e rt, 4/3 3 ξ 2 ) which can be written as (A r 2 I)ξ = (c) The eigenvalues are r 2 = and r2 2 = 4, with corresponding eigenvectors ( ) ( ) 3 3 ξ () = and ξ (2) = 2 4 (d) The linearly independent solutions are ( ) ( ) x () = C 3 e it and x (2) = 2 C 3 2 e 2it 4

305 76 35 in which C and C 2 are arbitrary complex coefficients In scalar form, x = 3c cos t + 3c 2 sin t + 3c 3 cos 2t + 3c 4 sin 2t x 2 = 2c cos t + 2c 2 sin t 4c 3 cos 2t 4c 4 sin 2t (e) Differentiating the above expressions, x = 3c sin t + 3c 2 cos t 6c 3 sin 2t + 6c 4 cos 2t x 2 = 2c sin t + 2c 2 cos t + 8c 3 sin 2t 8c 4 cos 2t It is evident that y= (x, x 2, x, x 2) T as in Equation (3) 3(a) The second order system is given by d 2 x dt 2 = 2x + x 2 d 2 x 2 dt 2 = x 2x 2 Let y = x, y 2 = x 2, y 3 = x and y 4 = x 2 In terms of the new variables, we have y = y 3 y 2 = y 4 y 3 = 2y + y 2 y 4 = y 2y 2 hence the coefficient matrix is A = 2 2 (b) The eigenvalues and corresponding eigenvectors of A are: r = i, r 2 = i, r 3 = 3 i, r 4 = 3 i, ξ () = (,, i, i) T ξ (2) = (,, i, i) T ξ (3) = (,, 3 i, 3 i) T ξ (4) = (,, 3 i, 3 i) T (c) Note that and ξ (3) e 3 it = ξ () e it = i (cos t + i sin t) i 3 i 3 i (cos 3 t + i sin 3 t)

306 36 Chapter 7 Systems of First Order Linear Equations Hence the general solution is cos t sin t cos 3 t y = c cos t sin t + c sin t 2 cos t + c cos 3 t 3 3 sin 3 t + c 4 sin t cos t 3 sin 3 t sin 3 t sin 3 t 3 cos 3 t 3 cos 3 t (d) The two modes have natural frequencies of ω = rad/sec and ω 2 = 3 rad/sec (e) For the initial condition y() = (, 3,, ) T, it is necessary that 3 = c + c 2 + c 3 + c 4 3, 3 resulting in the coefficients c =, c 2 =, c 3 = 2 and c 4 = The solutions are not periodic, since the two natural frequencies are incommensurate

307 (a) The eigenvalues and eigenvectors were found in Problem, Section 75 ( ) ( ) 2 r =, ξ () = ; r 2 = 2, ξ (2) = 2 The general solution is x = c ( e t 2 e t ) + c 2 ( 2 e 2t Hence a fundamental matrix is given by ( ) e t 2 e Ψ(t) = 2t 2 e t e 2t (b) We now have So that Ψ() = Φ(t) = Ψ(t)Ψ () = 3 e 2t ) ( ) 2 and Ψ () = ( ) 2, ( ) e t + 4e 2t 2e t 2e 2t 2e t + 2e 2t 4e t e 2t 3(a) The eigenvalues and eigenvectors were found in Problem 3, Section 75 The general solution of the system is ( ) ( ) e t e t x = c + c 2 e t 3e t Hence a fundamental matrix is given by ( ) e t e Ψ(t) = t e t 3e t (b) Given the initial conditions x() =e (), we solve the equations c + c 2 = c + 3c 2 =, to obtain c = 3/2, c 2 = /2 The corresponding solution is ( 3 2 x = et ) 2 e t 3 2 et 3 2 e t Given the initial conditions x() =e (2), we solve the equations c + c 2 = c + 3c 2 =,

308 38 Chapter 7 Systems of First Order Linear Equations to obtain c = /2, c 2 = /2 The corresponding solution is ( 2 x = et + ) 2 e t 2 et e t Therefore the fundamental matrix is Φ(t) = ( ) 3e t e t e t + e t 2 3e t 3e t e t + 3e t 5(a) The general solution, found in Problem 3, Section 76, is given by ( ) ( ) 5 cos t 5 sin t x = c + c 2 2 cos t + sin t cos t + 2 sin t Hence a fundamental matrix is given by ( ) 5 cos t 5 sin t Ψ(t) = 2 cos t + sin t cos t + 2 sin t (b) Given the initial conditions x() =e (), we solve the equations 5c = 2c c 2 =, resulting in c = /5, c 2 = 2/5 The corresponding solution is ( ) cos t + 2 sin t x = sin t Given the initial conditions x() =e (2), we solve the equations 5c = 2c c 2 =, resulting in c =, c 2 = The corresponding solution is ( ) 5 sin t x = cos t 2 sin t Therefore the fundamental matrix is ( ) cos t + 2 sin t 5 sin t Φ(t) = sin t cos t 2 sin t 7(a) The general solution, found in Problem 5, Section 75, is given by ( ) ( ) e 2t e 4t x = c 3e 2t + c 2 e 4t Hence a fundamental matrix is given by ( ) e 2t e Ψ(t) = 4t 3e 2t e 4t

309 77 39 (b) Given the initial conditions x() =e (), we solve the equations c + c 2 = 3c + c 2 =, resulting in c = /2, c 2 = 3/2 The corresponding solution is x = ( e 2t + 3e 4t ) 2 3e 2t + 3e 4t The initial conditions x() =e (2) require that c + c 2 = 3c + c 2 =, resulting in c = /2, c 2 = /2 The corresponding solution is x = ( e 2t e 4t ) 2 3e 2t e 4t Therefore the fundamental matrix is Φ(t) = ( ) e 2t + 3e 4t e 2t e 4t 2 3e 2t + 3e 4t 3e 2t e 4t 8(a) The general solution, found in Problem 5, Section 76, is given by ( ) ( ) x = c e t cos t + c 2 e t sin t 2 cos t + sin t cos t + 2 sin t Hence a fundamental matrix is given by ( e Ψ(t) = t cos t 2e t cos t + e t sin t ) e t sin t e t cos t + 2e t sin t (b) The specific solution corresponding to the initial conditions x() =e () is ( ) cos t + 2 sin t x = e t 5 sin t For the initial conditions x() =e (2), the solution is ( ) x = e t sin t cos t 2 sin t Therefore the fundamental matrix is ( ) Φ(t) = e t cos t + 2 sin t sin t 5 sin t cos t 2 sin t 9(a) The general solution, found in Problem 3, Section 75, is given by x = c 4e 2t 5e 2t + c 2 3e t 4e t + c 3 e 2t 7e 2t 2e t e 2t

310 3 Chapter 7 Systems of First Order Linear Equations Hence a fundamental matrix is given by Ψ(t) = 4e 2t 3e t 5e 2t 4e t e 2t 7e 2t 2e t e 2t (b) Given the initial conditions x() =e (), we solve the equations 4c + 3c 2 = 5c 4c 2 + c 3 = 7c 2c 2 c 3 =, resulting in c = /2, c 2 =, c 3 = 3/2 The corresponding solution is 2e 2t + 3e t x = 5e 2t /2 4e t + 3e 2t /2 7e 2t /2 2e t 3e 2t /2 The initial conditions x() =e (2), we solve the equations 4c + 3c 2 = 5c 4c 2 + c 3 = 7c 2c 2 c 3 =, resulting in c = /4, c 2 = /3, c 3 = 3/2 The corresponding solution is e 2t + e t x = 5e 2t /4 4e t /3 + 3e 2t /2 7e 2t /4 2e t /3 3e 2t /2 The initial conditions x() =e (3), we solve the equations 4c + 3c 2 = 5c 4c 2 + c 3 = 7c 2c 2 c 3 =, resulting in c = /4, c 2 = /3, c 3 = /2 The corresponding solution is e 2t + e t x = 5e 2t /4 4e t /3 + e 2t /2 7e 2t /4 2e t /3 e 2t /2 Therefore the fundamental matrix is Φ(t) = 24e 2t + 36e t 2e 2t + 2e t 2e 2t + 2e t 3e 2t 48e t + 8e 2t 5e 2t 6e t + 3e 2t 5e 2t 6e t + e 2t 2 42e 2t 24e t 8e 2t 2e 2t 8e t 3e 2t 2e 2t 8e t e 2t 2 The solution of the initial value problem is given by ( ) ( ) e x = Φ(t)x() = t cos 2t 2e t sin 2t 3 2 e t sin 2t e t cos 2t ( ) 3 cos 2t 2 sin 2t = e t 3 sin 2t + cos 2t 2 =

311 Let It follows that Ψ(t) = (t) x(n) (t) n (t) x (n) (t) x () x () x () (t ) x (n) (t ) Ψ(t ) = x () n (t ) x (n) n (t ) n is a scalar matrix, which is invertible, since the solutions are linearly independent Let Ψ (t ) = (c ij ) Then x () (t) x(n) (t) c c n Ψ(t)Ψ (t ) = x () n (t) x (n) n (t) c n c nn The j-th column of the product matrix is [ Ψ(t)Ψ (t ) ] (j) n = c kj x (k), which is a solution vector, since it is a linear combination of solutions Furthermore, the columns are all linearly independent, since the vectors x (k) are Hence the product is a fundamental matrix Finally, setting t = t, Ψ(t )Ψ (t ) =I This is precisely the definition of Φ(t) k = 4 The fundamental matrix Φ(t) for the system ( ) x = x 4 is given by Φ(t) = ( ) 2e 3t + 2e t e 3t e t 4 4e 3t 4e t 2e 3t + 2e t Direct multiplication results in Φ(t)Φ(s) = ( ) ( ) 2e 3t + 2e t e 3t e t 2e 3s + 2e s e 3s e s 6 4e 3t 4e t 2e 3t + 2e t 4e 3s 4e s 2e 3s + 2e s = 6 ( 8(e 3t+3s + e t s ) 4(e 3t+3s e t s ) 6(e 3t+3s e t s ) 8(e 3t+3s + e t s ) Hence Φ(t)Φ(s) = ( ) 2e 3(t+s) + 2e (t+s) e 3(t+s) e (t+s) 4 4e 3(t+s) 4e (t+s) 2e 3(t+s) + 2e (t+s) = Φ(t + s) )

312 32 Chapter 7 Systems of First Order Linear Equations 5(a) Let s be arbitrary, but fixed, and t variable Similar to the argument in Problem 3, the columns of the matrix Φ(t)Φ(s) are linear combinations of fundamental solutions Hence the columns of Φ(t)Φ(s) are also solution of the system of equations Further, setting t =, Φ()Φ(s) =I Φ(s) = Φ(s) That is, Φ(t)Φ(s) is a solution of the initial value problem Z =AZ, with Z() = Φ(s) Now consider the change of variable τ = t + s Let W(τ) =Z(τ s) The given initial value problem can be reformulated as d W = AW, with W(s) = Φ(s) dτ Since Φ(t) is a fundamental matrix satisfying Φ =AΦ, with Φ() =I, it follows that W(τ) = [ Φ(τ)Φ (s) ] Φ(s) = Φ(τ) That is, Φ(t + s) = Φ(τ) =W(τ) =Z(t) = Φ(t)Φ(s) (b) Based on part (a), Φ(t)Φ( t) = Φ(t + ( t)) = Φ() =I Hence Φ( t) = Φ (t) (c) It also follows that Φ(t s) = Φ(t + ( s)) = Φ(t)Φ( s) = Φ(t)Φ (s) 6 Let A be a diagonal matrix, with A= [ a e (), a 2 e (2),, a n e (n)] Note that for any positive integer k, [ A k = a k e (), a k 2 e (2),, a k n e (n)] It follows, from basic matrix algebra, that m k= m ak tk k! I + A k t k k! = m k= ak 2 tk k! k = m k= ak n tk k! It can be shown that the partial sums on the left hand side converge for all t Taking the limit as m on both sides of the equation, we obtain e at e At e a2t = e ant Alternatively, consider the system x =Ax Since the ODEs are uncoupled, the vectors x (j) = e ajt e (j), j =, 2, n, are a set of linearly independent solutions Hence the matrix x = [e at e (), e a2t e (2),, e ant e (n)] is a fundamental matrix Finally, since X() =I, it follows that [ e at e (), e a2t e (2),, e ant e (n)] = Φ(t) = e At

313 (a) Let x = u and x 2 = u ; then u = x 2 In terms of the new variables, we have x 2 + ω 2 x = with the initial conditions x () = u and x 2 () = v The equivalent first order system is x = x 2 x 2 = ω 2 x which can be expressed in the form ( ) ( ) ( ) x x = x 2 ω 2 ; x 2 ( ) ( ) x () u = x 2 () v (b) Setting it is easy to show that A = ( ) ω 2, A 2 = ω 2 I, A 3 = ω 2 A and A 4 = ω 4 I It follows inductively that and A 2k = ( ) k ω 2k I A 2k+ = ( ) k ω 2k A Hence and therefore e At = [( ) k ω2k t 2k (2k)! I + ( )k ω2k t 2k+ ] (2k + )! A k= [ ] [ = ( ) k ω2k t 2k I + ] ( ) k ω2k+ t 2k+ A (2k)! ω (2k + )! k= e At = cos ω ti + ω k= sin ω t A (c) From Equation (28), ( x x 2 ) = [ cos ω ti + ] ( ) ω sin ω t A u v = cos ωt ( u v ) + ω sin ω t ( v ω 2 u ) 8(a) Assuming that x= φ(t) is a solution, then φ =Aφ, with φ() =x Integrate both sides of the equation to obtain φ(t) φ() = t Aφ(s)ds

314 34 Chapter 7 Systems of First Order Linear Equations Hence φ(t) = x + t Aφ(s)ds (b) Proceed with the iteration φ (i+) (t) = x + t Aφ (i) (s)ds With φ () (t) =x, and noting that A is a constant matrix, φ () (t) = x + That is, φ () (t) = (I + At)x (c) We then have φ (2) (t) = x + t Now suppose that It follows that Therefore t Ax ds = x + Ax t A(I + At)x ds = x + Ax t + A 2 x t2 2 = (I + At + A2 t 2 2 )x φ (n) (t) = (I + At + A 2 t An t n t A(I + At + A 2 t An t n = A(It + A t2 2 + A2 t 3 = (At + A 2 t A3 t 3 n! )x n! )x ds = 3! + + An t n+ (n + )! )x 3! + + An+ t n n! )x φ (n+) (t) = (I + At + A 2 t An+ t n+ (n + )! )x By induction, the asserted form of φ (n) (t) is valid for all n (d) Define φ ( ) (t) = lim n φ (n) (t) It can be shown that the limit does exist In fact, φ ( ) (t) = e At x Term-by-term differentiation results in d dt φ( ) (t) = d dt (I + At + t 2 A t n An n! + )x = (A + A 2 t + + A n t n (n )! + )x = A(I + At + A 2 t An t n (n )! + )x

315 78 35 That is, d dt φ( ) (t) = Aφ ( ) (t) Furthermore, φ ( ) () =x Based on uniqueness of solutions, φ(t) = φ ( ) (t) 78 2(a) (b) All of the points on the line x 2 = 2x are equilibrium points Solutions starting at all other points become unbounded (c) Setting x= ξ t r results in the algebraic equations ( ) ( ) ( ) 4 r 2 ξ = 8 4 r The characteristic equation is r 2 =, with the single root r = Substituting r = reduces the system of equations to 2ξ ξ 2 = Therefore the only eigenvector is ξ = (, 2) T One solution is ( ) x () =, 2 which is a constant vector In order to generate a second linearly independent solution, we must search for a generalized eigenvector This leads to the system of equations ( ) ( ) ( ) 4 2 η = η 2 This system also reduces to a single equation, 2η η 2 = /2 Setting η = k, some arbitrary constant, we obtain η 2 = 2k /2 A second solution is ( ) ( ) ( ) ( ) ( ) x (2) k = t + = t + + k 2 2k /2 2 /2 2 ξ 2

316 36 Chapter 7 Systems of First Order Linear Equations Note that the last term is a multiple of x () and may be dropped Hence ( ) ( ) x (2) = t + 2 /2 The general solution is ( ) [( ) x = c + c 2 t (a) ( )] /2 (b) All trajectories converge to the origin (c) Solution of the ODE requires analysis of the algebraic equations ( ) ( ) ( ) 3 r 5 ξ 2 = 2 r 5 2 For a nonzero solution, we must have det(a ri) = r 2 + r + /4 = The only root is r = /2, which is an eigenvalue of multiplicity two Setting r = /2 is the coefficient matrix reduces the system to the single equation ξ + ξ 2 = Hence the corresponding eigenvector is ξ = (, ) T One solution is ( ) x () = e t/2 In order to obtain a second linearly independent solution, we find a solution of the system ( ) ( ) ( ) 5/2 5/2 η = 5/2 5/2 There equations reduce to 5η + 5η 2 = 2 Set η = k, some arbitrary constant Then η 2 = k + 2/5 A second solution is ( ) ( ) ( ) ( ) ( ) k x (2) = te t/2 + e t/2 = te t/2 + e t/2 + k e t/2 k + 2/5 2/5 Dropping the last term, the general solution is ( ) [( ) x = c e t/2 + c 2 te t/2 + η 2 ξ 2 ( ] )e t/2 2/5

317 The eigensystem is obtained from analysis of the equation r r ξ ξ 2 = r The characteristic equation of the coefficient matrix is r 3 3r 2 =, with roots r = 2 and r 2,3 = Setting r = 2, we have 2 2 ξ ξ 2 = 2 This system is reduced to the equations ξ 3 ξ 3 ξ ξ 3 = ξ 2 ξ 3 = A corresponding eigenvector is given by ξ () = (,, ) T system of equations is reduced to the single equation ξ + ξ 2 + ξ 3 = Setting r =, the An eigenvector vector is given by ξ (2) = (,, ) T Since the last equation has two free variables, a third linearly independent eigenvector (associated with r = ) is ξ (3) = (,, ) T Therefore the general solution may be written as x = c e 2t + c 2 e t + c 3 e t 7(a) Solution of the ODE requires analysis of the algebraic equations ( ) ( ) ( ) r 4 ξ = 4 7 r For a nonzero solution, we must have det(a ri) = r 2 + 6r + 9 = The only root is r = 3, which is an eigenvalue of multiplicity two Substituting r = 3 into the coefficient matrix, the system reduces to the single equation ξ ξ 2 = Hence the corresponding eigenvector is ξ = (, ) T One solution is ( ) x () = e 3t For a second linearly independent solution, we search for a generalized eigenvector Its components satisfy ( ) ( ) ( ) 4 4 η =, 4 4 η 2 that is, 4η 4η 2 = Let η 2 = k, some arbitrary constant Then η = k + /4 It follows that a second solution is given by ( ) ( ) ( ) ( ) ( ) k + /4 /4 x (2) = te 3t + e 3t = te 3t + e 3t + k e 3t k ξ 2

318 38 Chapter 7 Systems of First Order Linear Equations Dropping the last term, the general solution is ( ) [( ) ( ] /4 x = c e 3t + c 2 te 3t + )e 3t Imposing the initial conditions, we require that c + c 2 /4 = 3, c = 2, which results in c = 2 and c 2 = 4 Therefore the solution of the IVP is ( ) ( ) 3 4 x = e 3t + te 3t 2 4 (b) 8(a) Solution of the ODEs is based on the analysis of the algebraic equations ( 5 2 r ) ( ) ( ) 3 ξ r = ξ 2 The characteristic equation is r r + =, with a single root r = Setting r =, the two equations reduce to ξ + ξ 2 = The corresponding eigenvector is ξ = (, ) T One solution is ( ) x () = e t A second linearly independent solution is obtained by solving the system ( ) ( ) ( ) 3/2 3/2 η = 3/2 3/2 The equations reduce to the single equation 3η + 3η 2 = 2 Let η = k We obtain η 2 = 2/3 + k, and a second linearly independent solution is ( ) ( ) ( ) ( ) ( ) k x (2) = te t + e t = te t + e t + k e t 2/3 + k 2/3 Dropping the last term, the general solution is ( ) [( ) x = c e t + c 2 te t + η 2 ( ] )e t 2/3

319 78 39 Imposing the initial conditions, we find that c = 3, c + 2c 2 /3 =, so that c = 3 and c 2 = 6 Therefore the solution of the IVP is ( ) ( ) 3 6 x = e t te t 6 (b) (a) The eigensystem is obtained from analysis of the equation ( ) ( ) ( ) 3 r 9 ξ = 3 r The characteristic equation is r 2 =, with a single root r = Setting r =, the two equations reduce to ξ + 3ξ 2 = The corresponding eigenvector is ξ = ( 3, ) T Hence one solution is ( ) 3 x () =, which is a constant vector A second linearly independent solution is obtained from the system ( ) ( ) ( ) 3 9 η 3 = 3 η 2 The equations reduce to the single equation η + 3η 2 = Let η 2 = k We obtain η = 3k, and a second linearly independent solution is ( ) ( ) ( ) ( ) ( ) 3 3k 3 3 x (2) = t + = t + + k k Dropping the last term, the general solution is ( ) [( ) ( )] 3 3 x = c + c 2 t + Imposing the initial conditions, we require that 3c c 2 = 2, c = 4, which results in c = 4 and c 2 = 4 Therefore the solution of the IVP is ( ) ( ) 2 3 x = 4 t 4 ξ 2

