Introductory Differential Equations

Introductory Differential Equations Lecture Notes June 3, 208

Contents Introduction Terminology and Examples 2 Classification of Differential Equations 4 2 First Order ODEs 5 2 Separable ODEs 5 22 First order linear ODEs 8 23 Exact ODEs 0 24 Autonomous ODEs and Stability of Equilibrium 4 25 Existence and Uniqueness of Solutions 8 26 Euler s Method 20 27 Modeling with First Order ODEs 22 3 Second Order ODEs 24 3 Linear Homogeneous ODEs with Constant Coefficients 26 32 Linear Homogeneous ODEs and Wronskian 29 33 Linear Homogeneous ODEs : Repeated Real Roots 32 34 Linear Homogeneous ODEs : Complex Roots 34 35 Linear Nonhomogeneous ODEs : Undetermined Coefficients 37 36 Variation of Parameters 42 37 Modeling with Second Order ODEs 45 4 Higher Order ODEs 47 4 General Solutions 47 42 Higher Order ODEs : Undetermined Coefficients 50 5 Systems of Differential Equations 52 5 Introduction to Linear Algebra 52 52 Introduction to Systems of Linear ODEs 58 53 Solving Linear Systems with constant coefficients 60 54 Equilibrium Solutions and Phase Portraits 63 55 Complex Eigenvalues for Linear Systems 67 56 Repeated Eigenvalues for Linear Systems 70 6 Series Solutions of Differential Equations 72 6 Power Series 72 62 Series Solutions about an Ordinary Point 75

Introduction Terminology and Examples A differential equation is an equation containing a function and its derivatives Example 2 3 4 dy dx = y sin x dv dt = g γ v (motion of a falling object) m d 2 θ dt + g sin θ 2 L = 0 (motion of a pendulum) u t = k 2 u (one-dimensional heat equation) x2 Since derivative of a function means rate of change of the function, many physical phenomenon can be expressed or modelled as differential equations Motion of a falling object: Suppose an object of mass m is falling under the gravitational force Suppose v is its velocity at time t So its acceleration is dv If F is the force exerted on the object at time t, by dt Newton s second law of motion we have F = m dv dt Note that two external forces are acting on the object: mg, the weight of the object (g is the gravitational acceleration which is 98 m/s 2 ) 2 D, the drag or air resistance It is modelled to be proportional to v So D = γv where γ is the drag coefficient which depends on the shape and the velocity of the object and the density of the air Thus F = mg γv From (), we have () Dividing both sides by m, we get m dv dt = mg γv dv dt = g γ m v (2)

A particular case: Suppose a particle of mass 5 kg is dropped Suppose the drag coefficient is kg/m Then (2) becomes the following initial value problem (IVP) dv dt = 98 v, v(0) = 0 (3) 5 Solve the above differential equation, ie, find v as a function of t dv dt = 98 v 5 = 49 v 5 dv 49 v = dt 5 dv dt v 49 = 5 ln v 49 = t + c, where c is an integration constant 5 v 49 = e c t/5 = e c e t/5 v = 49 + ke t/5, where k = ±e c Checking the solution: For v = 49 + ke t/5, ( dv dt = ke t/5 ) ( = (v 49) ) = 98 v 5 5 5 Note that for each value of k, v = 49 + ke t/5 satisfies the following differential equation dv dt = 98 v 5 There are infinitely many solutions for differential equation (4) So v = 49 + ke t/5 is called the general solution of differential equation (4) Infinitely many solution curves corresponding to the general solution are called integral curves of the general solution Assigning a particular value of k in the general solution we get a particular solution to differential equation (4) v (4) v = 49 v = 49 49e t/5 Integral curves for dv dt = 98 v 5 t 2

We can find a particular value of k from v = 49+ke t/5 using the initial condition v(0) = 0: 0 = 49 + ke 0 = k = 49 So v = 49 49e t/5 is the solution of the initial value differential equation (3) End behavior of v: Note that as t, v = 49 + ke t/5 49 Equilibrium Solution and Direction Fields: An equilibrium solution is a particular solution to a differential equation that does not change with the independent variable, ie, the derivative is zero For the preceding problem, v = 49 is an equilibrium solution Clearly dv dt = 98 v 5 dv dt = 98 v 5 = 0 = v = 49 > 0 when v < 49 and dv dt = 98 v 5 < 0 when v > 49 In the t v graph, for a given value of v, we draw a vector with the slope dv For example, dt for v = 49, we have dv = 0 and we get horizontal vectors on the line v = 49 For v = 45, we dt have dv = 08 and we get vectors with slope 08 on the line v = 45 This vector plotting on dt the tv-plane gives the direction field or slope field of differential equation (4) 3