320 32 Chapter 7 Systems of First Order Linear Equations (b) 3 Setting x= ξ t r results in the algebraic equations ( ) ( ) ( ) 3 r 4 ξ = r The characteristic equation is r 2 2r + =, with a single root of r,2 = With r =, the system reduces to a single equation ξ 2 ξ 2 = An eigenvector is given by ξ = (2, ) T Hence one solution is ( ) 2 x () = t In order to find a second linearly independent solution, we search for a generalized eigenvector whose components satisfy ( ) ( ) ( ) 2 4 η 2 = 2 η 2 These equations reduce to η 2 η 2 = Let η 2 = k, some arbitrary constant Then η = + 2k (Before proceeding, note that if we set u = ln t, the original equation is transformed into a constant coefficient equation with independent variable u Recall that a second solution is obtained by multiplication of the first solution by the factor u This implies that we must multiply first solution by a factor of ln t) Hence a second linearly independent solution is ( ) ( ) ( ) ( ) ( ) 2 + 2k 2 2 x (2) = t ln t + t = t ln t + t + k t k Dropping the last term, the general solution is ( ) [( ) 2 2 x = c t + c 2 t ln t + ξ 2 ( ) ] t 6(a) Using the result in Problem 5, the eigenvalues are r,2 = 2RC ± The discriminant vanishes when L = 4R 2 C L2 4R 2 CL 2RCL

321 78 32 (b) The system of differential equations is ( ) ( ) ( d I = 4 I dt V V The associated eigenvalue problem is ( ) ( ) r ξ 4 = r ξ 2 ) ( ) The characteristic equation is r 2 + r + /4 =, with a single root of r,2 = /2 Setting r = /2, the algebraic equations reduce to 2ξ + ξ 2 = An eigenvector is given by ξ = (, 2) T Hence one solution is ( I V ) () ( ) = e t/2 2 A second solution is obtained from a generalized eigenvector whose components satisfy ( ) ( ) ( ) η 2 4 = 2 η 2 2 It follows that η = k and η 2 = 4 2k A second linearly independent solution is ( ) (2) ( ) ( ) ( ) ( ) ( ) I k = t e t/2 + e t/2 = t e t/2 + e t/2 + k e t/2 V 2 4 2k Dropping the last term, the general solution is ( ) ( ) [( ) I = c e t/2 + c 2 t e t/2 + V 2 2 ( ] )e t/2 4 Imposing the initial conditions, we require that c =, 2c + 4c 2 = 2, which results in c = and c 2 = Therefore the solution of the IVP is ( ) ( ) ( ) I = e t/2 + te t/2 V 2 2 9(a) The eigensystem is obtained from analysis of the equation 5 r r 4 ξ ξ 2 = r The characteristic equation of the coefficient matrix is r 3 3r 2 + 3r =, with a single root of multiplicity three, r = Setting r =, we have ξ ξ 2 = The system of algebraic equations reduces to a single equation ξ 3 ξ 3 4ξ 3ξ 2 2ξ 3 =

322 322 Chapter 7 Systems of First Order Linear Equations An eigenvector vector is given by ξ () = (,, 2) T Since the last equation has two free variables, a second linearly independent eigenvector (associated with r = ) is ξ (2) = (, 2, 3) T Therefore two solutions are obtained as x () = e t and x (2) = 2 e t 2 3 (b) It follows directly that x = ξte t + ξe t + ηe t Hence the coefficient vectors must satisfy ξte t + ξe t + ηe t =Aξte t +Aηe t Rearranging the terms, we have ξe t = (A I)ξte t + (A I)ηe t Given an eigenvector ξ, it follows that (A I)ξ = and (A I)η = ξ (c) Clearly, (A I) 2 η = (A I)(A I)η = (A I)ξ = Also, = (d) We get that ξ = (A I)η = = This is an eigenvector: = (e) Given the three linearly independent solutions, a fundamental matrix is given by e t 2t e t Ψ(t) = 2e t 4t e t 2e t 3e t 2t e t + e t (f) We construct the transformation matrix T = 2 4, 2 2 with inverse T = /2 /4 2 3/2 The Jordan form of the matrix A is J = T AT =

323 (a) Direct multiplication results in J 2 = λ2 λ 2 2λ, J 3 = λ3 λ 3 3λ 2, J 4 = λ4 λ 4 4λ 3 λ 2 λ 3 λ 4 (b) Suppose that J n = λn λ n nλ n λ n Then λ n λ λ λ n J n+ = λ n nλ n λ = λ λ n λ n + nλ λ n λ n λ λ λ n Hence the result follows by mathematical induction (c) Note that J is block diagonal Hence each block may be exponentiated Using the result in Problem 2, e λt e Jt = e λt te λt e λt (d) Setting λ =, and using the transformation matrix T in Problem 9, 2 e t e t 2e t 2t e t Te Jt = 4 e t te t = 4e t 4t e t 2 2 e t 2e t 2e t 2t e t e t Based on the form of J, e Jt is the fundamental matrix associated with the solutions y () = ξ () e t, y (2) = (2ξ () + 2ξ (2) )e t and y (3) = (2ξ () + 2ξ (2) )te t + ηe t Hence the resulting matrix is the fundamental matrix associated with the solution set { ξ () e t, (2ξ () + 2ξ (2) )e t, (2ξ () + 2ξ (2) )te t + ηe t}, as opposed to the solution set in Problem 9, given by { ξ () e t, ξ (2) e t, (2ξ () + 2ξ (2) )te t + ηe t} 22(a) Direct multiplication results in J 2 = λ2 2λ λ 2 2λ, J 3 = λ3 3λ 2 3λ λ 3 3λ 2, J 4 = λ4 4λ 3 6λ 2 λ 4 4λ 3 λ 2 λ 3 λ 4

324 324 Chapter 7 Systems of First Order Linear Equations (b) Suppose that Then 2 λ n 2 J n = λn nλ n n(n ) λ n nλ n λ n J n+ = λn nλ n n(n ) 2 λ n 2 λ n nλ n λ λ λ n λ = λ λn λ n + nλ λ n nλ n + n(n ) 2 λ λ n 2 λ λ n λ n + nλ λ n λ λ n The result follows by noting that nλ n + (c) We first observe that Therefore n = n(n ) λ λ n 2 = 2 n = n = [ n + λ n tn n! = eλt n tn nλ n! = t n(n ) n 2 tn λ 2 n! = t2 2 n = ] n(n ) 2 n = 2 λ n λ n 2 λ n = n2 + n λ n 2 t n (n )! = t eλt t n 2 e Jt = eλt te λt t 2 2 eλt e λt te λt e λt (n 2)! = t2 2 eλt (d) Setting λ = 2, and using the transformation matrix T in Problem 8, Te Jt = 2 e2t te 2t t 2 2 e2t e 2t te 2t + 2e 2t e 2t te 2t = e 2t te 2t + e 2t t2 3 e 2t 2 e2t + te 2t e 2t te 2t t2 2 e2t + 3e 2t 79 5 As shown in Problem 2, Section 78, the general solution of the homogeneous equation is ( ) ( ) t x c = c + c 2 2 2t 2

325 An associated fundamental matrix is ( ) t Ψ(t) = 2 2t 2 The inverse of the fundamental matrix is easily determined as ( ) Ψ 4t 3 2t + 2 (t) = 8t 8 4t + 5 We can now compute and Finally, where Ψ (t)g(t) = ( 2t 2 ) + 4t t 3, 2t 4 ( Ψ (t)g(t) dt = 2 t 2 + 4t ) 2 ln t 2t 2 2t v(t) = Ψ(t) Ψ (t)g(t) dt, v (t) = 2 t 2 + 2t 2 ln t 2, v 2 (t) = 5t 4 ln t 4 Note that the vector (2, 4) T is a multiple of one of the fundamental solutions Hence we can write the general solution as ( ) ( ) t x = c + c 2 2 2t ( ) /2 2 t 2 + ( ) ( ) 2 2 ln t t The eigenvalues of the coefficient matrix are r = and r 2 = 5 It follows that the solution of the homogeneous equation is ( ) ( ) 2e 5t x c = c + c 2 2 The coefficient matrix is symmetric Hence the system is diagonalizable Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are T = ( ) 2 T = ( ) Setting x=ty, and h(t) =T g(t), the transformed system is given, in scalar form, as y = 5 + 8t 5 t The solutions are readily obtained as e 5t y 2 = 5y y (t) = 5 ln t t + c and y 2 (t) = c 2 e 5t

326 326 Chapter 7 Systems of First Order Linear Equations Transforming back to the original variables, we have x=ty, with x = ( ) ( ) 2 y (t) = ( ) y 5 2 (t) + ( ) 2 y 2 (t) y 2 (t) Hence the general solution is ( ) ( 2e 5t x = k + k 2 2 e 5t ) + ( ) ln t The solution of the homogeneous equation is ( ) ( ) e t e 3t x c = c 2e t + c 2 2e 3t a 2 ( ) t ( ) 2 Based on the simple form of the right hand side, we use the method of undetermined coefficients Set v=a e t Substitution into the ODE yields ( ) ( ) ( ) ( ) a e t a 2 = e t + e t 4 In scalar form, after canceling the exponential, we have a 2 a = a + a a 2 = 4a + a 2, with a = /4 and a 2 = 2 Hence the particular solution is ( ) /4 v = e t, 2 so that the general solution is ( ) ( e t e 3t x = c 2e t + c 2 2e 3t ) + ( ) e t 4 8e t 9 Note that the coefficient matrix is symmetric Hence the system is diagonalizable The eigenvalues and eigenvectors are given by r = ( ) ( ) 2, ξ() = and r 2 = 2, ξ (2) = Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are T = ( ) T = ( ) 2 2 Setting x=ty, and h(t) =T g(t), the transformed system is given, in scalar form, as y = 2 y + 2 t + e t 2 y 2 = 2y t 2 e t

327 Using any elementary method for first order linear equations, the solutions are 2 y (t) = k e t/2 + 3 et t y 2 (t) = k 2 e 2t 3 2 et t Transforming back to the original variables, x=ty, the general solution is ( ) ( ) x = c e t/2 + c 2 e 2t ( ) 7 + ( ) 5 t + ( ) e t Since the coefficient matrix is symmetric, the differential equations can be decoupled The eigenvalues and eigenvectors are given by ( ) ( 2 r = 4, ξ () = and r 2 =, ξ (2) = 2 ) Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are T = ( ) 2 T = ( ) Setting x=ty, and h(t) =T g(t), the transformed system is given, in scalar form, as y = 4y + ( + 2)e t 3 The solutions are easily obtained as y 2 = y ( 2) e t y (t) = k e 4t ( + 2)e t, y 2 (t) = k 2 e t + 3 ( 2 )te t Transforming back to the original variables, the general solution is ( ) ( ) 2 x = c e 4t + c 2 2 e t + ( ) e t + ( 2 )te t Note that ( ) ( ) = ( 3 2 ) 2 The second vector is an eigenvector, hence the solution may be written as ( ) ( ) 2 x = c e 4t + c 2 2 e t + ( ) e t + ( 2 )te t Based on the solution of Problem 3 of Section 76, a fundamental matrix is given by ( ) 5 cos t 5 sin t Ψ(t) = 2 cos t + sin t cos t + 2 sin t

328 328 Chapter 7 Systems of First Order Linear Equations The inverse of the fundamental matrix is easily determined as Ψ (t) = ( ) cos t 2 sin t 5 sin t 5 2 cos t + sin t 5 cos t It follows that and Ψ (t)g(t) = Ψ (t)g(t) dt = ( ) cos t sin t cos 2, t ( ) 2 sin2 t 2 cos t sin t 2 t A particular solution is constructed as v(t) = Ψ(t) Ψ (t)g(t) dt, where v (t) = 5 2 cos t sin t cos2 t t +, v 2(t) = cos t sin t 2 cos2 t + t + 2 Hence the general solution is ( ) ( ) 5 cos t 5 sin t x = c + c 2 2 cos t + sin t cos t + 2 sin t ( ) ( ) ( ) 5/2 5/2 t sin t + t cos t cos t /2 3(a) As shown in Problem 25 of Section 76, the solution of the homogeneous system is ( (c)) ( ) ( ) x cos(t/2) sin(t/2) x (c) = c e t/2 + c 2 e t/2 4 sin(t/2) 4 cos(t/2) 2 Therefore the associated fundamental matrix is given by ( ) Ψ(t) = e t/2 cos(t/2) sin(t/2) 4 sin(t/2) 4 cos(t/2) (b) The inverse of the fundamental matrix is It follows that and Ψ (t) = et/2 4 ( 4 cos(t/2) sin(t/2) ) 4 sin(t/2) cos(t/2) Ψ (t)g(t) = 2 Ψ (t)g(t) dt = ( ) cos(t/2), sin(t/2) ( ) sin(t/2) cos(t/2) A particular solution is constructed as v(t) = Ψ(t) Ψ (t)g(t) dt,

329 where v (t) =, v 2 (t) = 4 e t/2 Hence the general solution is ( ) ( ) ( ) cos(t/2) sin(t/2) x = c e t/2 + c 2 e t/2 + 4 e t/2 4 sin(t/2) 4 cos(t/2) Imposing the initial conditions, we require that c =, 4c =, which results in c = and c 2 = Therefore the solution of the IVP is ( ) sin(t/2) x = e t/2 4 4 cos(t/2) 5 The general solution of the homogeneous problem is ( (c)) ( ) ( ) x 2 x (c) = c t + c 2 t 2, 2 2 which can be verified by substitution into the system of ODEs Since the vectors are linearly independent, a fundamental matrix is given by ( ) t 2t Ψ(t) = 2 2t t 2 The inverse of the fundamental matrix is Ψ (t) = ( ) t 2t 3 2t 2 t 2 Dividing both equations by t, we obtain ( ) 2 g(t) = t 3 t Proceeding with the method of variation of parameters, ( Ψ 2 3 (t)g(t) = t t 2 ) 3 3 t 4 3 t 2 +, 3 t 3 and ( 2 Ψ 5 (t)g(t) dt = t5 + 3 t2 2 3 t ) 6 t t 6 t 2 Hence a particular solution is obtained as ( ) 5 v = t4 + 3t t4 + 2t 3 2 The general solution is ( ) ( ) 2 x = c t + c 2 t Based on the hypotheses, ( ) 2 t 4 + ( ) ( ) 3 t 2 3/2 φ (t) = P(t)φ(t) + g(t) and v (t) = P(t)v(t) + g(t) Subtracting the two equations results in φ (t) v (t) = P(t)φ(t) P(t)v(t),

330 33 Chapter 7 Systems of First Order Linear Equations that is, [φ(t) v(t)] = P(t) [φ(t) v(t)] It follows that φ(t) v(t) is a solution of the homogeneous equation According to Theorem 742, φ(t) v(t) = c x () (t) + c 2 x (2) (t) + + c n x (n) (t) Hence φ(t) = u(t) + v(t), in which u(t) is the general solution of the homogeneous problem 7(a) Setting t = in Equation (34), x = Φ(t)x + Φ(t) t Φ (s)g(s)ds = Φ(t)x + t Φ(t)Φ (s)g(s)ds It was shown in Problem 5(c) in Section 77 that Φ(t)Φ (s) = Φ(t s) Therefore x = Φ(t)x + t Φ(t s)g(s)ds (b) The principal fundamental matrix is identified as Φ(t) = e At Hence x = e At x + t e A(t s) g(s)ds In Problem 27 of Section 36, the particular solution is given as y(t) = t t K(t s)g(s)ds, in which the kernel K(t) depends on the nature of the fundamental solutions 8 Similarly to Eq(43), here where G(s) = (si A)X(s) = G(s) + ( ) 2/(s + ) 3/s 2 ( α and si A = α 2 ), ( ) s + 2 s + 2 The transfer matrix is given by Eq(46): (si A) = From these equations we obtain that X(s) = (s + )(s + 3) ( 2(s+2) (s+) 2 (s+3) + 3 s 2 (s+)(s+3) + 2 (s+) 2 (s+3) + ( ) s + 2 s + 2 α(s+2) (s+)(s+3) + α 2 3(s+2) s 2 (s+)(s+3) + α (s+)(s+3) + (s+)(s+3) α2(s+2) (s+)(s+3) )

331 79 33 The inverse Laplace transform gives us that x(t) = ( 4+α+α 2 so α and α 2 should be chosen so that 2 e t + 4+3α 3α2 6 e 3t + t + te t α +α 2 2 e t + 4 3α+3α2 6 e 3t + 2t + te t α + α 2 2 = c and 4 + 3α 3α 2 6 = c This gives us α = ( 5 + 6c + 6c 2 )/6 and α 2 = c + c 2 3/6 ),

332 332 Chapter 7 Systems of First Order Linear Equations

333 C H A P T E R 8 Numerical Methods 8 2 The Euler formula for this problem is y n+ = y n + h(5 t n 3 y n ), in which t n = t + nh Since t =, we can also write y n+ = y n + 5nh 2 3h y n with y = 2 (a) Euler method with h = 5 : (b) Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n n = 4 n = 8 n = 2 n = 6 t n y n The backward Euler formula is y n+ = y n + h(5 t n+ 3 y n+ ) in which t n = t + nh Since t =, we can also write y n+ = y n + 5(n + )h 2 3h y n+, with y = 2 Solving for y n+, and choosing the positive root, we find that [ y n+ = 3 2 h + 2 ] 2 (2n + 29)h2 + 4y n 333

334 334 Chapter 8 Numerical Methods (c) Backward Euler method with h = 5 : n = 2 n = 4 n = 6 n = 8 t n y n (d) Backward Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The Euler formula for this problem is y n+ = y n + h(2 y n 3 t n ), in which t n = t + nh Since t =, y n+ = y n + 2hy n 3nh 2, with y = (a) Euler method with h = 5 : (b) Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n n = 4 n = 8 n = 2 n = 6 t n y n The backward Euler formula is y n+ = y n + h(2 y n+ 3 t n+ ), in which t n = t + nh Since t =, we can also write y n+ = y n + 2h y n+ 3(n + )h 2, with y = Solving for y n+, we find that y n+ = y n 3(n + )h 2 2h (c) Backward Euler method with h = 5 : n = 2 n = 4 n = 6 n = 8 t n y n (d) Backward Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n

335 The Euler formula is y n+ = y n + h [2 t n + e tn yn ] Since t n = t + nh and t =, we can also write y n+ = y n + 2nh 2 + h e nh yn, with y = (a) Euler method with h = 5 : n = 2 n = 4 n = 6 n = 8 t n y n (b) Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The backward Euler formula is y n+ = y n + h [2 t n+ + e tn+ yn+ ] Since t = and t n+ = (n + )h, we can also write y n+ = y n + 2h 2 (n + ) + h e (n+)h yn+, with y = This equation cannot be solved explicitly for y n+ At each step, given the current value of y n, the equation must be solved numerically for y n+ (c) Backward Euler method with h = 5 : n = 2 n = 4 n = 6 n = 8 t n y n (d) Backward Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The Euler formula for this problem is y n+ = y n + h(t 2 n y 2 n ) sin y n Here t = and t n = nh So that y n+ = y n + h(n 2 h 2 y 2 n) sin y n, with y = (a) Euler method with h = 5 : n = 2 n = 4 n = 6 n = 8 t n y n

336 336 Chapter 8 Numerical Methods (b) Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The backward Euler formula is y n+ = y n + h(t 2 n+ y 2 n+ ) sin y n+ Since t = and t n+ = (n + )h, we can also write y n+ = y n + h [ (n + ) 2 h 2 y 2 n+] sin yn+, with y = Note that this equation cannot be solved explicitly for y n+ Given y n, the transcendental equation must be solved numerically for y n+ y n+ + h y 2 n+ sin y n+ = y n + h(n + ) 2 h 2 (c) Backward Euler method with h = 5 : n = 2 n = 4 n = 6 n = 8 t n y n (d) Backward Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The Euler formula y n+ = y n + h(5 t n 3 y n ), in which t n = t + nh Since t =, we can also write y n+ = y n + 5nh 2 3h y n with y = 2 (a) Euler method with h = 25 : (b) Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n n = 4 n = 8 n = 2 n = 6 t n y n The backward Euler formula is y n+ = y n + h(5 t n+ 3 y n+ ), in which t n = t + nh Since t =, we can also write y n+ = y n + 5(n + )h 2 3h y n+ with

337 8 337 y = 2 Solving for y n+, and choosing the positive root, we find that [ y n+ = 3 2 h + 2 ] 2 (2n + 29)h2 + 4y n (c) Backward Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n (d) Backward Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The Euler formula is y n+ = y n + h [2 t n + e tn yn ] Since t n = t + nh and t =, we can also write y n+ = y n + 2nh 2 + h e nh yn, with y = (a) Euler method with h = 25 : (b) Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n n = 4 n = 8 n = 2 n = 6 t n y n The backward Euler formula is y n+ = y n + h [2 t n+ + e tn+ yn+ ] Since t = and t n+ = (n + )h, we can also write y n+ = y n + 2h 2 (n + ) + h e (n+)h yn+, with y = This equation cannot be solved explicitly for y n+ At each step, given the current value of y n, the equation must be solved numerically for y n+ (c) Backward Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n