2 Classification of Differential Equations Partial differential equations (PDE): A differential equation that contains partial derivatives is called a partial differential equation Note that a PDE has more than one independent variables For example, u t = k 2 u (heat equation) (5) x2 Ordinary differential equations (ODE): A differential equation that does not contain partial derivatives is called an ordinary differential equation Note that an ODE has only one independent variable For example, dv dt = g γ v (motion of falling objects) (6) m The order of a differential equation is the the order of the highest derivative in it Differential equation (6) has order Differential equation (5) has order 2 For an ODE, if the independent variable is t, then it is sometimes written as F (t, y(t), y, y,, y (n) ) = 0 or, y (n) = f(t, y(t), y, y,, y (n ) ) Linear ODE: An ODE F (t, y(t), y, y,, y (n) ) = 0 is linear if F is a linear function of t, y(t), y, y,, y (n) It can be written as a 0 (t)y (n) + a (t)y (n ) + + a n (t)y = g(t) For example, ty + y sin t = e t is a linear ODE of order 2 But the following 2nd order ODEs are nonlinear ty + yy = e t ty + y 2 = e t ty + sin(t + y) = e t 4

2 First Order ODEs 2 Separable ODEs A separable ODE is of the form Steps to solve: dy dx = M(x) N(y) N(y) dy = M(x) dx N(y) dy = M(x) dx + c Example Solve the following IVP (initial value problem) dy dx = 2x 3, y(0) = 3 (7) y 5 and find the valid interval of the solution Find the maximum value of the solution Solution dy dx = 2x 3 y 5 (y 5) dy = (2x 3) dx (y 5) dy = (2x 3) dx y 2 5y = 2x2 2 2 3x + c y 2 0y = 2(x 2 3x + c) y 2 0y 2x 2 + 6x 2c = 0 (8) Note that the initial condition is y(0) = 3 So we have From (8) we get 3 2 0 3 2c = 0 2c = 2 y 2 0y 2x 2 + 6x + 2 = 0 (implicit solution) y = 0 ± 00 4( 2x 2 + 6x + 2) 2 y = 0 ± 2 25 ( 2x 2 + 6x + 2) 2 y = 5 ± 2x 2 6x + 4 5

So we have two possible solutions: y = 5 + 2x 2 6x + 4 and y = 5 2x 2 6x + 4 Note that the first one does note satisfy the initial condition y(0) = 3 Thus the solution is y = 5 2x 2 6x + 4 (explicit solution) Obviously the maximum value of the solution is 5 because 2x 2 6x + 4 0 The solution is defined when 2x 2 6x + 4 0 2(x 2 3x + 2) 0 2(x )(x 2) 0 The domain of y is (, [2, ) Because of the initial condition y(0) = 3, the valid interval of the solution is (, Example Solve the following ODE Solution dy dx = e y x cos x (9) dy dx = x cos x e y e y dy = x cos x dx e y dy = x cos x dx e y = u dv u = x, dv = cos x dx = uv v du du = dx, v = dv = = x sin x sin x dx cos x dx = sin x = x sin x + cos x + c The general solution is y = ln x sin x + cos x + c 6

Substitution for homogeneous ODEs: The ODE dy dx = f(x, y) is called homogeneous if f(x, y) is a function of y x Steps to solve: Substitute v = y/x or, y = vx Then dy dx = xdv + v Write f(x, y) as a dx function of v Then it becomes a separable ODE of v and x Example Solve the following ODE Solution dy dx = x2 + 3y 2 2xy dy dx = (x2 + 3y 2 )/x 2 2xy/x 2 dy dx = + 3(y/x)2 2(y/x) (0) Let v = y/x or, y = vx Then dy dx = xdv + v So from (0) we get dx x dv dx + v = + 3v2 2v x dv dx = + 3v2 v = + v2 2v 2v 2v + v dv = dx 2 x 2v dx + v dv = 2 x (substitute u = + v2 ) ln + v 2 = ln x + ln c ln + v 2 = ln cx + v 2 = cx + y2 x 2 = cx x 2 + y 2 = cx 3 7

22 First order linear ODEs A first order linear ODE can be written as follows where p and g are functions of x dy + p(x)y = g(x), dx Steps to solve: Multiply both sides by the integrating factor e p(x) dx : dy dx e p dx + p(x)ye x dx = g(x)e p dx d ( ye ) p dx = g(x)e p dx dx ( d ye ) ( p dx = g(x)e p dx) dx ye ( p dx = g(x)e p dx) dx + c [ y = e ( p dx g(x)e p dx) dx + c Example Solve the following first order linear ODE x dy dx + (2x2 + )y = 2x, y() = 2 Find the valid interval of the solution Find the behavior of y when x Solution Let s write it in the standard form ( dy dx + 2x + ) y = 2, y() = 2 () x The IF is e (2x+ x) dx = e x2 +ln x = e x2 e ln x = xe x2 Multiply both sides of () by the IF xe x2 ( dy dx xex2 + y 2x + ) xe x2 = 2xe x2 x d ( yxe x2) = 2xe x2 dx ( d yxe x2) = 2xe x2 dx (substitute u = x 2 ) From the initial condition y() = 2, we get yxe x2 = e x2 + c 2 e = e + c = c = e 8

Thus we get yxe x2 = e x2 + e and the solution is y = ex2 + e xe x2 The domain of y is (, 0) (0, ) Because of the initial condition y() = 2, the valid interval of the solution is (0, ) Note that y = x + e 0 as x xe x2 Example Solve the following first order linear ODE dy dt + y cos t = cos t, y(0) = π Solution The IF is e cos t dt = e sin t Multiply both sides of the ODE by the IF e sin t dy dt esin t + y cos t e sin t = cos t e sin t d ( ) ye sin t = cos t e sin t dt d ( ye sin t) = cos t e sin t dt ye sin t = e u du u = sin t, du = cos t dt = e u + c = e sin t + c From the initial condition y(0) = π, we get π e 0 = e 0 + c = c = π Thus ye sin t = e sin t + π and hence the solution is y = esin t + π e sin t = + (π )e sin t 9

23 Exact ODEs An ODE of the form M(x, y) + N(x, y) dy dx = 0 (2) or, M(x, y) dx + N(x, y) dy = 0 is called exact if it can be written as (f(x, y)) = 0 x for some function f(x, y) Then the solution is f(x, y) = c Note that comparing with (2) we get Thus we have x (f(x, y)) = f x(x, y) + f y (x, y) dy dx = 0 f x (x, y) = M(x, y), f y (x, y) = N(x, y) M y (x, y) = f yx (x, y) = f xy (x, y) = N x (x, y) Theorem Let M, N, M y and N x be continuous functions on R = (a, b) (c, d) Then the ODE M(x, y) dx + N(x, y) dy = 0 is exact in R if and only if at each point in R M y (x, y) = N x (x, y) Example is exact since Example is non-exact since M y = y M y = y 2xy dx + (x 2 + 3y 2 ) dy = 0 ( (2xy) = 2x = x 2 + 3y 2) = N x x 2xy 2 dx + (x 2 + 3y 2 ) dy = 0 ( ) 2xy 2 = 4xy 2x = ( x 2 + 3y 2) = N x x 0

Steps to solve: Find a function f(x, y) such that f x (x, y) = M(x, y) f y (x, y) = N(x, y) 2 The solution is f(x, y) = c Example Solve the ODE Solution Note that M y = y 2xy + (x 2 + 3y 2 ) dy dx = 0 ( (2xy) = 2x = x 2 + 3y 2) = N x x So the given ODE is exact and there is a function f(x, y) such that f x (x, y) = 2xy (3) f y (x, y) = x 2 + 3y 2 (4) Integrating (3) partially with respect to x, we get f(x, y) = f x (x, y) dx = 2xy dx = x 2 y + h(y) (5) Differentiating it partially with respect to y, we get Comparing (4) and (6), we get h (y) = 3y 2 Thus h(y) = 3y 2 dy = y 3 f y (x, y) = x 2 + h (y) (6) From (5) we get Thus the solution is f(x, y) = x 2 y + y 3 x 2 y + y 3 = c

Non-exact to Exact: Sometimes a non-exact ODE can made exact when multiplied by an integrating factor µ(x, y) For example, (2x 2 + y 2 ) dx + xy dy = 0 is non-exact, but is exact where µ(x, y) = 2x 2x [(2x 2 + y 2 ) dx + xy dy = 0 ie, (4x 3 + 2xy 2 ) dx + 2x 2 y dy = 0 Finding such integrating factor µ(x, y) is complicated So we will concentrate on the IF that is a function of x, ie, µ(x, y) = µ(x) and µ y = 0 Suppose the following ODE is exact Then µ(x)m(x, y) + µ(x)n(x, y) dy dx = 0 (µm) y = (µn) x µ y M + µm y = µ x N + µn x µm y = µ x N + µn x since µ y = 0 µ x N = µm y µn x dµ dx = µm y N x N dµ µ = My N x dx N My N x ln µ = dx N µ = e My Nx N dx Example Find an integrating factor that makes the following non-exact ODE exact when multiplied by it (2x 2 + y 2 ) xy dy dx = 0 Solution Here M = 2x 2 + y 2 and N = xy So M y N x N = (2y) ( y) xy = 3 x 2