338 338 Chapter 8 Numerical Methods (d) Backward Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The Euler formula is y n+ = y n + h(4 t n y n )/( + y 2 n) Since t n = t + nh and t =, we can also write y n+ = y n + (4h nh 2 y n )/( + y 2 n), with y = 2 (a) Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n (b) Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The backward Euler formula is y n+ = y n + h(4 t n+ y n+ )/( + yn+) 2 Since t[ = and t n+ = (n] + )h, we can also write y n+ ( + yn+) 2 = y n ( + yn+) 2 + 4h (n + )h 2 y n+, with y = 2 This equation cannot be solved explicitly for y n+ At each step, given the current value of y n, the equation must be solved numerically for y n+ (c) Backward Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n (d) Backward Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The Euler formula is y n+ = y n + h(y 2 n + 2 t n y n )/(3 + t 2 n) Since t n = t + nh and t =, we can also write y n+ = y n + (h y 2 n + 2 nh 2 y n )/(3 + n 2 h 2 ), with y = 5

339 8 339 (a) Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n (b) Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The backward Euler formula is y n+ = y n + h(y 2 n+ + 2 t n+ y n+ )/(3 + t 2 n+) Since t = and t n+ = (n + )h, we can also write y n+ [ 3 + (n + ) 2 h 2] h y 2 n+ = y n [ 3 + (n + ) 2 h 2] + 2(n + )h 2 y n+, with y = 5 Note that although this equation can be solved explicitly for y n+, it is also possible to use a numerical equation solver At each time step, given the current value of y n, the equation may be solved numerically for y n+ (c) Backward Euler method with h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n (d) Backward Euler method with h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The Euler formula for this problem is y n+ = y n + h( t n + 4 y n ), in which t n = t + nh Since t =, we can also write y n+ = y n + h nh 2 + 4h y n, with y = With h =, a total number of 2 iterations is needed to reach t = 2 With h =, a total of 2 iterations are necessary 4 The backward Euler formula is y n+ = y n + h( t n+ + 4 y n+ ) Since the equation is linear, we can solve for y n+ in terms of y n : y n+ = y n + h h t n+ 4h Here t = and y = With h =, a total number of 2 iterations is needed to reach t = 2 With h =, a total of 2 iterations are necessary

340 34 Chapter 8 Numerical Methods 8 Let φ(t) be a solution of the initial value problem The local truncation error for the Euler method, on the interval t n t t n+, is given by e n+ = 2 φ ( t n )h 2, where t n < t n < t n+ Since φ (t) = t 2 + [φ(t)] 2, it follows that φ (t) = 2t + 2 φ(t)φ (t) = 2t + 2t 2 φ(t) + 2 [φ(t)] 3 and so e n+ = [t n + t 2 nφ(t n ) + φ 3 (t n )]h 2 where t n < t n < t n+ 2 Given that φ(t) is a solution of the initial value problem, the local truncation error for the Euler method, on the interval t n t t n+, is e n+ = 2 φ ( t n )h 2, where t n < t n < t n+ Based on the ODE, φ (t) = t + φ(t), and hence φ (t) = + φ (t) 2 t + φ(t) = 2 t + φ(t) + 2 and so e n+ = 4 [ + ] h 2 tn + φ( t n ) where t n < t n < t n+ 2 Let φ(t) be a solution of the initial value problem The local truncation error for the Euler method, on the interval t n t t n+, is given by e n+ = 2 φ ( t n )h 2, where t n < t n < t n+ Since φ (t) = 2t + e t φ(t), it follows that [ φ (t) = 2 [φ(t) + t φ (t)] e t φ(t) = 2 φ(t) + 2t 2 + t e t φ(t)] e t φ(t) Hence e n+ = h 2 h2 2 [ φ( t n ) + 2 t 2 n + t n e tnφ( tn)] e tnφ( tn) 22(a) Direct integration yields φ(t) = (/5π) sin 5πt +

341 8 34 (b) n = n = n = 2 n = 3 t n y n 2 2 (c) n = n = n = 2 n = 3 n = 4 t n y n

342 342 Chapter 8 Numerical Methods (d) Since φ (t) = 5π sin 5πt, the local truncation error for the Euler method, on the interval t n t t n+, is given by In order to satisfy it is necessary that e n+ = 5πh2 2 e n+ 5πh2 2 sin 5π t n < 5, h < 5π 8 26(a) (b) = () = = (6) = 6 (c) = ( 926) = Rounding to three digits, a(b c) 224 Likewise, to three digits, ab 72 and ac 477 It follows that ab ac The improved Euler formula is y n+ = y n + h 2 (5 t n 3 y n ) + h 2 (5 t n+ 3 K n ), in which K n = y n + h(5 t n 3 y n ) Since t n = t + nh and t =, we can also write y n+ = y n + h 2 (5 nh 3 y n ) + h [5 (n + )h 3 ] K n, 2 with y = 2 (a) h = 5 : n = 2 n = 4 n = 6 n = 8 t n y n

343 (b) h = 25 : (c) h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n n = 8 n = 6 n = 24 n = 32 t n y n The improved Euler formula for this problem is y n+ = y n + h 2 (4 y n 3 t n 3 t n+ ) + h 2 (2 y n 3 t n ) Since t n = t + nh and t =, we can also write with y = y n+ = y n + 2h y n + h2 2 (4 y n 3 6n) 3nh 3, (a) h = 5 : n = 2 n = 4 n = 6 n = 8 t n y n (b) h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n (c) h = 25 : 5 The improved Euler formula is n = 8 n = 6 n = 24 n = 32 t n y n in which y n+ = y n + h y2 n + 2 t n y n 2(3 + t 2 n) K n = y n + h y2 n + 2 t n y n 3 + t 2 n + h K2 n + 2 t n+ K n 2(3 + t 2 n+ ),

344 344 Chapter 8 Numerical Methods Since t n = t + nh and t =, we can also write with y = 5 (a) h = 5 : y n+ = y n + h y2 n + 2nh y n 2(3 + n 2 h 2 ) + h K2 n + 2(n + )hk n 2 [3 + (n + ) 2 h 2 ], (b) h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n (c) h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n n = 8 n = 6 n = 24 n = 32 t n y n The improved Euler formula for this problem is y n+ = y n + h 2 (t2 n y 2 n) sin y n + h 2 (t2 n+ K 2 n) sin K n, in which K n = y n + h (t 2 n y 2 n) sin y n Since t n = t + nh and t =, we can also write with y = (a) h = 5 : y n+ = y n + h 2 (n2 h 2 y 2 n) sin y n + h 2 [ (n + ) 2 h 2 Kn 2 ] sin Kn, (b) h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n n = 4 n = 8 n = 2 n = 6 t n y n

345 (c) h = 25 : n = 8 n = 6 n = 24 n = 32 t n y n The improved Euler formula for this problem is y n+ = y n + h 2 (4 y n t n t n+ + ) + h 2 (2 y n t n + 5) Since t n = t + nh and t =, we can also write with y = (a) h = 25 : y n+ = y n + h(2 y n + 5) + h 2 (2 y n n) nh 3, (b) h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n The improved Euler formula is n = 4 n = 8 n = 2 n = 6 t n y n y n+ = y n + h 2 (5 t n 3 y n ) + h 2 (5 t n+ 3 K n ), in which K n = y n + h(5 t n 3 y n ) Since t n = t + nh and t =, we can also write y n+ = y n + h 2 (5 nh 3 y n ) + h [5 (n + )h 3 ] K n, 2 with y = 2 (a) h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n

346 346 Chapter 8 Numerical Methods (b) h = 25 : n = 4 n = 8 n = 2 n = 6 t n y n The improved Euler formula for this problem is y n+ = y n + h tn + y n + h 2 2 tn+ + K n, in which K n = y n + h t n + y n Since t n = t + nh and t =, we can also write with y = 3 (a) h = 25 : y n+ = y n + h 2 nh + yn + h 2 (n + )h + Kn, (b) h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n The improved Euler formula is n = 4 n = 8 n = 2 n = 6 t n y n y n+ = y n + h 2 [ 2 tn + e tn yn] + h 2 [ 2 tn+ + e tn+kn], in which K n = y n + h [2 t n + e tn yn ] Since t n = t + nh and t =, we can also write y n+ = y n + h [ 2 nh + e nh y n ] h ] + [2(n + )h + e (n+)hkn, 2 2 with y = (a) h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n

347 (b) h = 25 : 2 The improved Euler formula is in which n = 4 n = 8 n = 2 n = 6 t n y n y n+ = y n + h y2 n + 2 t n y n 2(3 + t 2 n) K n = y n + h y2 n + 2 t n y n 3 + t 2 n Since t n = t + nh and t =, we can also write with y = 5 (a) h = 25 : + h K2 n + 2 t n+ K n 2(3 + t 2 n+ ), y n+ = y n + h y2 n + 2nh y n 2(3 + n 2 h 2 ) + h K2 n + 2(n + )hk n 2 [3 + (n + ) 2 h 2 ], (b) h = 25 : n = 2 n = 4 n = 6 n = 8 t n y n n = 4 n = 8 n = 2 n = 6 t n y n The exact solution of the initial value problem is φ(t) = e2t Based on the result in Problem 4(c), the local truncation error for a linear differential equation is e n+ = 6 φ ( t n )h 3, where t n < t n < t n+ Since φ (t) = 4 e 2t, the local truncation error is Furthermore, with t n, It also follows that for h =, e n+ = 2 3 e2 tn h 3 e n+ 2 3 e2 h 3 e 2 3 e2 () 3 = 5 e2

348 348 Chapter 8 Numerical Methods Using the improved Euler method, with h =, we have y The exact value is given by φ() = 74 7 The exact solution of the initial value problem is given by φ(t) = 2 t + e2t Using the modified Euler method, the local truncation error for a linear differential equation is e n+ = 6 φ ( t n )h 3, where t n < t n < t n+ Since φ (t) = 8 e 2t, the local truncation error is e n+ = 4 3 e2 tn h 3 Furthermore, with t n, the local error is bounded by It also follows that for h =, e n+ 4 3 e2 h 3 e 4 3 e2 () 3 = 75 e2 Using the improved Euler method, with h =, we have y 27 The exact value is given by φ() = Using the Euler method, y = + (5 + 2 ) = 25 Using the improved Euler method, y = + 5(5 + 2) + 5(5 + 25) = 27 The estimated error is e = 2 The step size should be adjusted by a factor of 25/2 354 Hence the required step size is estimated as h ()(36) = 36 2 Using the Euler method, y = = Using the improved Euler method, y = = The estimated error is e = 3858 The step size should be adjusted by a factor of 25/ Hence the required step size is estimated as h ()(85) = 85 2 Using the Euler method, Using the improved Euler method, y = (5) y = 5 + (5) = (58334)2 + 2()(58334) 3 + () 2 = 548 The estimated error is e = 8 The local truncation error is less than the given tolerance The step size can be adjusted by a factor of 25/8 785 Hence it is possible to use a step size of h ()(785) 7

349 Assuming that the solution has continuous derivatives at least to the third order, φ(t n+ ) = φ(t n ) + φ (t n )h + φ (t n ) 2! where t n < t n < t n+ Suppose that y n = φ(t n ) (a) The local truncation error is given by e n+ = φ(t n+ ) y n+ h 2 + φ (t n ) h 3, 3! The modified Euler formula is defined as [ y n+ = y n + h f t n + 2 h, y n + ] 2 h f(t n, y n ) Observe that φ (t n ) = f(t n, φ(t n )) = f(t n, y n ) It follows that e n+ = φ(t n+ ) y n+ = = h f(t n, y n ) + φ (t n ) 2! (b) As shown in Problem 4(b), h 2 + φ [ (t n ) h 3 h f t n + 3! 2 h, y n + ] 2 h f(t n, y n ) φ (t n ) = f t (t n, y n ) + f y (t n, y n )f(t n, y n ) Furthermore, [ f t n + 2 h, y n + ] 2 h f(t n, y n ) = f(t n, y n ) + f t (t n, y n ) h 2 + f y(t n, y n ) k+ + [ ] h 2 2! 4 f tt + hk f ty + k 2 f yy, t=ξ, y=η in which k = 2 h f(t n, y n ) and t n < ξ < t n + h/2, y n < η < y n + k Therefore e n+ = φ (t n ) h 3 h [ ] h 2 3! 2! 4 f tt + hk f ty + k 2 f yy t=ξ, y=η Note that each term in the brackets has a factor of h 2 Hence the local truncation error is proportional to h 3 (c) If f(t, y) is linear, then f tt = f ty = f yy =, and e n+ = φ (t n ) h 3 3! 23 The modified Euler formula for this problem is y n+ = y n + h {3 + t n + 2 [ h y n + ]} 2 h(3 + t n y n ) = y n + h(3 + t n y n ) + h2 2 (y n t n 2)

350 35 Chapter 8 Numerical Methods Since t n = t + nh and t =, we can also write y n+ = y n + h(3 + nh y n ) + h2 2 (y n nh 2), with y = Setting h = 5, we obtain the following values: n = 2 n = 4 n = 6 n = 8 t n y n The modified Euler formula is [ y n+ = y n + h 2y n 3t n 3 ] 2 h + h(2y n 3t n ) = y n + h(2y n 3t n ) + h2 2 (4y n 6t n 3) Since t n = t + nh and t =, we can also write y n+ = y n + h(2y n 3nh) + h2 2 (4y n 6 nh 3), with y = Setting h = 5, we obtain: n = 2 n = 4 n = 6 n = 8 t n y n The modified Euler formula for this problem is ] y n+ = y n + h [2t n + h + e (tn+ h 2 )Kn, in which K n = y n + (h/2) [2t n + e tnyn ] Now t n = t + nh, with t = and y = Setting h = 5, we obtain the following values : n = 2 n = 4 n = 6 n = 8 t n y n Let f(t, y) be linear in both variables The improved Euler formula is y n+ = y n + 2 h [f(t n, y n ) + f(t n + h, y n + hf(t n, y n ))] = y n + 2 hf(t n, y n ) + 2 hf(t n, y n ) + 2 hf [h, hf(t n, y n )] = y n + hf(h, y n ) + 2 hf [h, hf(t n, y n )] The modified Euler formula is [ y n+ = y n + hf t n + 2 h, y n + ] 2 hf(t n, y n ) = y n + hf(t n, y n ) + hf [ 2 h, ] 2 hf(t n, y n )

351 83 35 Since f(t, y) is linear in both variables, [ f 2 h, ] 2 hf(t n, y n ) = 2 f [h, hf(t n, y n )] 83 The ODE is linear, with f(t, y) = 3 + t y The Runge-Kutta algorithm requires the evaluations k n = f(t n, y n ) k n2 = f(t n + 2 h, y n + 2 hk n) k n3 = f(t n + 2 h, y n + 2 hk n2) k n4 = f(t n + h, y n + hk n3 ) The next estimate is given as the weighted average y n+ = y n + h 6 (k n + 2 k n2 + 2 k n3 + k n4 ) (a) For h = : (b) For h = 5 : n = n = 2 n = 3 n = 4 t n y n n = 2 n = 4 n = 6 n = 8 t n y n The exact solution of the IVP is y(t) = 2 + t e t 2 In this problem, f(t, y) = 5t 3 y At each time step, the Runge-Kutta algorithm requires the evaluations k n = f(t n, y n ) k n2 = f(t n + 2 h, y n + 2 hk n) k n3 = f(t n + 2 h, y n + 2 hk n2) k n4 = f(t n + h, y n + hk n3 )

352 352 Chapter 8 Numerical Methods The next estimate is given as the weighted average y n+ = y n + h 6 (k n + 2 k n2 + 2 k n3 + k n4 ) (a) For h = : (b) For h = 5 : n = n = 2 n = 3 n = 4 t n y n n = 2 n = 4 n = 6 n = 8 t n y n The exact solution of the IVP is given implicitly as 2 (2 y + 5t) 5 (t y ) 2 = 52 3 The ODE is linear, with f(t, y) = 2y 3t The Runge-Kutta algorithm requires the evaluations k n = f(t n, y n ) k n2 = f(t n + 2 h, y n + 2 hk n) k n3 = f(t n + 2 h, y n + 2 hk n2) k n4 = f(t n + h, y n + hk n3 ) The next estimate is given as the weighted average y n+ = y n + h 6 (k n + 2 k n2 + 2 k n3 + k n4 ) (a) For h = : (b) For h = 5 : n = n = 2 n = 3 n = 4 t n y n n = 2 n = 4 n = 6 n = 8 t n y n The exact solution of the IVP is y(t) = e 2t /4 + 3t/2 + 3/4

353 In this problem, f(t, y) = (y 2 + 2ty)/(3 + t 2 ) The Runge-Kutta algorithm requires the evaluations k n = f(t n, y n ) k n2 = f(t n + 2 h, y n + 2 hk n) k n3 = f(t n + 2 h, y n + 2 hk n2) k n4 = f(t n + h, y n + hk n3 ) The next estimate is given as the weighted average y n+ = y n + h 6 (k n + 2 k n2 + 2 k n3 + k n4 ) (a) For h = : (b) For h = 5 : n = n = 2 n = 3 n = 4 t n y n n = 2 n = 4 n = 6 n = 8 t n y n The exact solution of the IVP is y(t) = (3 + t 2 )/(6 t) 6 In this problem, f(t, y) = (t 2 y 2 ) sin y At each time step, the Runge-Kutta algorithm requires the evaluations k n = f(t n, y n ) k n2 = f(t n + 2 h, y n + 2 hk n) k n3 = f(t n + 2 h, y n + 2 hk n2) k n4 = f(t n + h, y n + hk n3 ) The next estimate is given as the weighted average y n+ = y n + h 6 (k n + 2 k n2 + 2 k n3 + k n4 ) (a) For h = : n = n = 2 n = 3 n = 4 t n y n

354 354 Chapter 8 Numerical Methods (b) For h = 5 : 7(a) For h = : n = 2 n = 4 n = 6 n = 8 t n y n (b) For h = 5 : n = 5 n = n = 5 n = 2 t n y n n = n = 2 n = 3 n = 4 t n y n The exact solution of the IVP is y(t) = e 2t + t/2 8 See Problem 2 for the exact solution (a) For h = : (b) For h = 5 : n = 5 n = n = 5 n = 2 t n y n (a) For h = : n = n = 2 n = 3 n = 4 t n y n (b) For h = 5 : n = 5 n = n = 5 n = 2 t n y n n = n = 2 n = 3 n = 4 t n y n

355 The exact solution is given implicitly as [ ] 2 ln + 2 t + y 2 arctanh t + y = t arctanh 3 y + t See Problem 4 (a) For h = : (b) For h = 5 : n = 5 n = n = 5 n = 2 t n y n n = n = 2 n = 3 n = 4 t n y n See Problem 5 for the exact solution (a) For h = : (b) For h = 5 : n = 5 n = n = 5 n = 2 t n y n n = n = 2 n = 3 n = 4 t n y n The ODE is linear, with f(t, y) = t + 4y The Runge-Kutta algorithm requires the evaluations k n = f(t n, y n ) k n2 = f(t n + 2 h, y n + 2 hk n) k n3 = f(t n + 2 h, y n + 2 hk n2) k n4 = f(t n + h, y n + hk n3 ) The next estimate is given as the weighted average y n+ = y n + h 6 (k n + 2 k n2 + 2 k n3 + k n4 ) The exact solution of the IVP is y(t) = 9e 4t /6 + t/4 3/6

356 356 Chapter 8 Numerical Methods (a) For h = : n = 5 n = n = 5 n = 2 t n y n (b) For h = 5 : n = n = 2 n = 3 n = 4 t n y n (a) (b) For the integral curve starting at (, ), the slope turns infinite near t M 5 Note that the exact solution of the IVP is defined implicitly as y 3 4y = t 3 (c) Using the classic Runge-Kutta algorithm with the given values of h, we obtain

357 the values h = h = 5 h = 25 h = n = 2 n = 3 n = 4 n = 5 n = 6 t n y n n = 26 n = 28 n = 3 n = 32 n = 34 t n y n n = 54 n = 56 n = 58 n = 6 n = 62 t n y n n = 42 n = 43 n = 44 n = 45 n = 46 t n y n Based on the direction field, the solution should decrease monotonically to the limiting value y = 2/ 3 In the following table, the value of t M corresponds to the approximate time in the iteration process that the calculated values begin to increase h t M (d) Numerical values will continue to be generated, although they will not be associated with the integral curve starting at (, ) These values are approximations to nearby integral curves (e) We consider the solution associated with the initial condition y() = The exact solution is given by y 3 4y = t 3 3

358 358 Chapter 8 Numerical Methods For the integral curve starting at (, ), the slope becomes infinite near t M 2 In the following table, the values of t M corresponds to the approximate time in the iteration process that the calculated values begin to increase h t M (a) Using the notation f n = f(t n, y n ), the predictor formula is y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) We use the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 2 3 y n n = 4 (pre) n = 4 (cor) n = 5 (pre) n = 5 (cor) t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) In this problem, f n+ = 3 + t n+ y n+ Since the ODE is linear, we can solve for y n+ = h [24 y n + 27 h + 9h t n+ + h(9 f n 5 f n + f n 2 )] n = 4 n = 5 t n 4 5 y n (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ]