dµ dx = µm y N x = µ 3 N x dµ dx µ = 3 x ln µ = 3 ln x = ln(x) 3 = ln µ = x 3 ( ) x 3 So the IF is µ = x 3 We can verify that the following ODE is exact x 3 [(2x2 + y 2 ) xy dy dx = 0 ( ) 2 x + y2 y dy x 3 x 2 dx = 0 Checking: M y = y ( ) 2 x + y2 = 2y x 3 x = ( y ) = N 3 x x 2 x The solution is 2 ln x y2 = c (verify) 2x2 3

24 Autonomous ODEs and Stability of Equilibrium An ODE of the form dy dx Example dy dx = y(y ) 2 y = e y sin y 3 dp dt = (00 P )P = f(y) is called autonomous Recall that an equilibrium solution to an ODE is a solution y = g(x) for which dy dx for the autonomous ODE dy dx For example, equilibrium solutions of dy dx = 0 So = f(y), equilibrium solutions are the solutions of f(y) = 0 = y(y ) are y = 0, Exponential population growth: Suppose P (t) is the population of a species at time t The simplest assumption may be that the population growth is proportional to the current population: dp dt = rp (7) where r is a constant representing growth when r > 0 and decline when r < 0 Solving (7) we get P = ce rt (verify) If the initial population is P (0) = P 0, then the solution to (7) is P P = P 0 e rt P = 2e t/3 P = 5e t/3 P = e t/3 2 P = 5e t/3 t P = 0 Integral curves for P 0 = 5, 5, 2 with r = /3 4

Note that P = 0 is the only equilibrium solution to (7) Since integral curves are going away from the equilibrium solution P = 0, this is called unstable equilibrium solution Logistic population growth: Suppose P (t) is the population of a species at time t A rational assumption may be that the population will grow when the current population is small and will decline when the current population is big because of the lack of resources There may be an ideal population size K (called carrying capacity) such that if P < K, then population will grow and if P > K, then population will decline In 838 Pierre-Francois Verhulst modeled it as follows: dp dt = rp ( P K ) (8) Equilibrium solutions: The ODE (8) has two equilibrium solutions P = 0, K These two equilibrium solutions have different flavors To see this, we draw integral curves of (8) from direction field using different initial population P 0 = P (0) When P 0 > K, dp dt < 0 and P K as t Also when 0 < P 0 < K, dp > 0 and P K as t So dt P = K is a stable equilibrium as arrows (integral curves) are moving toward P = K But arrows (integral curves) are moving away from P = 0 So P = 0 is a unstable equilibrium We can verify these observations after we solve (8) K dp dt = KrP K dp = r dt P (K P ( ) P + dp = r dt K P ln P ln(k P ) = rt + c ( P K ) = rp (K P ) 5

( ) P ln = rt + c K P P K P = ert+c = ce rt (where c = e c ) (9) P = ce rt (K P ) P ( + ce rt ) = Kce rt P = Kcert ( + ce rt ) If the initial population is P (0) = P 0, then from (9) we get Plugging c into (20), we get P = c = P 0 K P 0 KP 0 e rt K + P 0 (e rt ) KP 0 e rt Note that P = K + P 0 (e rt ) = K + (K P 0 )e rt P0 K as t Thus P = K is an (asymptotically) stable equilibrium Steps for classifications of stability: Suppose y = c is an equilibrium of y = f(y) If f(y) > 0 when y < c, and f(y) < 0 when y > c, then the equilibrium solution y = c is stable In this case arrows above and below y = c are moving toward y = c 2 If f(y) < 0 when y < c, and f(y) > 0 when y > c, then the equilibrium solution y = c is unstable In this case arrows above and below y = c are moving away from y = c 3 If f(y) has the same sign when y < c and y > c, then the equilibrium solution y = c is semistable In this case arrows above and below y = c are moving away from y = c from one side and moving toward y = c from another side (20) Y Y Y X y = 0 X y = 0 X y = 0 Stable equilibrium y = 0 Unstable equilibrium y = 0 6 Semistable equilibrium y = 0