359 In this problem, f n+ = 3 + t n+ y n+ Since the ODE is linear, we can solve for y n+ = h [36 h + 2h t n y n 36 y n + 6 y n 2 3 y n 3 ] n = 4 n = 5 t n 4 5 y n The exact solution of the IVP is y(t) = 2 + t e t 2(a) Using the notation f n = f(t n, y n ), the predictor formula is y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) We use the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 2 3 y n n = 4 (pre) n = 4 (cor) n = 5 (pre) n = 5 (cor) t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) In this problem, f n+ = 5t n+ 3 y n+ Since the ODE is nonlinear, an equation solver is needed to approximate the solution of y n+ = y n + h 24 [ 45tn+ 27 y n+ + 9 f n 5 f n + f n 2 ] at each time step We obtain the approximate values: n = 4 n = 5 t n 4 5 y n (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ]

360 36 Chapter 8 Numerical Methods Since the ODE is nonlinear, an equation solver is used to approximate the solution of y n+ = [ 48 yn 36 y n + 6 y n 2 3 y n 3 + 2h(5t n+ 3 y n+ ) ] 25 at each time step n = 4 n = 5 t n 4 5 y n The exact solution of the IVP is given implicitly by 2 (2 y + 5t) 5 (t y ) 2 = 52 3(a) The predictor formula is y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) Using the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 2 3 y n n = 4 (pre) n = 4 (cor) n = 5 (pre) n = 5 (cor) t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) In this problem, f n+ = 2 y n+ 3 t n+ Since the ODE is linear, we can solve for y n+ = 24 8 h [24 y n 27h t n+ + h(9 f n 5 f n + f n 2 )] n = 4 n = 5 t n 4 5 y n (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ]

361 84 36 In this problem, f n+ = 2 y n+ 3 t n+ Since the ODE is linear, we can solve for y n+ = h [48 y n 36 y n + 6 y n 2 3 y n 3 36h t n+ ] n = 4 n = 5 t n 4 5 y n The exact solution of the IVP is y(t) = e 2t /4 + 3t/2 + 3/4 5(a) The predictor formula is y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) Using the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 2 3 y n n = 4 (pre) n = 4 (cor) n = 5 (pre) n = 5 (cor) t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is In this problem, y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) f n+ = y2 n+ + 2 t n+ y n+ 3 + t 2 n+ Since the ODE is nonlinear, an equation solver is needed to approximate the solution of y n+ = y n + h [ ] 9 y2 n+ + 2 t n+ y n t f n 5 f n + f n 2 n+ at each time step n = 4 n = 5 t n 4 5 y n (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ]

362 362 Chapter 8 Numerical Methods Since the ODE is nonlinear, an equation solver is needed to approximate the solution of y n+ = [ ] 48 y n 36 y n + 6 y n 2 3 y n 3 + 2h y2 n+ + 2 t n+ y n t 2 n+ at each time step We obtain the approximate values: n = 4 n = 5 t n 4 5 y n The exact solution of the IVP is y(t) = (3 + t 2 )/(6 t) 6(a) The predictor formula is y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) We use the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 2 3 y n n = 4 (pre) n = 4 (cor) n = 5 (pre) n = 5 (cor) t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) In this problem, f n+ = (t 2 n+ y 2 n+) sin y n+ Since the ODE is nonlinear, we obtain the implicit equation y n+ = y n + h 24 [ 9(t 2 n+ y 2 n+) sin y n+ + 9 f n 5 f n + f n 2 ] n = 4 n = 5 t n 4 5 y n (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ]

363 Since the ODE is nonlinear, we obtain the implicit equation y n+ = 25 [ 48 yn 36 y n + 6 y n 2 3 y n 3 + 2h(t 2 n+ y 2 n+) sin y n+ ] 8(a) The predictor formula is n = 4 n = 5 t n 4 5 y n y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) We use the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 5 5 y n n = n = 2 n = 3 n = 4 t n y n (b) Since the ODE is nonlinear, an equation solver is needed to approximate the solution of y n+ = y n + h 24 [ 45tn+ 27 y n+ + 9 f n 5 f n + f n 2 ] at each time step We obtain the approximate values: n = n = 2 n = 3 n = 4 t n y n (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ] Since the ODE is nonlinear, an equation solver is needed to approximate the solution of y n+ = [ 48 yn 36 y n + 6 y n 2 3 y n 3 + 2h(5t n+ 3 y n+ ) ] 25

364 364 Chapter 8 Numerical Methods at each time step n = n = 2 n = 3 n = 4 t n y n The exact solution of the IVP is given implicitly by 2 (2 y + 5t) 5 (t y ) 2 = 52 9(a) The predictor formula is y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) Using the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 5 5 y n n = n = 2 n = 3 n = 4 t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) In this problem, f n+ = t n+ + y n+ Since the ODE is nonlinear, an equation solver must be implemented in order to approximate the solution of y n+ = y n + h [ 9 ] t n+ + y n+ + 9 f n 5 f n + f n 2 24 at each time step n = n = 2 n = 3 n = 4 t n y n (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ]

365 Since the ODE is nonlinear, an equation solver is needed to approximate the solution of y n+ = [48 y n 36 y n + 6 y n 2 3 y n 3 + 2h ] t n+ + y n+ 25 at each time step n = n = 2 n = 3 n = 4 t n y n The exact solution is given implicitly by [ ] 2 ln + 2 t + y 2 arctanh t + y = t arctanh 3 y + t (a) The predictor formula is y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) We use the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 5 5 y n n = n = 2 n = 3 n = 4 t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) In this problem, f n+ = 2 t n+ + e tn+ yn+ Since the ODE is nonlinear, an equation solver must be implemented in order to approximate the solution of y n+ = y n + h 24 at each time step { 9 [ 2 tn+ + e tn+ yn+] + 9 f n 5 f n + f n 2 } n = n = 2 n = 3 n = 4 t n y n

366 366 Chapter 8 Numerical Methods (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ] Since the ODE is nonlinear, we obtain the implicit equation y n+ = { 48 yn 36 y n + 6 y n 2 3 y n 3 + 2h [ 2 t n+ + e yn+]} tn+ 25 (a) The predictor formula is n = n = 2 n = 3 n = 4 t n y n y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) Using the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 5 5 y n n = n = 2 n = 3 n = 4 t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is In this problem, y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) f n+ = 4 t n+ y n+ + y 2 n+ Since the differential equation is nonlinear, an equation solver is used to approximate the solution of y n+ = y n + h [ 9 4 t ] n+ y n yn f n 5 f n + f n 2 at each time step n = n = 2 n = 3 n = 4 t n y n

367 (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ] Since the ODE is nonlinear, an equation solver must be implemented in order to approximate the solution of y n+ = [ 48 y n 36 y n + 6 y n 2 3 y n 3 + 2h 4 t ] n+ y n yn+ 2 at each time step n = n = 2 n = 3 n = 4 t n y n (a) The predictor formula is y n+ = y n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) With f n+ = f(t n+, y n+ ), the corrector formula is y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) We use the starting values generated by the Runge-Kutta method: n = n = n = 2 n = 3 t n 5 5 y n n = n = 2 n = 3 n = 4 t n y n (b) With f n+ = f(t n+, y n+ ), the fourth order Adams-Moulton formula is In this problem, y n+ = y n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) f n+ = y2 n+ + 2 t n+ y n+ 3 + t 2 n+ Since the ODE is nonlinear, an equation solver is needed to approximate the solution of y n+ = y n + h [ ] 9 y2 n+ + 2 t n+ y n t f n 5 f n + f n 2 n+

368 368 Chapter 8 Numerical Methods at each time step n = n = 2 n = 3 n = 4 t n y n (c) The fourth order backward differentiation formula is y n+ = 25 [48 y n 36 y n + 6 y n 2 3 y n 3 + 2hf n+ ] Since the ODE is nonlinear, we obtain the implicit equation y n+ = 25 [ 48 y n 36 y n + 6 y n 2 3 y n 3 + 2h y2 n+ + 2 t n+ y n+ 3 + t 2 n+ n = n = 2 n = 3 n = 4 t n y n The exact solution of the IVP is y(t) = (3 + t 2 )/(6 t) 3 Both Adams methods entail the approximation of f(t, y), on the interval [t n, t n+ ], by a polynomial Approximating φ (t) = P (t) A, which is a constant polynomial, we have φ(t n+ ) φ(t n ) = tn+ t n A dt = A(t n+ t n ) = Ah Setting A = λ f n + ( λ)f n, where λ, we obtain the approximation y n+ = y n + h [λ f n + ( λ)f n ] An appropriate choice of λ yields the familiar Euler formula Similarly, setting A = λ f n + ( λ)f n+, where λ, we obtain the approximation y n+ = y n + h [λ f n + ( λ)f n+ ] ] 4 For a third order Adams-Bashforth formula, we approximate f(t, y), on the interval [t n, t n+ ], by a quadratic polynomial using the points (t n 2, y n 2 ), (t n, y n ) and (t n, y n ) Let P 3 (t) = At 2 + Bt + C We obtain the system of equations At 2 n 2 + Bt n 2 + C = f n 2 At 2 n + Bt n + C = f n At 2 n + Bt n + C = f n

369 For computational purposes, assume that t =, and t n = nh It follows that We then have y n+ y n = which yields A = f n 2f n + f n 2 2 h 2 B = (3 2n)f n + (4n 4)f n + ( 2n)f n 2 2 h C = n2 3n tn+ t n f n + (2n n 2 )f n + n2 n f n 2 2 [ At 2 + Bt + C ] dt = Ah 3 (n 2 + n + 3 ) + Bh2 (n + 2 ) + Ch, y n+ y n = h 2 (23 f n 6 f n + 5 f n 2 ) 5 For a third order Adams-Moulton formula, we approximate f(t, y), on the interval [t n, t n+ ], by a quadratic polynomial using the points (t n, y n ), (t n, y n ) and (t n+, y n+ ) Let P 3 (t) = αt 2 + βt + γ This time we obtain the system of algebraic equations αt 2 n + βt n + γ = f n αt 2 n + βt n + γ = f n αt 2 n+ + βt n+ + γ = f n+ For computational purposes, again assume that t =, and t n = nh It follows that α = f n 2f n + f n+ 2 h 2 We then have y n+ y n = which results in β = (2n + )f n + 4nf n + ( 2n)f n+ 2 h γ = n2 + n 2 tn+ t n f n + ( n 2 )f n + n2 n f n+ 2 [ αt 2 + βt + γ ] dt = αh 3 (n 2 + n + 3 ) + βh2 (n + 2 ) + γh, y n+ y n = h 2 (5f n+ + 8f n f n ) 85 In vector notation, the initial value problem can be written as ( ) ( ) ( ) d x x + y + t =, x() = dt y 4x 2y

370 37 Chapter 8 Numerical Methods (a) The Euler formula is That is, ( xn+ y n+ ) = ( xn With h =, we obtain the values y n ) + h ( xn + y n + t n 4x n 2y n x n+ = x n + h(x n + y n + t n ) y n+ = y n + h(4x n 2y n ) ) n = 2 n = 4 n = 6 n = 8 n = t n x n y n (b) The Runge-Kutta method uses the following intermediate calculations: k n = (x n + y n + t n, 4x n 2y n ) T [ k n2 = x n + h 2 k n + y n + h 2 k2 n + t n + h 2, 4(x n + h 2 k n) 2(y n + h 2 k2 n) k n3 = [ x n + h 2 k n2 + y n + h 2 k2 n2 + t n + h 2, 4(x n + h 2 k n2) 2(y n + h ] T 2 k2 n2) k n4 = [ x n + hk n3 + y n + hk 2 n3 + t n + h, 4(x n + hk n3) 2(y n + hk 2 n3) ] T With h = 2, we obtain the values: n = n = 2 n = 3 n = 4 n = 5 t n x n y n (c) With h =, we obtain n = 2 n = 4 n = 6 n = 8 n = t n x n y n The exact solution of the IVP is ] T 3(a) The Euler formula is ( ) xn+ = y n+ x(t) = e 2t e 3t 3 t 2 9 y(t) = e 2t 8 9 e 3t 2 3 t 9 ( xn y n ) + h ( tn x n y n x n )

371 85 37 That is, x n+ = x n + h( t n x n y n ) y n+ = y n + h(x n ) With h =, we obtain the values n = 2 n = 4 n = 6 n = 8 n = t n x n y n (b) The Runge-Kutta method uses the following intermediate calculations: k n = ( t n x n y n, x n ) T [ k n2 = (t n + h 2 )(x n + h 2 k n) (y n + h 2 k2 n ), x n + h ] T 2 k n [ k n3 = (t n + h 2 )(x n + h 2 k n2) (y n + h 2 k2 n2), x n + h ] T 2 k n2 k n4 = [ (t n + h)(x n + hk n3) (y n + hk 2 n3 ), x n + hk n3] T With h = 2, we obtain the values: n = n = 2 n = 3 n = 4 n = 5 t n x n y n (c) With h =, we obtain n = 2 n = 4 n = 6 n = 8 n = t n x n y n (a) The Euler formula gives With h =, we obtain the values x n+ = x n + h(x n y n + x n y n ) y n+ = y n + h(3x n 2y n x n y n ) n = 2 n = 4 n = 6 n = 8 n = t n x n y n

372 372 Chapter 8 Numerical Methods (b) Given f(t, x, y) = x y + x y g(t, x, y) = 3x 2y x y, the Runge-Kutta method uses the following intermediate calculations: k n = [f(t n, x n, y n ), g(t n, x n, y n )] T [ k n2 = f(t n + h 2, x n + h 2 k n, y n + h 2 k2 n), g(t n + h 2, x n + h 2 k n, y n + h 2 k2 n) k n3 = [ f(t n + h 2, x n + h 2 k n2, y n + h 2 k2 n2), g(t n + h 2, x n + h 2 k n2, y n + h ] T 2 k2 n2) k n4 = [ f(t n + h, x n + hk n3, y n + hk 2 n3), g(t n + h, x n + hk n3, y n + hk 2 n3) ] T With h = 2, we obtain the values: ] T n = n = 2 n = 3 n = 4 n = 5 t n x n y n (c) With h =, we obtain n = 2 n = 4 n = 6 n = 8 n = t n x n y n (a) The Euler formula gives x n+ = x n + h [x n ( 5 x n 5 y n )] y n+ = y n + h [y n ( x n )] With h =, we obtain the values n = 2 n = 4 n = 6 n = 8 n = t n x n y n (b) Given f(t, x, y) = x( 5 x 5 y) g(t, x, y) = y( x),

373 the Runge-Kutta method uses the following intermediate calculations: k n = [f(t n, x n, y n ), g(t n, x n, y n )] T [ k n2 = f(t n + h 2, x n + h 2 k n, y n + h 2 k2 n), g(t n + h 2, x n + h 2 k n, y n + h 2 k2 n) k n3 = [ f(t n + h 2, x n + h 2 k n2, y n + h 2 k2 n2), g(t n + h 2, x n + h 2 k n2, y n + h ] T 2 k2 n2) k n4 = [ f(t n + h, x n + hk n3, y n + hk 2 n3), g(t n + h, x n + hk n3, y n + hk 2 n3) ] T With h = 2, we obtain the values: ] T n = n = 2 n = 3 n = 4 n = 5 t n x n y n (c) With h =, we obtain n = 2 n = 4 n = 6 n = 8 n = t n x n y n (a) The Euler formula gives With h =, we obtain the values x n+ = x n + h [ e xn+yn cos x n ] y n+ = y n + h [sin(x n 3 y n )] n = 2 n = 4 n = 6 n = 8 n = t n x n y n (b) The Runge-Kutta method uses the following intermediate calculations: k n = [f(t n, x n, y n ), g(t n, x n, y n )] T [ k n2 = f(t n + h 2, x n + h 2 k n, y n + h 2 k2 n), g(t n + h 2, x n + h 2 k n, y n + h 2 k2 n) k n3 = [ f(t n + h 2, x n + h 2 k n2, y n + h 2 k2 n2), g(t n + h 2, x n + h 2 k n2, y n + h ] T 2 k2 n2) k n4 = [ f(t n + h, x n + hk n3, y n + hk 2 n3), g(t n + h, x n + hk n3, y n + hk 2 n3) ] T ] T

374 374 Chapter 8 Numerical Methods With h = 2, we obtain the values: n = n = 2 n = 3 n = 4 n = 5 t n x n y n (c) With h =, we obtain n = 2 n = 4 n = 6 n = 8 n = t n x n y n The Runge-Kutta method uses the following intermediate calculations: k n = [x n 4 y n, x n + y n ] T [ k n2 = x n + h 2 k n 4(y n + h 2 k2 n), (x n + h 2 k n) + y n + h ] T 2 k2 n [ k n3 = x n + h 2 k n2 4(y n + h 2 k2 n2), (x n + h 2 k n2) + y n + h ] T 2 k2 n2 k n4 = [ x n + hk n3 4(y n + hk 2 n3), (x n + hk n3) + y n + hk 2 n3] T Using h = 5, we obtain the following values: n = 4 n = 8 n = 2 n = 6 n = 2 t n x n y n Using h = 25, we obtain the following values: n = 8 n = 6 n = 24 n = 32 n = 4 t n x n y n The exact solution is given by φ(t) = e t + e 3t 2 and the associated tabulated values:, ψ(t) = e t e 3t, 4 n = 5 n = n = 5 n = 2 n = 25 t n φ(t n ) ψ(t n )

375 The predictor formulas are x n+ = x n + h 24 (55 f n 59 f n + 37 f n 2 9 f n 3 ) y n+ = y n + h 24 (55 g n 59 g n + 37 g n 2 9 g n 3 ) With f n+ = x n+ 4 y n+ and g n+ = x n+ + y n+, the corrector formulas are x n+ = x n + h 24 (9 f n+ + 9 f n 5 f n + f n 2 ) y n+ = y n + h 24 (9 g n+ + 9 g n 5 g n + g n 2 ) We use the starting values from the exact solution: n = n = n = 2 n = 3 t n 2 3 x n y n One time step using the predictor-corrector method results in the approximate values: n = 4 (pre) n = 4 (cor) t n 4 4 x n y n The solution of the initial value problem is φ(t) = e t + t (a,b) Based on the exact solution, the local truncation error for both of the Euler methods is e loc 4 2 e tn h 2 Hence e n 5 h 2, for all < t n < Furthermore, the local truncation error is greatest near t = Therefore e 5 h 2 < 5 for h < 3 Now the truncation error accumulates at each time step Therefore the actual time step should be much smaller than h 3 For example, with h = 25, we obtain the data Euler BEuler φ(t) t = t =

376 376 Chapter 8 Numerical Methods Note that the total number of time steps needed to reach t = is N = 4 (c) Using the Runge-Kutta method, comparisons are made for several values of h; h = 5 is sufficient h = : h = 5 : φ(t) y n y n φ(t n ) t = t = φ(t) y n y n φ(t n ) t = t = (a) Using the method of undetermined coefficients, it is easy to show that the general solution of the ODE is y(t) = c e λt + t 2 Imposing the initial condition, it follows that c = and hence the solution of the IVP is φ(t) = t 2 (b) Using the Runge-Kutta method, with h =, numerical solutions are generated for various values of λ : λ = : λ = : φ(t) y n y n φ(t n ) t = t = t = t = λ = 2 : φ(t) y n y n φ(t n ) t = t = t = t = φ(t) y n y n φ(t n ) t = t = t = t =

377 λ = 5 : φ(t) y n y n φ(t n ) t = t = t = t = The following table shows the calculated value, y, at the first time step: φ(t) y (λ = ) y (λ = ) y (λ = 2) y (λ = 5) (c) Referring back to the exact solution given in part (a), if a nonzero initial condition, say y() = ε, is specified, the solution of the IVP becomes φ ε (t) = ε e λt + t 2 We then have φ(t) φ ε (t) = ε e λt It is evident that for any t >, lim φ(t) φ ε(t) = λ This implies that virtually any error introduced early in the calculations will be magnified as λ The initial value problem is inherently unstable

378 378 Chapter 8 Numerical Methods

379 C H A P T E R 9 Nonlinear Differential Equations and Stability 9 2(a) Setting x= ξ e rt results in the algebraic equations ( ) ( ) ( ) 5 r ξ = 3 r For a nonzero solution, we must have det(a r I) = r 2 6 r + 8 = The roots of the characteristic equation are r = 2 and r 2 = 4 For r = 2, the system of equations reduces to 3ξ = ξ 2 The corresponding eigenvector is ξ () = (, 3) T Substitution of r = 4 results in the single equation ξ = ξ 2 A corresponding eigenvector is ξ (2) = (, ) T (b) The eigenvalues are real and positive, hence the critical point is an unstable node ξ 2 379

380 38 Chapter 9 Nonlinear Differential Equations and Stability (c,d) 3(a) Solution of the ODE requires analysis of the algebraic equations ( ) ( ) ( ) 2 r ξ = 3 2 r For a nonzero solution, we must have det(a r I) = r 2 = The roots of the characteristic equation are r = and r 2 = For r =, the system of equations reduces to ξ = ξ 2 The corresponding eigenvector is ξ () = (, ) T Substitution of r = results in the single equation 3 ξ ξ 2 = A corresponding eigenvector is ξ (2) = (, 3) T (b) The eigenvalues are real, with r r 2 < Hence the critical point is an unstable saddle point (c,d) ξ 2 5(a) The characteristic equation is given by r 5 3 r = r2 + 2 r + 2 = The equation has complex roots r = + i and r 2 = i For r = + i, the components of the solution vector must satisfy ξ (2 + i)ξ 2 = Thus the corresponding eigenvector is ξ () = (2 + i, ) T Substitution of r = i results in the single equation ξ (2 i)ξ 2 = A corresponding eigenvector is ξ (2) = (2 i, ) T

381 9 38 (b) The eigenvalues are complex conjugates, with negative real part Hence the origin is an asymptotically stable spiral (c,d) 6(a) Solution of the ODEs is based on the analysis of the algebraic equations ( ) ( ) ( ) 2 r 5 ξ = 2 r For a nonzero solution, we require that det(a r I) = r 2 + = The roots of the characteristic equation are r = ±i Setting r = i, the equations are equivalent to ξ (2 + i)ξ 2 = The eigenvectors are ξ () = (2 + i, ) T and ξ (2) = (2 i, ) T (b) The eigenvalues are purely imaginary Hence the critical point is a stable center (c,d) ξ 2 8(a) The characteristic equation is given by r /4 r = (r + )(r + /4) =, with roots r = and r 2 = /4 For r =, the components of the solution vector must satisfy ξ 2 = Thus the corresponding eigenvector is ξ () = (, ) T Substitution of r = /4 results in the single equation 3ξ /4 + ξ 2 = A corresponding eigenvector is ξ (2) = (4, 3) T

382 382 Chapter 9 Nonlinear Differential Equations and Stability (b) The eigenvalues are real and both negative asymptotically stable node Hence the critical point is an (c,d) 9(a) Solution of the ODEs is based on the analysis of the algebraic equations ( ) ( ) ( ) 3 r 4 ξ = r For a nonzero solution, we require that det(a r I) = r 2 2 r + = The single root of the characteristic equation is r = Setting r =, the components of the solution vector must satisfy ξ 2 ξ 2 = A corresponding eigenvector is ξ = (2, ) T (b) Since there is only one linearly independent eigenvector, the critical point is an unstable, improper node (c,d) ξ 2 (a) The characteristic equation is (r + ) 2 =, with double root r = It is easy to see that the two linearly independent eigenvectors are ξ () = (, ) T and ξ (2) = (, ) T (b) Since there are two linearly independent eigenvectors, the critical point is an asymptotically stable proper node

383 9 383 (c,d) 2(a) Setting x= ξ e rt results in the algebraic equations ( ) ( ) 2 r 5/2 ξ = 9/5 r ξ 2 ( ) For a nonzero solution, we require that det(a r I) = r 2 r + 5/2 = The roots of the characteristic equation are r = /2 ± 3i/2 Substituting r = /2 3i/2, the equations reduce to (3 + 3i)ξ 5 ξ 2 = Therefore the two eigenvectors are ξ () = (5, 3 + 3i) T and ξ (2) = (5, 3 3i) T (b) Since the eigenvalues are complex, with positive real part, the critical point is an unstable spiral (c,d) 4 Setting x =, that is, ( ) ( ) 2 2 x =, 2 we find that the critical point is x = (, ) T With the change of dependent variable, x=x +u, the differential equation can be written as ( ) du 2 dt = u 2

384 384 Chapter 9 Nonlinear Differential Equations and Stability The critical point for the transformed equation is the origin Setting u= ξ e rt results in the algebraic equations ( 2 ) ( ) ( ) r ξ = 2 r For a nonzero solution, we require that det(a r I) = r 2 + 4r + 3 = The roots of the characteristic equation are r = 3, Hence the critical point is an asymptotically stable node 5 Setting x =, that is, ξ 2 ( ) ( ) x =, 2 5 we find that the critical point is x = ( 2, ) T With the change of dependent variable, x=x +u, the differential equation can be written as ( ) du dt = u 2 The characteristic equation is det(a r I) = r 2 + 2r + 3 =, with complex conjugate roots r = ± i 2 Since the real parts of the eigenvalues are negative, the critical point is an asymptotically stable spiral 6 The critical point x satisfies the system of equations ( ) ( ) β α x = δ γ It follows that x = γ/δ and y = α/β Using the transformation, x=x +u, the differential equation can be written as ( ) du β dt = u δ The characteristic equation is det(a r I) = r 2 + β δ = Since β δ >, the roots are purely imaginary, with r = ± i βδ Hence the critical point is a stable center 2(a) If q > and p <, then the roots are either complex conjugates with negative real parts, or both real and negative (b) If q > and p =, then the roots are purely imaginary (c) If q <, then the roots are real, with r r 2 < If p >, then either the roots are real, with r > or the roots are complex conjugates with positive real parts

385 The differential equations can be combined to obtain a related ODE The equation is separable, with dy dx = 2y x dy y = 2 dx x The solution is given by y = C x 2 Note that the system is uncoupled, and hence we also have x = x e t and y = y e 2t Matching the initial conditions, for the first case we obtain x(t) = 4e t and y(t) = 2e 2t, for the second case we obtain x(t) = 4e t and y(t) = In order to determine the direction of motion along the trajectories, observe that for positive initial conditions, x will decrease, whereas y will increase 4 The trajectories of the system satisfy the ODE dy dx = bx ay The equation is separable, with ay dy = bx dx Hence the trajectories are given by b x 2 + a y 2 = C 2, in which C is arbitrary Evidently, the trajectories are ellipses Invoking the initial condition, we find that C 2 = ab The system of ODEs can also be written as dx dt = ( ) a x b Using the methods in Chapter 7, it is easy to show that x = a cos ab t, y = b sin ab t

386 386 Chapter 9 Nonlinear Differential Equations and Stability Note that for positive initial conditions, x will increase, whereas y will decrease 6(a) The critical points are solutions of the equations + 2y = 3x 2 = There are two critical points, ( / 3, /2) and (/ 3, /2) (b) (c) Locally, the trajectories near the point ( / 3, /2) resemble the behavior near a saddle Hence the critical point is unstable Near the point (/ 3, /2), the solutions are periodic Therefore the second critical point is stable 7(a) The critical points are solutions of the equations 2x x 2 xy = 3y 2y 2 3xy = There are four critical points, (, ), (, 3/2), (2, ), and (, 3)

387 (b) (c) Examining the phase plot we can conclude that (, ) is an unstable node, (, 3/2) is a saddle point (hence unstable), (2, ) is an asymptotically stable node, and (, 3) is an asymptotically stable node (d) Again, the phase plot shows us that the basin of (2, ) is the (open) first and fourth quadrants and the basin of (, 3) is the (open) second quadrant 8(a) The critical points are solutions of the equations (2 + y)(x + y) = y( x) = There are three critical points, (, ), (, ), and (, 2) (b) (c) Examining the phase plot we can conclude that (, ) is an asymptotically stable node, (, ) is a saddle point (hence unstable), and (, 2) is an asymptotically stable spiral (d) The phase plot suggests that the basin of (, ) is the whole plane except for a subset of the fourth quadrant that is the basin of (, 2)

388 388 Chapter 9 Nonlinear Differential Equations and Stability 9(a) The critical points are given by the solution set of the equations y(2 x y) = x y 2xy = Clearly, (, ) is a critical point If x = 2 y, then it follows that y(y 2) = The additional critical points are ( 2, + 2 ) and ( + 2, 2 ) (b) (c) The behavior near the origin is that of a stable spiral Hence the point (, ) is asymptotically stable At the critical point ( 2, + 2 ), the trajectories resemble those near a saddle Hence the critical point is unstable Near the point ( + 2, 2 ), the trajectories resemble those near a saddle Hence the critical point is also unstable (d) Observing the direction field and the trajectories in (b), we can see that the basin of attraction of the origin is a complicated region including portions of all four quadrants (a) The critical points are solutions of the equations (2 + x)(y x) = y(2 + x x 2 ) = The origin is evidently a critical point If x = 2, then y = If x = y, then either y = or x = y = or x = y = 2 Hence the other critical points are ( 2, ), (, ) and (2, 2)

389 (b) (c) Based on the global phase portrait, the critical points (, ) and ( 2, ) have the characteristics of a saddle Hence these points are unstable The behavior near the remaining two critical points resembles those near a stable spiral Hence the critical points (, ) and (2, 2) are asymptotically stable (d) The basin of (2, 2) is the part of the upper half plane where x > 2, the basin of (, ) is the part of the lower half plane where x > 2 (a) The critical points are given by the solution set of the equations x( 2y) = y x 2 y 2 = If x =, then either y = or y = If y = /2, then x = ±/2 Hence the critical points are at (, ), (, ), ( /2, /2) and (/2, /2) (b) (c) The trajectories near the critical points ( /2, /2) and (/2, /2) are closed curves Hence the critical points have the characteristics of a center, which is stable The trajectories near the critical points (, ) and (, ) resemble those near a saddle Hence these critical points are unstable

390 39 Chapter 9 Nonlinear Differential Equations and Stability (d) As the two stable critical points are centers, they have no basins of attraction Trajectories near the critical points are ovals around those points 3(a) The critical points are solutions of the equations (2 + x)(y x) = (4 x)(y + x) = If y = x, then either x = y = or x = y = 4 If x = 2, then y = 2 If x = y, then y = 2 or y = Hence the critical points are at (, ), (4, 4) and ( 2, 2) (b) (c) The critical point at (4, 4) is evidently a stable spiral, which is asymptotically stable Closer examination of the critical point at (, ) reveals that it is a saddle, which is unstable The trajectories near the critical point ( 2, 2) resemble those near an unstable node (d) The basin of attraction for (4, 4) consists of the points for which x > 2 and which lie over a curve (inferred from part (b)) passing through the origin and ( 2, 2) 4(a) The critical points are given by the solution set of the equations (2 x)(y x) = y(2 x x 2 ) = If x = 2, then y = If y =, then x = Also, x = and x = 2 are roots of the second equation, and then y = x from the first equation Hence the critical points are at (2, ), (, ), (, ) and ( 2, 2)

391 92 39 (b) (c) The critical points (, ) and (2, ) are saddles, hence unstable The other two critical points are asymptotically stable spirals (d) The basin of (, ) is the part of the upper half plane where x < 2 (so all points (x, y) such that x < 2 and y > ), the basin of ( 2, 2) is all points (x, y) such that x < 2 and y < 5(a) The critical points are given by the solution set of the equations x(2 x y) = x + 3y 2xy = If x =, then y = The other critical points can be found by setting y = 2 x and substituting this into the second equation This gives us x = 3 and x = Hence the critical points are at (, ), (, ) and (3, ) (b) (c) The critical point (, ) is a saddle, hence unstable (, ) is an unstable node and (3, ) is an asymptotically stable spiral (d) The basin of (3, ) consists of all points (x, y) for which x > and x > y

392 392 Chapter 9 Nonlinear Differential Equations and Stability 6(a) The critical points are given by the solution set of the equations x(2 x y) = ( y)(2 + x) = If x =, then y = Also, when x = 2, then y = 4 The last critical point is given by y =, x = Hence the critical points are at (, ), (, ) and ( 2, 4) (b) (c) The critical point (, ) is a saddle, hence unstable ( 2, 4) is an unstable spiral and (, ) is an asymptotically stable node (d) The basin of (, ) is the right half plane, ie all the points (x, y) for which x > 8(a) The trajectories are solutions of the differential equation dy dx = 4x y, which can also be written as 4x dx + y dy = Integrating, we obtain Hence the trajectories are ellipses (b) 4x 2 + y 2 = C 2

393 Based on the differential equations, the direction of motion on each trajectory is clockwise 9(a) The trajectories of the system satisfy the ODE dy dx = 2x + y, y which can also be written as (2x + y)dx ydy = This differential equation is homogeneous Setting y = x v(x), we obtain that is, v + x dv dx = 2 v +, x dv dx = 2 + v v2 v The resulting ODE is separable, with solution x 3 (v + )(v 2) 2 = C Reverting back to the original variables, the trajectories are level curves of H(x, y) = (x + y)(y 2x) 2 (b) The origin is a saddle Along the line y = 2x, solutions increase without bound Along the line y = x, solutions converge toward the origin 2(a) The trajectories are solutions of the differential equation dy dx = x + y x y, which is homogeneous Setting y = x v(x), we obtain that is, v + x dv dx = x + xv x xv, x dv dx = + v2 v

394 394 Chapter 9 Nonlinear Differential Equations and Stability The resulting ODE is separable, with solution arctan(v) = ln x + v 2 Reverting back to the original variables, the trajectories are level curves of H(x, y) = arctan(y/x) ln x 2 + y 2 (b) The origin is a stable spiral 22(a) The trajectories are solutions of the differential equation dy dx = 2xy2 + 6 xy 2x 2 y 3x 2 4y, which can also be written as (2xy 2 6 xy)dx + (2x 2 y 3x 2 4y)dy = The resulting ODE is exact, with H x = 2xy2 6 xy and H y = 2x2 y 3x 2 4y Integrating the first equation, we find that H(x, y) = x 2 y 2 3x 2 y + f(y) It follows that H y = 2x2 y 3x 2 + f (y) Comparing the two partial derivatives, we obtain f(y) = 2y 2 + c Hence H(x, y) = x 2 y 2 3x 2 y 2y 2 (b) The associated direction field shows the direction of motion along the trajectories

395 (a) The trajectories are solutions of the differential equation dy dx 6 x + x3 =, 6 y which can also be written as (6 x x 3 )dx + 6 ydy = The resulting ODE is exact, with H x = 6 x x3 and H y = 6 y Integrating the first equation, we have H(x, y) = 3x 2 x 4 /4 + f(y) that H y = f (y) It follows Comparing the two partial derivatives, we conclude that f(y) = 3y 2 + c Hence (b) H(x, y) = 3x 2 x y2

396 396 Chapter 9 Nonlinear Differential Equations and Stability 93 Write the system in the form x =Ax+g(x) In this case, it is evident that ( ) ( ) ( ) ( ) d x x y 2 = + dt y 2 y x 2 That is, g(x) = ( y 2, x 2 ) T Using polar coordinates, g(x) = r 2 sin 4 θ + cos 4 θ and x = r Hence g(x) lim = lim r sin 4 θ + cos r x 4 θ =, r and the system is locally linear The origin is an isolated critical point of the linear system ( ) ( ) ( ) d x x = dt y 2 y The characteristic equation of the coefficient matrix is r 2 + r 2 =, with roots r = and r 2 = 2 Hence the critical point is a saddle, which is unstable 2 The system can be written as ( ) ( ) ( d x x = dt y 4 y ) + ( ) 2xy x 2 y 2 Following the discussion in Example 3, we note that F (x, y) = x + y + 2xy and G(x, y) = 4x y + x 2 y 2 Both of the functions F and G are twice differentiable, hence the system is locally linear Furthermore, F x = + 2y, F y = + 2x, G x = 4 + 2x, G y = 2y The origin is an isolated critical point, with ( ) ( ) Fx (, ) F y (, ) = G x (, ) G y (, ) 4 The characteristic equation of the associated linear system is r r + 5 =, with complex conjugate roots r,2 = ± 2i The origin is a stable spiral, which is asymptotically stable 5(a) The critical points consist of the solution set of the equations (2 + x)(y x) = (4 x)(y + x) = As shown in Problem 3 of Section 92, the only critical points are at (, ), (4, 4) and ( 2, 2) (b,c) First note that F (x, y) = (2 + x)(y x) and G(x, y) = (4 x)(y + x) The Jacobian matrix of the vector field is ( ) ( ) Fx (x, y) F J = y (x, y) 2 2x + y 2 + x = G x (x, y) G y (x, y) 4 y 2x 4 x

397 At the origin, the coefficient matrix of the linearized system is ( ) 2 2 J(, ) =, 4 4 with eigenvalues r = 7 and r 2 = + 7 The eigenvalues are real, with opposite sign Hence the critical point is a saddle, which is unstable At the point ( 2, 2), the coefficient matrix of the linearized system is ( ) 4 J( 2, 2) =, 6 6 with eigenvalues r = 4 and r 2 = 6 The eigenvalues are real, unequal and positive, hence the critical point is an unstable node At the point (4, 4), the coefficient matrix of the linearized system is ( ) 6 6 J(4, 4) =, 8 with complex conjugate eigenvalues r,2 = 3 ± i 39 The critical point is a stable spiral, which is asymptotically stable Based on Table 93, the nonlinear terms do not affect the stability and type of each critical point (d) 7(a) The critical points are solutions of the equations y = (x y)(x + y) = The first equation requires that y = Based on the second equation, x = ± Hence the critical points are (, ) and (, ) (b,c) F (x, y) = y and G(x, y) = x 2 y 2 The Jacobian matrix of the vector field is ( ) ( ) Fx (x, y) F J = y (x, y) = G x (x, y) G y (x, y) 2x 2y

398 398 Chapter 9 Nonlinear Differential Equations and Stability At the critical point (, ), the coefficient matrix of the linearized system is ( ) J(, ) =, 2 2 with eigenvalues r = 3 and r 2 = + 3 The eigenvalues are real, with opposite sign Hence the critical point is a saddle, which is unstable At the point (, ), the coefficient matrix of the linearized system is ( ) J(, ) =, 2 2 with complex conjugate eigenvalues r,2 = ± i The critical point is a stable spiral, which is asymptotically stable (d) Based on Table 93, the nonlinear terms do not affect the stability and type of each critical point 8(a) The critical points are given by the solution set of the equations x( x y) = y(2 y 3x) = If x =, then either y = or y = 2 If y =, then x = or x = If y = x, then either x = /2 or x = If y = 2 3x, then x = or x = /2 Hence the critical points are at (, ), (, 2), (, ) and (/2, /2) (b,c) Note that F (x, y) = x x 2 xy and G(x, y) = (2y y 2 3xy)/4 The Jacobian matrix of the vector field is ( ) ( ) Fx (x, y) F J = y (x, y) 2x y x = G x (x, y) G y (x, y) 3y/4 /2 y/2 3x/4 At the origin, the coefficient matrix of the linearized system is ( ) J(, ) =, 2

399 with eigenvalues r = and r 2 = /2 The eigenvalues are real and both positive Hence the critical point is an unstable node At the point (, 2), the coefficient matrix of the linearized system is ( ) J(, 2) = 3 2, 2 with eigenvalues r = and r 2 = /2 The eigenvalues are both negative, hence the critical point is a stable node At the point (, ), the coefficient matrix of the linearized system is ( ) J(, ) =, 4 with eigenvalues r = and r 2 = /4 Both of the eigenvalues are negative, and hence the critical point is a stable node At the critical point (/2, /2), the coefficient matrix of the linearized system is ( ) /2 /2 J(/2, /2) =, 3/8 /8 with eigenvalues r = 5/6 57 /6 and r 2 = 5/ /6 The eigenvalues are real, with opposite sign Hence the critical point is a saddle, which is unstable (d) Based on Table 93, the nonlinear terms do not affect the stability and type of each critical point 9(a) The critical points are given by the solution set of the equations (2 + y)(y x/2) = (2 x)(y + x/2) = If y = 2, then either x = 2 or x = 4 If x = 2, then y = 2 or y = Also, x = y = is a solution Hence the critical points are at (, ), (2, 2), (4, 2) and (2, )

400 4 Chapter 9 Nonlinear Differential Equations and Stability (b,c) Note that F (x, y) = 2y + y 2 x xy/2 and G(x, y) = 2y xy + x x 2 /2 The Jacobian matrix of the vector field is ( ) ( ) Fx (x, y) F J = y (x, y) y/ y x/2 = G x (x, y) G y (x, y) y + x 2 x At the origin, the coefficient matrix of the linearized system is ( ) 2 J(, ) =, 2 with eigenvalues r = ( + 7)/2 and r 2 = ( 7)/2 The eigenvalues are real, with opposite sign Hence the critical point is a saddle, which is unstable At the point (2, 2), the coefficient matrix of the linearized system is ( ) 3 J(2, 2) =, with eigenvalues r = 3i and r 2 = 3i The eigenvalues are purely imaginary, hence the critical point is either a center or a spiral point At the point (4, 2), the coefficient matrix of the linearized system is ( ) 4 J(4, 2) =, 2 with eigenvalues r = + 5 and r 2 = 5 The eigenvalues are real, with opposite sign Hence the critical point is a saddle, which is unstable At the critical point (2, ), the coefficient matrix of the linearized system is ( ) 3/2 3 J(2, ) =, 2 with eigenvalues r = 3/ i/4 and r 2 = 3/4 87 i/4 point is a stable spiral, which is asymptotically stable The critical (d) We observe that the point (2, 2) is a spiral point

401 93 4 (a) The critical points are solutions of the equations 2x + y + xy 3 = x 2y xy = Substitution of y = x/(x + 2) into the first equation results in 3x 4 + 3x x 2 + 2x = One root of the resulting equation is x = The only other real root of the equation is x = [ ( ) /3 83( ] 29 ) /3 3 9 Hence the critical points are (, ) and ( 9345, 4797) (b,c) F (x, y) = 2x + y + xy 3 and G(x, y) = x 2y xy The Jacobian matrix of the vector field is ( ) ( ) Fx (x, y) F J = y (x, y) 2 + y 3 + 3xy = 2 G x (x, y) G y (x, y) y 2 x At the origin, the coefficient matrix of the linearized system is ( ) 2 J(, ) =, 2 with eigenvalues r = 5 and r 2 = 5 The eigenvalues are real and of opposite sign Hence the critical point is a saddle, which is unstable At the point ( 9345, 4797), the coefficient matrix of the linearized system is ( ) J( 9345, 4797) =, with complex conjugate eigenvalues r,2 = 232 ± 425 i The critical point is a stable spiral, which is asymptotically stable (d)