Example Find the equilibrium solutions of y = y 2 (y )(y 2) and classify the stability of the equilibrium solutions Solution y = y 2 (y )(y 2) = 0 = y = 0,, 2 So equilibrium solutions are y = 0,, 2 Note that y = y 2 (y )(y 2) > 0 when y > 2 and y = y 2 (y )(y 2) < 0 when < y < 2 Thus y = 2 is an unstable solution Also y = y 2 (y )(y 2) < 0 when < y < 2 and y = y 2 (y )(y 2) > 0 when y < Thus y = is a stable solution But y = y 2 (y )(y 2) > 0 when y < which implies y = 0 is a semistable solution These are also clear from the following direction field 7

25 Existence and Uniqueness of Solutions We ask two questions for the following ODE dy dx = f(x, y), y(x 0) = y 0 (2) Is there any solution to (2)? (existence of solution?) 2 Is there a unique solution to (2)? (uniqueness of solution?) The answers are given by the Existence and Uniqueness Theorem: Theorem If f is a continuous function on a neighborhood (a, b) (c, d) of (x 0, y 0 ), then there is a solution y = φ(x) to (2) in a neighborhood N of x 0 where N is inside (a, b) 2 If f and f y are continuous functions on a neighborhood (a, b) (c, d) of (x 0, y 0 ), then there is a unique solution y = φ(x) to (2) in a neighborhood N of x 0 where N is inside (a, b) d Y y 0 y = φ(x) c a x 0 b X N Example y = 2y x, y() = 3 (a) Determine if the above ODE has a solution (b) Find the largest interval in which the following ODE has a unique solution Solution (a) Since f(x, y) = 2y is continuous function for all values of (x, y) where x 0, x the give ODE has a solution y = φ(x) in any neighborhood of not containing 0 (b) f(x, y) = 2y f and x y = 2 are continuous functions for all values of (x, y) where x 0 x Because of the initial condition y() = 3, the given ODE has a unique solution y = φ(x) in the interval (0, ) (which is the largest possible neighborhood of for y = φ(x)) Note 8

If the initial condition were y( ) = 3, then the largest interval would be (, 0) for the unique solution y = φ(x) 2 Solving the ODE y = 2y, we get y = x cx2 So y = cx 2 is a solution of y = 2y x value of c for each (a) If the initial condition were y(0) = 3, then there is no solution which makes sense because f(x, y) = 2y is not continuous at (0, 3) x (b) If the initial condition were y(0) = 0, then there is infinitely many solutions of the form y = cx 2 It is justified because f(x, y) = 2y f and x y = 2 are not continuous x at (0, 0) (c) If the initial condition were y(x 0 ) = y 0 where x 0 0, then there is a unique solution y = y 0 x 2 It is justified because f(x, y) = 2y f and x 2 0 x y = 2 are continuous at x and near (x 0, y 0 ) where x 0 0 Y 4 y = x 2 /3 3 y = x 2 /5 y = x 2 /7 2 4 3 2 0 2 3 4 X Example Find the largest interval in which the following ODE has a unique solution Justify your answer (t )(t 3)y 2y = 0, y(2) = 5 Solution y = 2y, y(2) = 5 (t )(t 3) 2y f f(t, y) = and (t )(t 3) y = 2 are continuous functions for all values of (t )(t 3) (t, y) where t, 3 Because of the initial condition y(2) = 5, the given ODE has a unique solution y = φ(t) in the interval (, 3) (which is the largest possible neighborhood of 2 for y = φ(t)) 9

26 Euler s Method Can we numerically find an approximate solution to the following ODE? dy dx = f(x, y), y(x 0) = y 0 Euler s Method: The tangent line to a solution y = φ(x) at the point (x 0, y 0 ) has slope dy = f(x 0, y 0 ) So an equation of the tangent line would be dx (x 0,y 0 ) y = y 0 + (x x 0 )f(x 0, y 0 ) If x is close to x 0, then y = y 0 + (x x 0 )f(x 0, y 0 ) would be a good approximation to y = φ(x ) Similarly if x 2 is close to x, then y 2 = y + (x 2 x )f(x, y ) would be a good approximation to y = φ(x 2 ) Thus a general formula for this tangent line expression would be y n+ = y n + (x n+ x n )f(x n, y n ), n = 0,, 2, (22) If we take x 0, x, x 3, with a fixed difference h (ie x n+ = x n + h), then (22) becomes y n+ = y n + hf(x n, y n ), n = 0,, 2, (23) Example Use Euler s method with step size h = 05 to approximate y(4) where y = φ(x) is a solution to the following ODE dy dx = x y, y(0) = 2 Solution We have h = 05, x 0 = 0, y 0 = 2 and f(x, y) = x y So from (23) we have x = x 0 + 05 = 05, y = y 0 + 05(x 0 y 0 ) = 2 + 05(0 2) = x 2 = x + 05 =, y 2 = y + 05(x y ) = + 05(05 ) = 075 n x n y n 0 0 2 05 2 075 3 5 0875 4 2 875 5 25 59375 6 3 2046875 7 35 25234375 8 4 307875 So y(4) y 8 = 307875, ie, y(4) is approximately y 8 = 307875 20

4 Y y = (x ) + 3e x 3 2 0 2 3 4 X Euler s tangent line approximation to dy dx = x y, y(0) = 2 with h = 05 Checking solution: Solving dy dx = x y, y(0) = 2, we get y = (x ) + 3e x The exact value of y(4) is (4 ) + 3e 4 = 3083 which was approximated by 307 (pretty close!) 2

27 Modeling with First Order ODEs Radioactive Decay: Let N(t) be the mass of a radioactive element at time t The rate of change (decay) of N is proportional to its current value dn dt = kn, where k > 0 depends on the element Solving this ODE we get N = ce kt Suppose the initial amount is N 0 = N(0) Then we get c = N 0 Thus the solution is N(t) = N 0 e kt Carbon Dating: There are 2 types of carbon atoms: 2 C (the stable nuclide) and radioactive 4 C with a halflife of about 5,730 years The ratio 4 C : 2 C is approximately constant in nature A living creature taking carbon from nature is made up with this ratio After its death 4 C starts to decay Example Suppose 25% of the original amount of 4 C remained in a fossil Find the age of the fossil Solution First we find k for 4 C in N(t) = N 0 e kt We know that N 0 2 = N 0 e 5730k Solving we get k = ln 2/5730 and hence N(t) = N 0 e t ln 2/5730 For the fossil we want to find t for which N(t) = 25N 0 /00 Plugging this into the preceding equation we get t ln 2/5730 25N 0 /00 = N 0 e = t =, 460 years Newton s Law of Cooling: Let T (t) be the temperature of an object at time t Then dt dt = k(t T a), where k > 0 is a constant and T a is the constant ambient temperature Solving this ODE we get T = T a + ce kt Suppose the initial temperature is T 0 = T (0) Then we get c = T 0 T a Thus the solution is T (t) = T a + (T 0 T a )e kt Note that T (t) = T a + (T 0 T a )e kt T a as t 22

Kirchhoff s Circuit Law: Consider an electric circuit with a capacitor, resistor, and battery Let Q(t) be the charge of the capacitor at time t Then R dq dt + Q C = V, where R is the constant resistant, C is the constant capacitance, and V is the constant voltage supplied Solving this ODE we get Q = CV + ke t/rc Suppose the initial charge is Q(0) = 0 Then we get k = CV Thus the solution is Q(t) = CV ( e t/rc ) Note that Q(t) = CV ( e t/rc ) CV as t Mixing Rate Problems: Suppose a tank contains 200 gal of salt water containing 00 lb salt Then salt water with 3 lb salt/gal is pumped into the tank at a rate 2 gal/min and the well-stirred salt water is pumped out of the tank at a rate 2 gal/min Suppose Q(t) denotes the amount of salt in lb at time t min Then dq dt = 2 3 2 Q, Q(0) = 00 200 Solving this ODE we get Q = 600 + ce t/00 Using the initial condition Q(0) = 00 we get c = 500 Thus the solution is Q(t) = 600 500e t/00 Note that Q(t) = 600 500e t/00 600 as t 23

3 Second Order ODEs A second order ODE has the following form d 2 y dx 2 = f ( x, y, dy ) dx A second order ODE is called linear if it can be written as d 2 y dx = p(x)dy q(x)y + g(x) or, 2 dx d 2 y dx + p(x)dy + q(x)y = g(x) (24) 2 dx The linear ODE (24) is called homogeneous if g(x) = 0 The general solution of a linear homogeneous second order ODE has two constants which are found using two initial conditions Example The general solution of y + y 2y = 0 is y = c e x + c 2 e 2x Suppose the initial conditions are y(0) =, y (0) = 4 Using y(0) =, we get c + c 2 = (25) Then we find y = c e x 2c 2 e 2x Using y (0) = 4, we get c 2c 2 = 4 (26) Solving (25) and (26), we get c = 2 and c 2 = Thus the solution is y = 2e x e 2x In chapter 3, we mostly focus on linear second order ODE with constant coefficients: d 2 y dx + ady + by = g(x) 2 dx 24