402 42 Chapter 9 Nonlinear Differential Equations and Stability In both cases, the nonlinear terms do not affect the stability and type of the critical point 2(a) The critical points are given by the solution set of the equations ( + x) sin y = x cos y = If x =, then we must have cos y = 2, which is impossible Therefore sin y =, which implies that y = nπ, n =, ±, 2, Based on the second equation, x = cos nπ It follows that the critical points are located at (, 2kπ) and (2, (2k + )π), where k =, ±, ± 2, (b,c) Given that F (x, y) = ( + x) sin y and G(x, y) = x cos y, the Jacobian matrix of the vector field is ( ) sin y ( + x) cos y J = sin y At the critical points (, 2kπ), the coefficient matrix of the linearized system is ( ) J(, 2kπ) =, with purely complex eigenvalues r,2 = ± i The critical points of the associated linear systems are centers, which are stable Note that Theorem 932 does not provide a definite conclusion regarding the relation between the nature of the critical points of the nonlinear systems and their corresponding linearizations At the points (2, (2k + )π), the coefficient matrix of the linearized system is ( ) 3 J [2, (2k + )π] =, with eigenvalues r = 3 and r 2 = 3 The eigenvalues are real, with opposite sign Hence the critical points of the associated linear systems are saddles, which are unstable (d)

403 93 43 As asserted in Theorem 932, the trajectories near the critical points (2, (2k + )π) resemble those near a saddle Upon closer examination, the critical points (, 2kπ) are indeed centers 3(a) The critical points are solutions of the equations x y 2 = y x 2 = Substitution of y = x 2 into the first equation results in x x 4 =, with real roots x =, Hence the critical points are at (, ) and (, ) (b,c) In this problem, F (x, y) = x y 2 and G(x, y) = y x 2 The Jacobian matrix of the vector field is ( ) 2y J = 2x At the origin, the coefficient matrix of the linearized system is ( ) J(, ) =, with repeated eigenvalues r = and r 2 = It is easy to see that the corresponding eigenvectors are linearly independent Hence the critical point is an unstable proper node Theorem 932 does not provide a definite conclusion regarding the relation between the nature of the critical point of the nonlinear system and the corresponding linearization At the critical point (, ), the coefficient matrix of the linearized system is ( ) 2 J(, ) =, 2 with eigenvalues r = 3 and r 2 = The eigenvalues are real, with opposite sign Hence the critical point is a saddle, which is unstable (d)

404 44 Chapter 9 Nonlinear Differential Equations and Stability Closer examination reveals that the critical point at the origin is indeed a proper node 4(a) The critical points are given by the solution set of the equations xy = x y 3 = After multiplying the second equation by y, it follows that y = ± Hence the critical points of the system are at (, ) and (, ) (b,c) Note that F (x, y) = xy and G(x, y) = x y 3 The Jacobian matrix of the vector field is ( ) y x J = 3y 2 At the critical point (, ), the coefficient matrix of the linearized system is ( ) J(, ) =, 3 with eigenvalues r = 2 and r 2 = 2 The eigenvalues are real and equal It is easy to show that there is only one linearly independent eigenvector Hence the critical point is a stable improper node Theorem 932 does not provide a definite conclusion regarding the relation between the nature of the critical point of the nonlinear system and the corresponding linearization At the point (, ), the coefficient matrix of the linearized system is ( ) J(, ) =, 3 with eigenvalues r = + 5 and r 2 = 5 The eigenvalues are real, with opposite sign Hence the critical point of the associated linear system is a saddle, which is unstable (d) Closer examination reveals that the critical point at (, ) is indeed a stable improper node, which is asymptotically stable

405 (a) The critical points are given by the solution set of the equations 2x y x(x 2 + y 2 ) = x y + y(x 2 + y 2 ) = It is clear that the origin is a critical point Solving the first equation for y, we find that y = ± 8x 2 4x 4 2x Substitution of these relations into the second equation results in two equations of the form f (x) = and f 2 (x) = Plotting these functions, we note that only f (x) = has real roots given by x ± 3376 It follows that the additional critical points are at ( 3376, 924) and ( 3376, 924) (b,c) Given that F (x, y) = 2x y x(x 2 + y 2 ) G(x, y) = x y + y(x 2 + y 2 ), the Jacobian matrix of the vector field is ( ) 2 3x J = 2 y 2 2xy + 2xy + x 2 + 3y 2 At the critical point (, ), the coefficient matrix of the linearized system is ( ) 2 J(, ) =, with complex conjugate eigenvalues r,2 = ( 3 ± i 3 )/2 Hence the critical point is a stable spiral, which is asymptotically stable At the point ( 3376, 924), the coefficient matrix of the linearized system is J( 3376, 924) = ( with eigenvalues r = 3592 and r 2 = 2677 The eigenvalues are real, with opposite sign Hence the critical point of the associated linear system is a saddle, which is unstable Identical results hold for the point at ( 3376, 924) (d) ),

406 46 Chapter 9 Nonlinear Differential Equations and Stability 6(a) The critical points are solutions of the equations y + x( x 2 y 2 ) = x + y( x 2 y 2 ) = Multiply the first equation by y and the second equation by x The difference of the two equations gives x 2 + y 2 = Hence the only critical point is at the origin (b,c) With F (x, y) = y + x( x 2 y 2 ) and G(x, y) = x + y( x 2 y 2 ), the Jacobian matrix of the vector field is ( ) 3x J = 2 y 2 2xy 2xy x 2 3y 2 At the origin, the coefficient matrix of the linearized system is ( ) J(, ) =, with complex conjugate eigenvalues r,2 = ± i Hence the origin is an unstable spiral (d) 7(a) The critical points are given by the solution set of the equations 4 y 2 = (x + 3/2)(y x) = Clearly, y = ±2 The second equation tells us that x = 3/2 or x = y Hence the critical points are at ( 3/2, 2), ( 3/2, 2), (2, 2) and ( 2, 2) (b,c) Note that F (x, y) = 4 y 2 and G(x, y) = xy + 3y/2 x 2 3x/2 The Jacobian matrix of the vector field is ( ) ( ) Fx (x, y) F J = y (x, y) 2y = G x (x, y) G y (x, y) y 2x 3/2 x + 3/2

407 93 47 At ( 3/2, 2), the coefficient matrix of the linearized system is ( ) 4 J( 3/2, 2) =, 7/2 with eigenvalues r = 4i and r 2 = 4i The eigenvalues are purely imaginary, hence the critical point is either a center or a spiral point At the point ( 3/2, 2), the coefficient matrix of the linearized system is ( ) 4 J( 3/2, 2) =, /2 with eigenvalues r = 2i and r 2 = 2i The eigenvalues are purely imaginary, hence the critical point is either a center or a spiral point At the point (2, 2), the coefficient matrix of the linearized system is ( ) 4 J(2, 2) =, 7/2 7/2 with eigenvalues r = ( )/4 and r 2 = (7 273)/4 The eigenvalues are real, with opposite sign Hence the critical point is a saddle, which is unstable At the critical point ( 2, 2), the coefficient matrix of the linearized system is ( ) 4 J( 2, 2) =, /2 /2 with eigenvalues r = ( + 33 )/4 and r 2 = 33 )/4 Hence the critical point is a saddle, which is unstable (d) Further observation indicates that ( 3/2, 2) and ( 3/2, 2) are unstable spiral points 9(a) The Jacobian matrix of the vector field is ( ) J = + 6x 2 At the origin, the coefficient matrix of the linearized system is ( ) J(, ) =,

408 48 Chapter 9 Nonlinear Differential Equations and Stability with eigenvalues r = and r 2 = The eigenvalues are real, with opposite sign Hence the critical point is a saddle point (b) The trajectories of the linearized system are solutions of the differential equation dy dx = x y, which is separable Integrating both sides of the equation x dx y dy =, the solution is x 2 y 2 = C The trajectories consist of a family of hyperbolas It is easy to show that the general solution is given by x(t) = c e t + c 2 e t and y(t) = c e t c 2 e t The only bounded solutions consist of those for which c = In that case, x(t) = c 2 e t = y(t) (c) The trajectories of the given system are solutions of the differential equation dy dx = x + 2x3, y which can also be written as (x + 2x 3 )dx y dy = The resulting ODE is exact, with H x = x + 2x3 and H y = y Integrating the first equation, we find that H(x, y) = x 2 /2 + x 4 /2 + f(y) It follows that H y = f (y) Comparing the partial derivatives, we obtain f(y) = y 2 /2 + c Hence the solutions are level curves of the function H(x, y) = x 2 /2 + x 4 /2 y 2 /2 (The equation is also separable Separation of variables yields the same H(x, y)) The trajectories approaching to, or diverging from, the origin are no longer straight lines

409 (a) The solutions of the system of equations y = ω 2 sin x = consist of the points (± nπ, ), n =,, 2, The functions F (x, y) = y and G(x, y) = ω 2 sin x are analytic on the entire plane It follows that the system is locally linear near each of the critical points (b) The Jacobian matrix of the vector field is ( ) J = ω 2 cos x At the origin, the coefficient matrix of the linearized system is J(, ) = ( ) ω 2, with purely complex eigenvalues r,2 = ± iω Hence the origin is a center Since the eigenvalues are purely complex, Theorem 932 gives no definite conclusion about the critical point of the nonlinear system Physically, the critical point corresponds to the state θ =, θ = That is, the rest configuration of the pendulum (c) At the critical point (π, ), the coefficient matrix of the linearized system is J(π, ) = ( ) ω 2, with eigenvalues r,2 = ± ω The eigenvalues are real and of opposite sign Hence the critical point is a saddle Theorem 932 asserts that the critical point for the nonlinear system is also a saddle, which is unstable This critical point corresponds to the state θ = π, θ = That is, the upright rest configuration (d) Let ω 2 = The following is a plot of the phase curves near (, )

410 4 Chapter 9 Nonlinear Differential Equations and Stability The local phase portrait shows that the origin is indeed a center (e) It should be noted that the phase portrait has a periodic pattern, since θ = x mod 2π 22(a) The trajectories of the system in Problem 2 are solutions of the differential equation dy dx = ω2 sin x, y which can also be written as ω 2 sin x dx + y dy = The resulting ODE is exact, with H x = ω2 sin x and H y = y Integrating the first equation, we find that H(x, y) = ω 2 cos x + f(y) It follows that H y = f (y) Comparing the partial derivatives, we obtain f(y) = y 2 /2 + C Hence the solutions are level curves of the function H(x, y) = ω 2 cos x + y 2 /2

411 93 4 Adding an arbitrary constant, say ω 2, to the function H(x, y) does not change the nature of the level curves Hence the trajectories are can be written as 2 y2 + ω 2 ( cos x) = c, in which c is an arbitrary constant (b) Multiplying by ml 2 and reverting to the original physical variables, we obtain 2 ml2 ( dθ dt )2 + ml 2 ω 2 ( cos θ) = ml 2 c Since ω 2 = g/l, the equation can be written as in which E = ml 2 c 2 ml2 ( dθ dt )2 + mgl( cos θ) = E, (c) The absolute velocity of the point mass is given by v = L dθ/dt The kinetic energy of the mass is T = mv 2 /2 Choosing the rest position as the datum, that is, the level of zero potential energy, the gravitational potential energy of the point mass is V = mgl( cos θ) It follows that the total energy, T + V, is constant along the trajectories 23(a) A = 25 Since the system is undamped, and y() =, the amplitude is 25 The period is estimated at τ 36

412 42 Chapter 9 Nonlinear Differential Equations and Stability (b) R τ A = A = 335 A = A = (c) Since the system is conservative, the amplitude is equal to the initial amplitude On the other hand, the period of the pendulum is a monotone increasing function of the initial position A It appears that as A, the period approaches π, the period of the corresponding linear pendulum (2π/ω)

413 93 43 (d) The pendulum is released from rest, at an inclination of 4 π radians from the vertical Based on conservation of energy, the pendulum will swing past the lower equilibrium position (θ = 2π) and come to rest, momentarily, at a maximum rotational displacement of θ max = 3π (4 π) = 4π 4 The transition between the two dynamics occurs at A = π, that is, once the pendulum is released beyond the upright configuration 26(a) It is evident that the origin is a critical point of each system Furthermore, it is easy to see that the corresponding linear system, in each case, is given by dx dt = y dy dt = x The eigenvalues of the coefficient matrix are r,2 = ± i Hence the critical point of the linearized system is a center (b) Using polar coordinates, it is also easy to show that g(x) lim = r x Alternatively, the nonlinear terms are analytic in the entire plane systems are locally linear near the origin Hence both (c) For system (ii), note that x dx dt + y dy dt = xy x2 (x 2 + y 2 ) xy y 2 (x 2 + y 2 ) Converting to polar coordinates, and differentiating the equation r 2 = x 2 + y 2 with respect to t, we find that r dr dt = xdx dt + y dy dt = (x2 + y 2 ) 2 = r 4 That is, r = r 3 It follows that r 2 = /(2t + c), where c = /r 2 Since r as t and dr/dt <, regardless of the value of r, the origin is an asymptotically stable critical point

414 44 Chapter 9 Nonlinear Differential Equations and Stability On the other hand, for system (i), r dr dt = xdx dt + y dy dt = (x2 + y 2 ) 2 = r 4 That is, r = r 3 Solving the differential equation results in r 2 = c 2t (2t c) 2 Imposing the initial condition r() = r, we obtain a specific solution r 2 r 2 = 2 r 2 t Since the solution becomes unbounded as t /2r 2, the critical point is unstable 27 The characteristic equation of the coefficient matrix is r 2 + =, with complex roots r,2 = ± i Hence the critical point at the origin is a center The characteristic equation of the perturbed matrix is r 2 2 ɛ r + + ɛ 2 =, with complex conjugate roots r,2 = ɛ ± i As long as ɛ, the critical point of the perturbed system is a spiral point Its stability depends on the sign of ɛ 28 The characteristic equation of the coefficient matrix is (r + ) 2 =, with roots r = r 2 = Hence the critical point is an asymptotically stable node On the other hand, the characteristic equation of the perturbed system is r 2 + 2r + + ɛ =, with roots r,2 = ± ɛ If ɛ >, then r,2 = ± i ɛ are complex roots The critical point is a stable spiral If ɛ <, then r,2 = ± ɛ are real and both negative ( ɛ < ) The critical point remains a stable node 94 (a)

415 94 45 (b) The critical points are solutions of the system of equations x(5 x 5 y) = y(2 y 75 x) = The four critical points are (, ), (, 2), (5, ) and (8, 4) (c) The Jacobian matrix of the vector field is ( ) 3/2 2x y/2 x/2 J = 3y/4 2 3x/4 2y At the critical point (, ), the coefficient matrix of the linearized system is ( ) 3/2 J(, ) = 2 The eigenvalues and eigenvectors are ( ) r = 3/2, ξ () = ; r 2 = 2, ξ (2) = ( ) The eigenvalues are positive, hence the origin is an unstable node At the critical point (, 2), the coefficient matrix of the linearized system is ( ) /2 J(, 2) = 3/2 2 The eigenvalues and eigenvectors are ( ) r = /2, ξ () = ; r 2 = 2, ξ (2) = 6 ( ) The eigenvalues are of opposite sign Hence the critical point is a saddle, which is unstable At the critical point (5, ), the coefficient matrix of the linearized system is J(5, ) = ( The eigenvalues and eigenvectors are ( ) ( ) 3579 r = 5, ξ () = ; r 2 = 875, ξ (2) = The eigenvalues are of opposite sign Hence the critical point is also a saddle, which is unstable At the critical point (8, 4), the coefficient matrix of the linearized system is ( ) 8 4 J(8, 4) = 5 4 The eigenvalues and eigenvectors are r = ( ) 5 +, ξ() = ) ; r 2 = ( ) 5, ξ(2) =

416 46 Chapter 9 Nonlinear Differential Equations and Stability The eigenvalues are both negative Hence the critical point is a stable node, which is asymptotically stable (d,e) (f) Except for initial conditions lying on the coordinate axes, almost all trajectories converge to the stable node at (8, 4) Thus the species can coexist 2(a) (b) The critical points are the solution set of the system of equations x(5 x 5 y) = y(2 5 y 5 x) = The four critical points are (, ), (, 4), (5, ) and (, ) (c) The Jacobian matrix of the vector field is ( ) 3/2 2x y/2 x/2 J = 3y/2 2 3x/2 y At the origin, the coefficient matrix of the linearized system is ( ) 3/2 J(, ) = 2

417 94 47 The eigenvalues and eigenvectors are ( ) r = 3/2, ξ () = ; r 2 = 2, ξ (2) = ( ) The eigenvalues are positive, hence the origin is an unstable node At the critical point (, 4), the coefficient matrix of the linearized system is ( ) /2 J(, 4) = 6 2 The eigenvalues and eigenvectors are ( ) r = /2, ξ () = ; r 2 = 2, ξ (2) = 4 ( ) The eigenvalues are both negative, hence the critical point (, 4) is a stable node, which is asymptotically stable At the critical point (3/2, ), the coefficient matrix of the linearized system is ( ) 3/2 3/4 J(3/2, ) = /4 The eigenvalues and eigenvectors are ( ) r = 3/2, ξ () = ; r 2 = /4, ξ (2) = ( ) 3 5 The eigenvalues are both negative, hence the critical point is a stable node, which is asymptotically stable At the critical point (, ), the coefficient matrix of the linearized system is ( ) /2 J(, ) = 3/2 /2 The eigenvalues and eigenvectors are r = 3 + ( ) 3, ξ () = ; r 2 = , ξ (2) = ( ) The eigenvalues are of opposite sign, hence (, ) is a saddle, which is unstable (d,e)

418 48 Chapter 9 Nonlinear Differential Equations and Stability (f) Trajectories approaching the critical point (, ) form a separatrix Solutions on either side of the separatrix approach either (, 4) or (5, ) Thus depending on the initial conditions, one species will drive the other to extinction 4(a) (b) The critical points are solutions of the system of equations x(5 5 x y) = y(75 y 25 x) = The four critical points are (, ), (, 3/4), (3, ) and (2, /2) (c) The Jacobian matrix of the vector field is ( ) 3/2 x y x J = y/8 3/4 x/8 2y At the origin, the coefficient matrix of the linearized system is ( ) 3/2 J(, ) = 3/4 The eigenvalues and eigenvectors are ( ) r = 3/2, ξ () = ; r 2 = 3/4, ξ (2) = ( ) The eigenvalues are positive, hence the origin is an unstable node At the critical point (, 3/4), the coefficient matrix of the linearized system is ( ) 3/4 J(, 3/4) = 3/32 3/4 The eigenvalues and eigenvectors are ( ) 6 r = 3/4, ξ () = ; r 2 = 3/4, ξ (2) = ( ) The eigenvalues are of opposite sign, hence the critical point (, 3/4) is a saddle, which is unstable

419 94 49 At the critical point (3, ), the coefficient matrix of the linearized system is ( ) 3/2 3 J(3, ) = 3/8 The eigenvalues and eigenvectors are ( ) r = 3/2, ξ () = ; r 2 = 3/8, ξ (2) = ( ) 8 5 The eigenvalues are of opposite sign, hence the critical point (, 3/4) is a saddle, which is unstable At the critical point (2, /2), the coefficient matrix of the linearized system is ( ) 2 J(2, /2) = /6 /2 The eigenvalues and eigenvectors are r = 3 + ( ) 3, ξ () = ; r 2 = , ξ (2) = ( ) The eigenvalues are negative, hence the critical point (2, /2) is a stable node, which is asymptotically stable (d,e) (f) Except for initial conditions along the coordinate axes, almost all solutions converge to the stable node (2, /2) Thus the species can coexist 7 It follows immediately that (σ X + σ 2 Y ) 2 4σ σ 2 XY = σ 2 X 2 + 2σ σ 2 XY + σ 2 2Y 2 4σ σ 2 XY = (σ X σ 2 Y ) 2, so (σ X + σ 2 Y ) 2 4(σ σ 2 α α 2 )XY = (σ X σ 2 Y ) 2 + 4α α 2 XY Since all parameters and variables are positive, it follows that (σ X + σ 2 Y ) 2 4(σ σ 2 α α 2 )XY Hence the radicand in Eq(39) is nonnegative

420 42 Chapter 9 Nonlinear Differential Equations and Stability (a) The critical points consist of the solution set of the equations x(ɛ σ x α y) = y(ɛ 2 σ 2 y α 2 x) = If x =, then either y = or y = ɛ 2 /σ 2 If ɛ σ x α y =, then solving for x results in x = (ɛ α y)/σ Substitution into the second equation yields (σ σ 2 α α 2 )y 2 (σ ɛ 2 ɛ α 2 )y = Based on the hypothesis, it follows that (σ ɛ 2 ɛ α 2 )y = So if σ ɛ 2 ɛ α 2, then y =, and the critical points are located at (, ), (, ɛ 2 /σ 2 ) and (ɛ /σ, ) For the case σ ɛ 2 ɛ α 2 =, y can be arbitrary From the relation x = (ɛ α y)/σ, are critical points, in addi- we conclude that all points on the line σ x + α y = ɛ tion to the point (, ) (b) The Jacobian matrix of the vector field is ( ) ɛ 2σ J = x α y α x α 2 y ɛ 2 2σ 2 y α 2 x At the origin, the coefficient matrix of the linearized system is ( ) ɛ J(, ) =, ɛ 2 with eigenvalues r = ɛ and r 2 = ɛ 2 Since both eigenvalues are positive, the origin is an unstable node At the point (, ɛ 2 /σ 2 ), the coefficient matrix of the linearized system is J(, ɛ 2 /σ 2 ) = ( ) (ɛ σ 2 α ɛ 2 )/σ 2 = ɛ 2 α 2 /σ 2 ɛ 2 ( ) (ɛ α 2 σ ɛ 2 )/α 2, ɛ 2 α 2 /σ 2 ɛ 2 since σ σ 2 α α 2 = implies that α /σ 2 = σ /α 2 Thus the matrix has eigenvalues r = (ɛ α 2 σ ɛ 2 )/α 2 and r 2 = ɛ 2 If σ ɛ 2 ɛ α 2 >, then both eigenvalues are negative Hence the point (, ɛ 2 /σ 2 ) is a stable node, which is asymptotically stable If σ ɛ 2 ɛ α 2 <, then the eigenvalues are of opposite sign Hence the point (, ɛ 2 /σ 2 ) is a saddle, which is unstable At the point (ɛ /σ, ), the coefficient matrix of the linearized system is ( ) ɛ ɛ J(ɛ /σ, ) = α /σ, (σ ɛ 2 ɛ α 2 )/σ with eigenvalues r = (σ ɛ 2 ɛ α 2 )/σ and r 2 = ɛ If σ ɛ 2 ɛ α 2 >, then the eigenvalues are of opposite sign Hence the point (ɛ /σ, ) is a saddle, which is unstable If σ ɛ 2 ɛ α 2 <, then both eigenvalues are negative In that case the point (ɛ /σ, ) is a stable node, which is asymptotically stable (c) As shown in part (a), when σ ɛ 2 ɛ α 2 =, the set of critical points consists of (, ) and all the points on the straight line σ x + α y = ɛ Based on part (b), the

421 94 42 origin is still an unstable node Setting y = (ɛ σ x)/α, the Jacobian matrix of the vector field, along the given straight line, is ( ) σ J(ɛ /σ, ) = x α x α 2 (ɛ σ x)/α ɛ 2 2σ 2 (ɛ σ x)/α α 2 x ( ) σ = x α x, α 2 (ɛ σ x)/α α 2 x ɛ α 2 /σ since σ σ 2 α α 2 = implies that σ 2 /α = α 2 /σ and since σ ɛ 2 ɛ α 2 = implies that ɛ 2 = ɛ α 2 /σ The characteristic equation of the matrix is [ r 2 ɛ α 2 α 2 σ x + σ 2 ] + x r = σ Since ɛ 2 = ɛ α 2 /σ, we have that (ɛ α 2 α 2 σ x + σ 2 x)/σ = ɛ 2 α 2 x + σ x Hence the characteristic equation can be written as r 2 + [ɛ 2 α 2 x + σ x] r = First note that x ɛ /σ Since the coefficient in the quadratic equation is linear, and { ɛ 2 at x = ɛ 2 α 2 x + σ x = ɛ at x = ɛ /σ, it follows that the coefficient is positive for x ɛ /σ Therefore, along the straight line σ x + α y = ɛ, one eigenvalue is zero and the other one is negative Hence the continuum of critical points consists of stable nodes, which are asymptotically stable (a) The critical points are solutions of the system of equations Assume solutions of the form x( x y) + δa = y(75 y 5 x) + δb = x = x + x δ + x 2 δ 2 + y = y + y δ + y 2 δ 2 + Substitution of the series expansions results in x ( x y ) + (x 2x x x y x y + a)δ + = y (75 y 5 x ) + (75 y 2y y x y /2 x y /2 + b)δ + = (b) Taking a limit as δ, the equations reduce to the original system of equations It follows that x = y = 5 (c) Setting the coefficients of the linear terms equal to zero, we find that y /2 x /2 + a = x /4 y /2 + b =, with solution x = 4a 4b and y = 2a + 4b

422 422 Chapter 9 Nonlinear Differential Equations and Stability (d) Consider the ab -parameter space The collection of points for which b < a represents an increase in the level of species At points where b > a, x δ < Likewise, the collection of points for which 2b > a represents an increase in the level of species 2 At points where 2b < a, y δ < It follows that if b < a < 2b, the level of both species will increase This condition is represented by the wedge-shaped region on the graph Otherwise, the level of one species will increase, whereas the level of the other species will simultaneously decrease Only for a = b = will both populations remain the same 2(a) The critical points consist of the solution set of the equations y = γ y x(x 5)(x 2) = Setting y =, the second equation becomes x(x 5)(x 2) =, with roots x =, 5 and 2 Hence the critical points are located at (, ), (5, ) and (2, ) The Jacobian matrix of the vector field is ( J = 3x x 3 γ ) At the origin, the coefficient matrix of the linearized system is ( ) J(, ) =, 3 γ with eigenvalues r,2 = γ 2 ± 25 γ2 + 3 Regardless of the value of γ, the eigenvalues are real and of opposite sign Hence (, ) is a saddle, which is unstable At the critical point (5, ), the coefficient matrix of the linearized system is with eigenvalues J(5, ) = ( 2775 γ ), r,2 = γ 2 ± γ2 2

423 If γ 2, then the eigenvalues are real Furthermore, since r r 2 = 2775, both eigenvalues will have the same sign Therefore the critical point is a node, with its stability dependent on the sign of γ If γ 2 <, the eigenvalues are complex conjugates In that case the critical point (5, ) is a spiral, with its stability dependent on the sign of γ At the critical point (2, ), the coefficient matrix of the linearized system is with eigenvalues J(2, ) = ( 37 γ ), r,2 = γ 2 ± 25 γ Regardless of the value of γ, the eigenvalues are real and of opposite sign Hence (2, ) is a saddle, which is unstable (b) For γ = 8, the critical point (5, ) is a stable spiral Closer examination shows that for γ = 5, the critical point (5, ) is a stable node (c) Based on the phase portraits in part (b), it is apparent that the required value of γ satisfies 8 < γ < 5 Using the initial condition x() = 2 and y() =, it is possible to solve the initial value problem for various values of γ A reasonable first guess is γ = This value marks the change in qualitative behavior of

424 424 Chapter 9 Nonlinear Differential Equations and Stability the critical point (5, ) Numerical experiments show that the solution remains positive for γ 2 4(a) Nullclines: (a) α = 3 (b) α = 8/3 (c) α = 2 (b) The critical points are solutions of the algebraic system 3 2 α y = 4x + y + x 2 = which are (2 ± α, 3 2 α), and exist for α 8/3 (c) For α = 2, the critical points are at (, 3) and (3, 3) The Jacobian matrix of the vector field is ( ) J = x At the critical point (, 3), the coefficient matrix of the linearized system is ( ) J(, 3) = 2

425 The eigenvalues of the Jacobian, r = and 2, are real and opposite in sign; hence the critical point is a saddle point At the critical point (3, 3), the coefficient matrix of the linearized system is J(3, 3) = The eigenvalues of the Jacobian are ( 2 r = ± i 7, 2 hence the critical point is an unstable spiral ) (d) The bifurcation value is α = 8/3 The coincident critical points are at (2, 4) The coefficient matrix of the linearized system is ( ) J(2, 4) = The eigenvalues of the Jacobian are r = and

426 426 Chapter 9 Nonlinear Differential Equations and Stability (e) 5(a) Nullclines: (a) α = 3 (b) α = 9/4 (c) α = 2 (b) The critical points are solutions of the algebraic system which are and x = α 2 x = 3 9 4α 2 These critical points exist for α 9/4 4x + y + x 2 = α x + y =, y = α α 2, y = α α 2 (c) For α = 2, the critical points are at (, 3) and (2, 4) The Jacobian matrix of the vector field is ( ) 2x 4 J = At the critical point (, 3), the coefficient matrix of the linearized system is ( ) 2 J(, 3) =

427 The eigenvalues of the Jacobian are r = ± 5, 2 hence the critical point is a saddle point At the critical point (2, 4), the coefficient matrix of the linearized system is ( ) J(2, 4) = The eigenvalues of the Jacobian are r = ± i 3 2 hence the critical point is an unstable spiral, (d) A bifurcation occurs for α = 9/4 The coincident critical points are at (3/2, 5/4) The coefficient matrix of the linearized system is ( ) J(3/2, 5/4) =, with eigenvalues both equal to zero

428 428 Chapter 9 Nonlinear Differential Equations and Stability (e) 6(a) Nullclines: (a) α = 3 (b) α = 9/4 (c) α = 2 (b) The critical points are solutions of the algebraic system which are and x = α 2 x = 3 9 4α 2 These critical points exist for α 9/4 α x + y = 4x + y + x 2 =, y = α α 2, y = α α 2 (c) For α = 2, the critical points are at (, 3) and (2, 4) The Jacobian matrix of the vector field is ( ) J = 2x 4 At the critical point (, 3), the coefficient matrix of the linearized system is ( ) J(, 3) = 2

429 The eigenvalues of the Jacobian are r = ± i, hence the critical point is a center At the critical point (2, 4), the coefficient matrix of the linearized system is ( ) J(2, 4) = The eigenvalues of the Jacobian, r = and, are real and opposite in sign; hence the critical point is a saddle point (d) The bifurcation value is α = 8/3 The coincident critical point is at (3/2, 4/4) The coefficient matrix of the linearized system is ( ) J(3/2, 5/4) = The eigenvalues are both equal to zero

430 43 Chapter 9 Nonlinear Differential Equations and Stability (e) 8(a) Nullclines: (a) α = /2 (b) α = 3/4 (c) α = (b) The critical points are (, ), (, ), ((4α 3)/(4α 2), /(4α 2)), and (, 3/(4α)) The third critical point is in the first quadrant as long as α 3/4 (c) The third and fourth critical points will coincide (see part (b)) when α = 3/4 (d,e) The Jacobian is J = ( ) 2x y x y/2 3/4 2αy x/2 This means that at the origin J(, ) = ( ), 3/4 so the origin is an unstable node (The eigenvalues are clearly and 3/4) At the critical point (, ) the Jacobian is ( ) J(, ) =, /4 which means that this critical point is a saddle

431 94 43 At the critical point (, 3/(4α)) the Jacobian is J(, 3/(4α)) = ( ) 3/(4α), 3/(8α) 3/4 which implies that this critical point is a saddle when α > 3/4 and an asymptotically stable node when < α < 3/4 At the critical point ( 4α 3 4α 2, 4α 2 ) the Jacobian is given by ( ) J( 4α 3 4α 2, 3 4α 3 4α 4α 2 ) = 4α 2 4α 2 /2 4α 2 α 4α 2 It can be shown that this is an asymptotically stable node when α > 3/4 (f) Phase portraits: (a) α = /2 (b) α = 3/4 (c) α = 9(a) Nullclines: (a) α = 3/4 (b) α = (c) α = 5/4 (b) The critical points are (, ), (, ), (, α) and (/2, /2) Also, when α =, then all points on the line x + y = are critical points (c) Clearly, α = is the bifurcation value

432 432 Chapter 9 Nonlinear Differential Equations and Stability (d,e) The Jacobian is J = This means that at the origin ( ) 2x y x (2α )y α 2y (2α )x J(, ) = ( ), α so the origin is an unstable node when α > At the critical point (, ) the Jacobian is ( ) J(, ) =, α which means that this critical point is a saddle when < α < and an asymptotically stable node when α > At the critical point (, α) the Jacobian is ( ) α J(, α)) =, α(2α ) α which implies that this critical point is a saddle when < α < and an asymptotically stable node when α > At the critical point (/2, /2) the Jacobian is given by ( ) /2 /2 J(/2, /2) = /2 α /2 The eigenvalues of J(/2, /2) are ( ± + 2α)/2 Thus the critical point is an asymptotically stable spiral for < α < /2, an asymptotically stable node for /2 α <, and a saddle for α > (f) Phase portraits: (a) α = 3/4 (b) α = (c) α = 5/4

433 (a) (b) The critical points are solutions of the system of equations x(5 5 y) = y( 5 + x) = The two critical points are (, ) and (5, 3) (c) The Jacobian matrix of the vector field is ( ) 3/2 y/2 x/2 J = y /2 + x At the critical point (, ), the coefficient matrix of the linearized system is ( ) 3/2 J(, ) = /2 The eigenvalues and eigenvectors are ( ) r = 3/2, ξ () = ; r 2 = /2, ξ (2) = ( ) The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable At the critical point (5, 3), the coefficient matrix of the linearized system is ( ) /4 J(5, 3) = 3 The eigenvalues and eigenvectors are ( ) 3 r = i 2, ξ() = 2 i 3 ; r 2 = i ( ) 3 2, ξ(2) = 2 i 3 The eigenvalues are purely imaginary Hence the critical point is a center, which is stable

434 434 Chapter 9 Nonlinear Differential Equations and Stability (d,e) (f) Except for solutions along the coordinate axes, almost all trajectories are closed curves about the critical point (5, 3) 2(a) (b) The critical points are the solution set of the system of equations x( 5 y) = y( x) = The two critical points are (, ) and (5, 2) (c) The Jacobian matrix of the vector field is ( ) y/2 x/2 J = y/2 /4 + x/2 At the critical point (, ), the coefficient matrix of the linearized system is ( ) J(, ) = /4 The eigenvalues and eigenvectors are ( ) r =, ξ () = ; r 2 = /4, ξ (2) = ( )

435 The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable At the critical point (5, 2), the coefficient matrix of the linearized system is ( ) /4 J(5, 2) = The eigenvalues and eigenvectors are ( ) r = i/2, ξ () = ; r 2 = i/2, ξ (2) = 2 i ( ) 2 i The eigenvalues are purely imaginary Hence the critical point is a center, which is stable (d,e) (f) Except for solutions along the coordinate axes, almost all trajectories are closed curves about the critical point (5, 2) 4(a) (b) The critical points are the solution set of the system of equations x(9/8 x y/2) = y( + x) = The three critical points are (, ), (9/8, ) and (, /4)

436 436 Chapter 9 Nonlinear Differential Equations and Stability (c) The Jacobian matrix of the vector field is ( ) 9/8 2x y/2 x/2 J = y + x At the critical point (, ), the coefficient matrix of the linearized system is ( ) 9/8 J(, ) = The eigenvalues and eigenvectors are ( ) r = 9/8, ξ () = ; r 2 =, ξ (2) = ( ) The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable At the critical point (9/8, ), the coefficient matrix of the linearized system is ( ) 9/8 9/6 J(9/8, ) = /8 The eigenvalues and eigenvectors are r = 9 ( ) 8, ξ() = ; r 2 = 8, ξ(2) = ( ) 9 2 The eigenvalues are of opposite sign, hence the critical point (9/8, ) is a saddle, which is unstable At the critical point (, /4), the coefficient matrix of the linearized system is J(, /4) = The eigenvalues and eigenvectors are r = 2 + ( ) , ξ () = 4 ( /2 /4 ) ; r 2 = ( ) 2 2, ξ (2) = The eigenvalues are both negative Hence the critical point is a stable node, which is asymptotically stable (d,e)

437 (f) Except for solutions along the coordinate axes, all solutions converge to the critical point (, /4) 5(a) (b) The critical points are solutions of the system of equations x( + 25 x 3 y x 2 ) = y( 5 + x) = The four critical points are (, ), (/2, ), (2, ) and (3/2, 5/3) (c) The Jacobian matrix of the vector field is ( ) + 5x 3x J = 2 3y/ 3x/ y 3/2 + x At the critical point (, ), the coefficient matrix of the linearized system is ( ) J(, ) = 3/2 The eigenvalues and eigenvectors are ( ) r =, ξ () = ; r 2 = 3/2, ξ (2) = ( ) The eigenvalues are both negative, hence the critical point (, ) is a stable node, which is asymptotically stable At the critical point (/2, ), the coefficient matrix of the linearized system is J(/2, ) = ( 3/4 3/2 ) The eigenvalues and eigenvectors are r = 3 ( ) 4, ξ() = ; r 2 =, ξ (2) = ( ) 3 35 The eigenvalues are of opposite sign, hence the critical point (/2, ) is a saddle, which is unstable

438 438 Chapter 9 Nonlinear Differential Equations and Stability At the critical point (2, ), the coefficient matrix of the linearized system is ( ) 3 3/5 J(2, ) = /2 The eigenvalues and eigenvectors are ( ) r = 3, ξ () = ; r 2 = /2, ξ (2) = ( ) 6 35 The eigenvalues are of opposite sign, hence the critical point (2, ) is a saddle, which is unstable At the critical point (3/2, 5/3), the coefficient matrix of the linearized system is J(3/2, 5/3) = The eigenvalues and eigenvectors are r = 3 + i ( 39 9+i 3 39 ), ξ () = 4 8 ( 3/4 9/2 5/3 ) ; r 2 = 3 i 39 8, ξ (2) = ( 9 i 3 39 ) 4 The eigenvalues are complex conjugates Hence the critical point (3/2, 5/3) is a stable spiral, which is asymptotically stable (d,e) (f) The single solution curve that converges to the node at (/2, ) is a separatrix Except for initial conditions on the coordinate axes, trajectories on either side of the separatrix converge to the node at (, ) or the stable spiral at (3/2, 5/3) 6 Given that t is measured from the time that x is a maximum, we have x = c γ + ck γ cos( ac t) y = a α + K a c α α sin( ac t) (a) Since sin θ reaches a maximum at θ = π/2, y reaches a maximum at ac t = π/2, or t = π/(2 ac) = T/4, where T = 2π/ ac is the period of oscillation

439 (b) Note that dx ac dt = ck sin( ac t) γ Since sin θ reaches a minimum of at θ = 3π/2, 7π/2,, x is increasing most rapidly at ac t = 3π/2, 7π/2,, or t = 3T/4, 7T/4, Since sin θ reaches a maximum of at θ = π/2, 5π/2,, x is decreasing most rapidly at ac t = π/2, 5π/2,, or t = T/4, 5T/4, Since cos θ reaches a minimum at θ = π, 3π,, x reaches a minimum at ac t = π, 3π,, or t = T/2, 3T/2, (c) Note that dy ( a ) 3/2 dt = Kc cos( ac t) α Since cos θ reaches a minimum of at θ = π, 3π,, y is decreasing most rapidly at ac t = π, 3π,, or t = T/2, 3T/2, Since cos θ reaches a maximum of at θ =, 2π,, y is increasing most rapidly at ac t =, 2π,, or t =, T, Since sin θ reaches a minimum at θ = 3π/2, 7π/2,, y reaches a minimum at ac t = 3π/2, 7π/2,, or t = 3T/4, 7T/4, (d) In the following example, the system in Problem 2 is solved numerically with the initial conditions x() = 7 and y() = 2 The critical point of interest is at (5, 2) Since a = and c = /4, it follows that the period of oscillation is T = 4 π 8(a) The period of oscillation for the linear system is T = 2π/ ac In system (2), a = and c = 75 Hence the period is estimated as T = 2π/ (b) The estimated period appears to agree with the graphic in Figure 953 (c) The critical point of interest is at (3, 2) The system is solved numerically, with y() = 2 and x() = 35, 4, 45, 5 The resulting periods are shown in the table: x() = 35 x() = 4 x() = 45 x() = 5 T The actual period steadily increases as x() increases

440 44 Chapter 9 Nonlinear Differential Equations and Stability 9 The system dx dt = a x( y 2 ) dy dt = b y( + x 3 ) is solved numerically for various values of the parameters The initial conditions are x() = 5, y() = 2 (a) a = and b = : The period is estimated by observing when the trajectory becomes a closed curve In this case, T 645 (b) a = 3 and a = /3, with b = : For a = 3, T 369 For a = /3, T 44 (c) b = 3 and b = /3, with a = :

441 95 44 For b = 3, T 382 For b = /3, T 6 (d) It appears that if one of the parameters is fixed, the period varies inversely with the other parameter Hence one might postulate the relation T = k f(a, b) (a) Since T = 2π/ ac, we first note that Hence A+T A cos( ac t + φ)dt = A+T A sin( ac t + φ)dt = x = T A+T A c γ dt = c γ and y = T A+T A a α dt = a α (b) One way to estimate the mean values is to find a horizontal line such that the area above the line is approximately equal to the area under the line From Figure 953, it appears that x 325 and y 2 In Example, a =, c = 75, α = 5 and γ = 25 Using the result in part (a), x = 3 and y = 2 (c) The system dx dt = x( y 2 ) dy dt = y( x 4 ) is solved numerically for various initial conditions x() = 3 and y() = 25 : x() = 3 and y() = 3 :

442 442 Chapter 9 Nonlinear Differential Equations and Stability x() = 3 and y() = 35 : x() = 3 and y() = 4 : It is evident that the mean values increase as the amplitude increases That is, the mean values increase as the initial conditions move farther from the critical point 2(a) The critical points are the solutions of the system x(a σx αy) = y( c + γx) = If x =, then y = If y =, then x = a/σ The third solution is found by substituting x = c/γ into the first equation This implies that y = a/α σc/(γα)