Reduction of order: Solve y + y = e x Solution Substituting u = y, we get the first order ODE Multiplying by the IF e x, we get u + u = e x u e x + ue x = e x e x d ) = dx(uex d(ue x ) = dx ue x = x + c dy = u = (x + c)e x dx dy = (x + c)e x dx y = (x + c + )e x + d 25

3 Linear Homogeneous ODEs with Constant Coefficients In this section, we learn how to solve a linear homogeneous second order ODE with constant coefficients a d2 y dx + bdy + cy = 0, where a 0 (27) 2 dx Suppose y = e rx is a solution of (27) for some values of r Then y = re rx and y = r 2 e rx Plugging these into (27), we get ar 2 e rx + bre rx + ce rx = 0 e rx (ar 2 + br + c) = 0 Since e rx 0, we get a quadratic equation in r, called the characteristic equation of (27): There are three possibilities for the roots r and r 2 of (28): r and r 2 are real and distinct 2 r and r 2 are real and the same (see Section 33) 3 r and r 2 are complex conjugates (see Section 34) ar 2 + br + c = 0 (28) For this section assume that r and r 2 are real and distinct Then we have two solutions y = e rx and y = e r2x for (27) It can be verified that any of their linear combinations would also be a solution of (27) (Principle of superposition) Thus we get the general solution: y = c e rx + c 2 e r 2x Now c and c 2 can be found using two initial conditions as discussed before Steps to solve: Find the characteristic equation ar 2 + br + c = 0 2 Find the (distinct) roots r and r 2 3 The general solution is y = c e r x + c 2 e r 2x 4 Find c and c 2 when two initial conditions are given 26

Example Find the general solution of the following ODE Solution The characteristic equation is d 2 y dx 2 2dy dx 3y = 0 r 2 2r 3 = 0 (r + )(r 3) = 0 r =, 3 So the general solution is y = c e x + c 2 e 3x Example Solve the following IVP 2y + 5y 3y = 0, y(0) = /7, y (0) = 26/7 Solution The characteristic equation is 2r 2 + 5r 3 = 0 (2r )(r + 3) = 0 r = 3, /2 So the general solution is y = c e 3x + c 2 e x/2 Using the initial condition y(0) = /7, we get c + c 2 = /7 (29) Note that y = 3c e 3x + c 2 2 e x/2 Using the initial condition y (0) = 26/7, we get 3c + c 2 /2 = 26/7 (30) Solving (29) and (30), we get c = 9/7 and c 2 = 2/7 Thus the solution of the IVP is y = 9 7 e 3x + 2 7 ex/2 27

Y 4 3 y = e x/2 2 y = 9 7 e 3x + 2 7 ex/2 y = e 3x 0 2 3 4 5 X 28

32 Linear Homogeneous ODEs and Wronskian Consider the linear homogeneous second order ODE y + p(x)y + q(x)y = 0 (3) Principle of superposition: If y and y 2 are solutions of (3), then c y + c 2 y 2 is also a solution of (3) (verify) Question If y is a solution of (3), is it true that y = c y + c 2 y 2 for some scalars c and c 2? Answer Yes if y and y 2 are linearly independent functions Definition y, y 2,, y n are linearly independent functions if c y + c 2 y 2 + + c n y n = 0 = c = c 2 = = c n = 0 Theorem y, y 2,, y n are linearly dependent functions if and only if at least one of them is a linear combination of others Example y = x + and y 2 = x are linearly independent Because if c y + c 2 y 2 = 0, then c (x + ) + c 2 (x ) = 0 = (c + c 2 )x + (c c 2 ) = 0 = c + c 2 = 0, c c 2 = 0 = c = c 2 = 0 2 y = e x and y 2 = e 2x are linearly independent 3 y = x and y 2 = 2x are linearly dependent since 2y + y 2 = 0 Wronskian of y and y 2 is denoted by W (y, y 2 ) and defined by W (y, y 2 ) = y y 2 y y 2 = y y 2 y 2 y Example W (x+, x ) = x + x (x + ) (x ) = x + x = (x+) (x ) = 2 0 2 W (e x, e 2x ) = ex e 2x (e x ) (e 2x ) = ex e 2x e x 2e 2x = e x 2e 2x e 2x e x = e 3x 0 29