443 So the critical points are (, ), ( a σ, ) and ( c γ, a α σc γα ) When σ is increasing, the critical point ( a σ, ) moves to the left and the critical point ( c γ, a α σc ) moves down The assumption a > σc γ quadrant (When a = σc γ is necessary for the third critical point to be in the first, then the two nonzero critical points coincide) (b,c) The Jacobian of the system is ( ) a 2σx αy αx J = γy c + γx This implies that at the origin J(, ) = ( ) a, c which implies that the origin is a saddle point (a > and c > by our assumption) At the critical point ( a σ, ) J( a ( ) a αa/σ σ, ) =, c + γa/σ which implies that this critical point is also a saddle as long as our assumption a > σc γ is valid At the critical point ( c γ, a α σc γα ) J( c γ, a α σc γα ) = The eigenvalues of the matrix are ( ) σc/γ αc/γ γa/α σc/α cσ ± c 2 σ 2 + 4c 2 γσ 4acγ 2 2γ We set the discriminant equal to zero and find that the greater solution is σ = 2γ + 2γ ac + c 2 c First note that σ >, since ac + c 2 > c Next we note that σ < aγ/c Since we see that σ = 2γ + 2γ ac + c 2 c ac + c2 < a 2 < 2γ + 2γ c 4 + ac + c2 = a 2 + c, γα ( a ) 2 + c = 2γ + aγ c + 2γ = aγ c For < σ < σ, the eigenvalues will be complex conjugates with negative real part, so the critical point will be an asymptotically stable spiral point For σ = σ, the eigenvalues will be repeated and negative, so the critical point will be an asymptotically stable spiral point or node For σ < σ < ac/γ, the eigenvalues will be distinct and negative, so the critical point will be an asymptotically stable node

444 444 Chapter 9 Nonlinear Differential Equations and Stability (d) Since the third critical point is asymptotically stable for < σ < ac/γ, and the other critical points are saddle points, the populations will coexist for all such values of σ 3(a) The critical points are the solutions of the system x( x 5 2y x + 6 ) = y( 4 + x x + 6 ) = If x =, then y = If y =, then x = 5 The third critical point can be found by setting /4 = x/(x + 6), which gives x = 2 and then y = 24 So the critical points are (, ), (5, ) and (2, 24) (b) The Jacobian of the system is ( 2x J = This implies that at the origin 5 2y (x+6) 2 2x x+6 6y (x+6) x x+6 J(, ) = ( ), /4 which implies that the origin is a saddle point At the critical point (5, ) ( ) / J(5, ) =, 9/44 which implies that this critical point is also a saddle point At the critical point (2, 24) ( ) /4 /2 J(2, 24) =, 9/4 whose eigenvalues are complex with negative real part, which implies that this critical point is an asymptotically stable spiral )

445 (a) Solving for the equilibrium of interest we obtain x = E 2 + c γ y = a α σ α E2 + c γ E α So if E > and E 2 =, then we have the same amount of prey and fewer predators (b) If E = and E 2 >, then we have more prey and fewer predators (c) If E > and E 2 >, then we have more prey and even fewer predators 6(b) The equilibrium solutions are given by the solutions of the system x( y 2 ) = H y( x 4 ) = H 2 Now if H 2 =, then x = 3 and the first equation gives y = 2 2H /3 This means we have the same amount of prey and fewer predators (c) If H =, then y = 2 and the second equation gives x = 3 + 2H 2 This means we have the same amount of predators and more prey (d) If H > and H 2 >, then the second equation gives (x 3)y = 4H 2 and using this we obtain from the first equation that x( 2H2 x 3 ) = H This gives the quadratic equation x 2 (3 + 2H 2 + H )x + 3H = Now at the old value x = 3 this expression is 6H 2, so there is a root which is bigger than x = 3 The other root of the quadratic equation is closer to, so the equilibrium increases here: we have more prey (Check this with eg H = H 2 = : the roots are 3 ± 6, so both of the original roots get bigger) A similar analysis shows that we will have fewer predators in this case 96 2 We consider the function V (x, y) = a x 2 + c y 2 The rate of change of V along any trajectory is dx V = V x dt + V dy y dt = 2ax( 2 x3 + 2xy 2 ) + 2cy( y 3 ) = ax 4 + 4ax 2 y 2 2cy 4 Let us complete the square now the following way: V = ax 4 + 4ax 2 y 2 2cy 4 = a(x 4 4x 2 y 2 ) 2cy 4 = = a(x 2 2y 2 ) 2 + 4ay 4 2cy 4 = a(x 2 2y 2 ) 2 + (4a 2c)y 4 If a > and c >, then V (x, y) is positive definite Clearly, if 4a 2c <, ie when c > 2a, then V (x, y) is negative definite One such example is V (x, y) =

446 446 Chapter 9 Nonlinear Differential Equations and Stability x y 2 It follows from Theorem 96 that the origin is an asymptotically stable critical point 4 Given V (x, y) = a x 2 + c y 2, the rate of change of V along any trajectory is Setting a = 2c, dx V = V x dt + V dy y dt = 2ax(x3 y 3 ) + 2cy(2xy 2 + 4x 2 y + 2y 3 ) = = 2a x 4 + (4c 2a)xy 3 + 8c x 2 y 2 + 4c y 4 V = 4c x 4 + 8c x 2 y 2 + 4c y 4 4c x 4 + 4c y 4 As long as a = 2c >, the function V (x, y) is positive definite and V (x, y) is also positive definite It follows from Theorem 962 that (, ) is an unstable critical point 5 Given V (x, y) = c(x 2 + y 2 ), the rate of change of V along any trajectory is dx V = V x dt + V dy y dt = 2c x [y xf(x, y)] + 2cy [ x yf(x, y)] = 2c(x2 + y 2 )f(x, y) If c >, then V (x, y) is positive definite Furthermore, if f(x, y) is positive in some neighborhood of the origin, then V (x, y) is negative definite Theorem 96 asserts that the origin is an asymptotically stable critical point On the other hand, if f(x, y) is negative in some neighborhood of the origin, then V (x, y) and V (x, y) are both positive definite It follows from Theorem 962 that the origin is an unstable critical point 9(a) Letting x = u and y = u, we obtain the system of equations dx dt = y dy dt = g(x) y Since g() =, it is evident that (, ) is a critical point of the system Consider the function V (x, y) = 2 y2 + x g(s)ds It is clear that V (, ) = Since g(u) is an odd function in a neighborhood of u =, and x x g(s)ds = g(s)ds > for x >, x g(s)ds > for x < Therefore V (x, y) is positive definite The rate of change of V along any trajectory is dx V = V x dt + V dy y dt = g(x) (y) + y [ g(x) y] = y2

447 It follows that V (x, y) is only negative semidefinite Hence the origin is a stable critical point (b) Given V (x, y) = 2 y2 + x 2 y sin(x) + sin(s)ds, It is easy to see that V (, ) = The rate of change of V along any trajectory is dx V = V x dt + V dy [ y dt = sin x + y ] 2 cos x (y) + [y + 2 ] sin x [ sin x y] = = 2 y2 cos x 2 sin2 x y 2 sin x y2 For π/2 < x < π/2, we can write sin x = x α x 3 /6 and cos x = β x 2 /2, in which α = α(x), β = β(x) Note that < α, β < Then V (x, y) = y2 2 Using polar coordinates, V (r, θ) = r2 2 It is easy to show that β x2 ( 2 ) α x3 (x 2 6 )2 y α x3 (x 2 6 ) y2 [ + sin θ cos θ + h(r, θ)] = r2 2 h(r, θ) 2 r r4 [ + 2 ] sin 2θ + h(r, θ) So if r is sufficiently small, then h(r, θ) < /2 and 2 sin 2θ + h(r, θ) < Hence V (x, y) is negative definite Now we show that V (x, y) is positive definite Since g(u) = sin u, This time we set V (x, y) = 2 y2 + y sin(x) + cos x 2 cos x = x2 2 + γ x4 24 Note that < γ < for π/2 < x < π/2 Converting to polar coordinates, ] V (r, θ) = [ r2 + sin θ cos θ r2 2 2 sin θ cos3 θ γ r2 24 cos4 θ ] = [ r2 2 r2 + sin 2θ 2 2 sin θ cos3 θ γ r2 24 cos4 θ Now It follows that when r >, r2 2 sin θ cos3 θ γ r2 24 cos4 θ > 8 for r < V (r, θ) > r2 2 [ ] sin 2θ 3 r2 2 6 >

448 448 Chapter 9 Nonlinear Differential Equations and Stability Therefore V (x, y) is indeed positive definite, and by Theorem 96, the origin is an asymptotically stable critical point 2(a) We consider the linear system ( ) ( ) ( ) x a a = 2 x y a 2 a 22 y Let V (x, y) = Ax 2 + Bxy + Cy 2, in which A = a2 2 + a (a a 22 a 2 a 2 ) 2 B = a 2a 22 + a a 2 C = a2 + a (a a 22 a 2 a 2 ) 2 and = (a + a 22 )(a a 22 a 2 a 2 ) Based on the hypothesis, the coefficients A and B are negative Therefore, except for the origin, V (x, y) is negative on each of the coordinate axes Along each trajectory, V = (2Ax + By)(a x + a 2 y) + (2Cy + Bx)(a 2 x + a 22 y) = x 2 y 2, Hence V (x, y) is negative definite unstable critical point Theorem 962 asserts that the origin is an (b) We now consider the system ( ) ( ) ( x a a = 2 x y a 2 a 22 y ) + ( ) F (x, y), G (x, y) in which F (x, y)/r and G (x, y)/r as r Let in which V (x, y) = Ax 2 + Bxy + Cy 2, A = a2 2 + a (a a 22 a 2 a 2 ) 2 B = a 2a 22 + a a 2 C = a2 + a (a a 22 a 2 a 2 ) 2 and = (a + a 22 )(a a 22 a 2 a 2 ) Based on the hypothesis, A, B > Except for the origin, V (x, y) is positive on each of the coordinate axes Along each trajectory, V = x 2 + y 2 + (2Ax + By)F (x, y) + (2Cy + Bx)G (x, y) Converting to polar coordinates, for r, V = = r 2 + r(2a cos θ + B sin θ) F + r(2c sin θ + B cos θ) G [ = r 2 + r 2 (2A cos θ + B sin θ) F r + (2C sin θ + B cos θ) G ] r,

449 Since the system is almost linear, there is an R such that (2A cos θ + B sin θ) F r + (2C sin θ + B cos θ) G r < 2, and hence for r < R It follows that (2A cos θ + B sin θ) F r + (2C sin θ + B cos θ) G r > 2 V > 2 r2 as long as < r < R Hence V is positive definite on the domain D = { (x, y) x 2 + y 2 < R 2} By Theorem 962, the origin is an unstable critical point 97 3 The critical points of the ODE dr dt = r(r )(r 3) are given by r =, r 2 = and r 3 = 3 Note that dr dt > for < r < and r > 3 ; dr dt < for < r < 3 r = corresponds to an unstable critical point The critical point r 2 = is asymptotically stable, whereas the critical point r 3 = 3 is unstable Since the critical values are isolated, a limit cycle is given by r =, θ = t + t which is asymptotically stable Another periodic solution is found to be which is unstable 5 The critical points of the ODE r = 3, θ = t + t dr dt = sin πr are given by r = n, n =,, 2, Based on the sign of r in the neighborhood of each critical value, the critical points r = 2k, k =, 2, correspond to unstable periodic solutions, with θ = t + t The critical points r = 2k +, k =,, 2, correspond to stable limit cycles, with θ = t + t The solution r = represents an unstable critical point

450 45 Chapter 9 Nonlinear Differential Equations and Stability 6 The critical points of the ODE dr = r r 2 (r 3) dt are given by r =, r 2 = 2 and r 3 = 3 Note that dr dt < for < r < 3 ; dr dt > for r > 3 r = corresponds to an asymptotically stable critical point The critical points r 2 = 2 is semistable, whereas the critical point r 3 = 3 is unstable Since the critical values are isolated, a semistable limit cycle is given by Another periodic solution is found to be which is unstable r = 2, θ = t + t r = 3, θ = t + t Given F (x, y) = a x + a 2 y and G(x, y) = a 2 x + a 22 y, it follows that F x + G y = a + a 22 Based on the hypothesis, F x + G y is either positive or negative on the entire plane By Theorem 972, the system cannot have a nontrivial periodic solution 2 Given that F (x, y) = 2x 3y xy 2 and G(x, y) = y + x 3 x 2 y, F x + G y = x 2 y 2 Since F x + G y < on the entire plane, Theorem 972 asserts that the system cannot have a nontrivial periodic solution 4(a) Based on the given graphs, the following table shows the estimated values: µ = 2 T 629 µ = T 666 µ = 5 T 6 (b) The initial conditions were chosen as x() = 2, y() = For µ = 5, T 638 :

451 97 45 For µ = 2, T 765 : For µ = 3, T 886 : For µ = 5, T 25 : (c) The period, T, appears to be a quadratic function of µ

452 452 Chapter 9 Nonlinear Differential Equations and Stability 5(a) Setting x = u and y = u, we obtain the system of equations dx dt = y dy dt = x + µ( 3 y2 )y (b) Evidently, y = It follows that x = Hence the only critical point of the system is at (, ) The components of the vector field are infinitely differentiable everywhere Therefore the system is almost linear The Jacobian matrix of the vector field is ( ) J = µ µy 2 At the critical point (, ), the coefficient matrix of the linearized system is ( ) J(, ) =, µ with eigenvalues r,2 = µ 2 ± 2 µ2 4 If µ =, the equation reduces to the ODE for a simple harmonic oscillator For the case < µ < 2, the eigenvalues are complex, and the critical point is an unstable spiral For µ 2, the eigenvalues are real, and the origin is an unstable node (c) The initial conditions were chosen as x() = 2, y() = µ = : A 26 and T 665

453 (d) µ = 2: A 2 and T 63 µ = 5: A 24 and T 638 µ = 2: A 26 and T 762 µ = 5: A 437 and T 6

454 454 Chapter 9 Nonlinear Differential Equations and Stability (e) A T µ = µ = µ = µ = µ = (a) The critical points are solutions of the algebraic system µ x + y x(x 2 + y 2 ) = x + µ y y(x 2 + y 2 ) = Multiply the first equation by y and the second equation by x to obtain µ xy + y 2 xy(x 2 + y 2 ) = x 2 + µ xy xy(x 2 + y 2 ) = Subtraction of the two equations results in which is satisfied only for x = y = x 2 + y 2 =, (b) The Jacobian matrix of the vector field is ( ) µ 3x J = 2 y 2 2xy 2xy µ x 2 3y 2 At the critical point (, ), the coefficient matrix of the linearized system is ( ) µ J(, ) =, µ resulting in the linear system x = µ x + y y = x + µ y

455 The characteristic equation for the coefficient matrix is λ 2 2µ λ + µ 2 + =, with solution λ = µ ± i For µ <, the origin is a stable spiral When µ =, the origin is a center For µ >, the origin is an unstable spiral (c) Introduce polar coordinates r and θ, so that x = r cos θ and y = r sin θ for r Multiply the first of Eqns (i) by x and the second equation by y to obtain Addition of the two equations results in Since r 2 = x 2 + y 2 x x = µ x 2 + xy x 2 (x 2 + y 2 ) y y = xy + µ y 2 y 2 (x 2 + y 2 ) x x + y y = µ(x 2 + y 2 ) (x 2 + y 2 ) 2 and r r = x x + y y, it follows that r r = µ r 2 r 4 and dr dt = µ r r3 for r > Multiply the first of Eqns (i) by y and the second equation by x, the difference of the two equations results in y x x y = x 2 + y 2 Since y x x y = r 2 θ, the above equation reduces to dθ dt = (d) From r = r(µ r 2 ) and θ =, it follows that one solution of the system is given by r = µ and θ = t + t valid for µ > This corresponds to a periodic solution with a circular trajectory Since r, observe that dr dt = r(µ r2 ) < for r > µ = r(µ r 2 ) > for µ > r > Hence solutions with initial condition r() µ are attracted to the limit cycle 7(a) The critical points are solutions of the algebraic system y = x + µ ( x 2 )y = Clearly, y =, and this implies that x = So the origin is the only critical point The Jacobian matrix of the vector field is ( ) J = 2µxy µ( x 2 )

456 456 Chapter 9 Nonlinear Differential Equations and Stability At the critical point (, ), the coefficient matrix of the linearized system is J(, ) = µ The characteristic equation is λ2 µλ + =, and the roots are p µ ± µ2 4 λ= 2 This implies that the origin is an asymptotically stable node for µ < 2, an asymptotically stable spiral point for 2 < µ <, an unstable spiral point for < µ < 2 and an unstable node for µ > 2 (b) (c) (a) µ = 25 (b) µ = 2 (c) µ = 5 (d) µ = 5

457 (d) (a) µ = 2 (b) µ = (c) µ = (d) µ = 2 9(a) The critical points are solutions of the algebraic system x(a x 5 2y x + 6 ) = y( 4 + x x + 6 ) = If x =, then y = If y =, then x = 5a from the first equation The third critical point comes from x = 2, and then the first equation implies that y = 4a 8/5 So the critical points are (, ), (5a, ) and (2, 4a 8/5) (b) The Jacobian matrix of the vector field is ( a 2x 5 J = 2y (x+6) 2 2x x+6 6y (x+6) x x+6 At the critical point (2, 4a 8/5), the coefficient matrix of the linearized system is ( ) a/4 /2 /2 J(2, 4a 8/5) = 3a/8 3/2 The characteristic equation is λ 2 + λ(/2 a/4) + (3a/8 3/2)/2 =, and the roots are λ = a 8 4 ± a a + 2 We can conclude that a = 2 )

458 458 Chapter 9 Nonlinear Differential Equations and Stability (c) 2(a) The critical points are solutions of the algebraic system (b + )x + x 2 y/4 = b x x 2 y/4 = Add the two equations to obtain x =, with solution x = The second of the above equations yields y = 4 b (b) The Jacobian matrix of the vector field is J = ( ) 4(b + ) + 2xy x 2 4 4b 2xy x 2 At the critical point (, 4b), the coefficient matrix of the linearized system is ( ) b /4 J(, 4b) =, b /4 with characteristic equation and eigenvalues λ 2 + ( 5 4 b)λ + 4 = λ = 5 4b ± 9 4 b + 6 b 2 8 = (5/4 b) ± (5/4 b) 2 2 (c) Let L = 5/4 b The eigenvalues can be expressed as λ = L ± L 2 2 We find that the eigenvalues are real if L 2 and are complex if L 2 < For the complex case, that is, < L <, the critical point is a stable spiral if < L < ; it is an unstable spiral if < L < That is, the critical point is an asymptotically stable spiral if /4 < b < 5/4 and is an unstable spiral if 5/4 < b < 9/4 When L 2 >, the critical point is a node The critical point is a stable node if L > ; it is an unstable node if L < That is, the critical point is an asymptotically stable node if < b < /4 and is an unstable node if b > 9/4

459 (d) From part (c), the critical point changes from an asymptotically stable spiral to an unstable spiral when b = 5/4 (e,f) (a) b = (b) b = 5 (c) b = 75 (d) b = r = 28, with initial point (5, 5, 5): r = 28, with initial point (5, 5, 5):

460 46 Chapter 9 Nonlinear Differential Equations and Stability 7 r = 28 : 9(a) r =, initial point ( 5, 3, 55 ) : The period appears to be T 2 (b) r = 9994, initial point ( 5, 3, 55 ) : The periodic trajectory appears to have split into two strands, indicative of a perioddoubling Closer examination reveals that the peak values of z(t) are slightly different r = 997, initial point ( 5, 3, 55 ) :

461 98 46 (c) r = 996, initial point ( 5, 3, 55 ) : The strands again appear to have split Closer examination reveals that the peak values of z(t) are different (a) r = 5, initial point ( 5, 3, 55 ) : r = 7, initial point ( 5, 3, 55 ) :

462 462 Chapter 9 Nonlinear Differential Equations and Stability (b) r = 8, initial point ( 5, 3, 55 ) : r = 8, initial point ( 5, 3, 55 ) : The strands of the periodic trajectory are beginning to split apart r = 82, initial point ( 5, 3, 55 ) :

463 r = 83, initial point ( 5, 3, 55 ) : r = 84, initial point ( 5, 3, 55 ) : 2 The system is given by x = y z y = x + y/4 z = /2 + z(x c) (a) We obtain that the critical points (when c 2 > /2) are ( c 2 + 4c2 2, 2c 4 4c 2 2, 2c + 4c 2 2), ( c 2 4 4c2 2, 2c + 4c 2 2, 2c 4c 2 2) The Jacobian of the system is J = /4 z x c In the following computations, we approximate the values to 4 decimal points When c = 3, the two critical points are (954, 4787, 4787) and (46, 483, 483)

464 464 Chapter 9 Nonlinear Differential Equations and Stability The corresponding eigenvalues of the Jacobian are 893, 29 ± 247i and 963, 8 ± 652i (b) (c) (d) T 59

465 (b) Using the eigenvalue idea: c Critical Point Complex Eigenvalue Pair 2 (52, 469, 469) 63 ± 859i 25 (96, 4384, 4384) ± 747i Clearly, somewhere in between we have a Hopf bifurcation Similar computations show that the bifurcation value is c 243 4(a) When c = 3, the two critical points are (29577, 83, 83) and (423, 69, 69) The corresponding eigenvalues of the Jacobian are 2273, 98 ± 3582i and 2953, 988 ± 9969i (b) (c) T 2 8 5(a) When c = 38, the two critical points are (37668, 5673, 5673) and (332, 327, 327)

466 466 Chapter 9 Nonlinear Differential Equations and Stability The corresponding eigenvalues of the Jacobian are 2324, 78 ± 478i and 37335, 83 ± 994i (b) T (c)