Theorem y and y 2 are linearly independent functions if and only if W (y, y 2 ) 0 In other words, y and y 2 are linearly dependent functions if and only if W (y, y 2 ) = 0 Proof (= ) Assume that y and y 2 are linearly dependent functions Then one of y and y 2 is a multiple of the other, say y 2 = cy for some scalar c Then W (y, y 2 ) = W (y, cy ) = y c y cy y = 0 ( =) Assume that W (y, y 2 ) = y y 2 y 2 y = 0 Note that ( ) d y2 = y y 2 y 2 y = 0 dx y y 2 Then y 2 y = c, constant Thus y 2 = cy and consequently y and y 2 are linearly dependent functions Theorem Suppose that y and y 2 are solutions of the IVP y + p(x)y + q(x)y = 0, y(x 0 ) = r, y (x 0 ) = s Then c y +c 2 y 2 is also a solution for some scalars c and c 2 if and only if W (y, y 2 )(x 0 ) 0 Proof ( =) Assume that W (y, y 2 )(x 0 ) 0 We will find c and c 2 such that c y + c 2 y 2 is also a solution of the given IVP First note that since y and y 2 are solutions of y + p(x)y + q(x)y = 0, c y + c 2 y 2 is also a solution of y + p(x)y + q(x)y = 0 by the principle of superposition Now we will find c and c 2 such that c y +c 2 y 2 satisfies the initial conditions y(x 0 ) = r = c y (x 0 ) + c 2 y 2 (x 0 ) = r y(x 0 ) = s = c y (x 0 ) + c 2 y 2(x 0 ) = s Since W (y, y 2 )(x 0 ) 0, solving for c and c 2 we get (= ) Similar c = c 2 = ry 2(x 0 ) sy 2 (x 0 ) y (x 0 )y 2(x 0 ) y 2 (x 0 )y (x 0 ) = ry 2(x 0 ) sy 2 (x 0 ), W (y, y 2 )(x 0 ) ry (x 0 ) + sy (x 0 ) y (x 0 )y 2(x 0 ) y 2 (x 0 )y (x 0 ) = ry (x 0 ) + sy (x 0 ) W (y, y 2 )(x 0 ) Theorem Suppose that y and y 2 are solutions of y + p(x)y + q(x)y = 0 Then any solution y can be written as y = c y + c 2 y 2 for some scalars c and c 2 if and only if W (y, y 2 )(x 0 ) 0 for some x 0 Proof ( =) Assume W (y, y 2 )(x 0 ) 0 Consider an arbitrary solution y = φ(x) of the IVP y + p(x)y + q(x)y = 0, y(x 0 ) = φ(x 0 ), y (x 0 ) = φ (x 0 ) By the preceding theorem y = c y + c 2 y 2 is a solution for some scalars c and c 2 Then φ = y = c y + c 2 y 2 by the uniqueness of the solution of the preceding IVP (= ) It follows from the linear independence of the fundamental solutions y and y 2 30

Note: It can be shown for y and y 2 in the preceding theorem that W (y, y 2 )(x) = 0 either for all values of x or no values of x (see Abel s theorem) 2 When W (y, y 2 ) 0, y and y 2 are linearly independent and any solution y can be written as y = c y + c 2 y 2 for some scalars c and c 2 That is why they are called fundamental solutions of y + p(x)y + q(x)y = 0 and the general solution of y + p(x)y + q(x)y = 0 is y = c y + c 2 y 2 for arbitrary scalars c and c 2 Example Recall from Section 3 that y 2y 3y = 0 has the general solution y = c e x + c 2 e 3x Here y = e x and y 2 = e 3x are fundamental solutions because W (e x, e 3x ) = e x e 3x (e x ) (e 3x ) = e x e 3x e x 3e 3x = e x 3e 3x e 3x ( e x ) = 4e 2x 0 Example Consider y = e r x and y 2 = e r 2x, r r 2 W (e r x, e r 2x ) = er x e r 2x (e r x ) (e r 2x ) = erx e r2x r e r x r 2 e r 2x = e r x r 2 e r 2x e r 2x r e r x ) = (r 2 r )e (r +r 2 )x 0 For ay + by + cy = 0, if the roots r and r 2 of the characteristic equation ar 2 + br + c = 0 are real and distinct, then y = e r x and y = e r 2x are fundamental solutions and the general solution is y = c e r x + c 2 e r 2x 3

33 Linear Homogeneous ODEs : Repeated Real Roots Consider the linear homogeneous second order ODE ay + by + cy = 0, a 0 (32) We know that if the roots r and r 2 of the characteristic equation ar 2 +br+c = 0 are real and distinct, then y = e r x and y = e r 2x are fundamental solutions Now assume b 2 4ac = 0 Then r = r 2 = b/2a and we get only one fundamental solution y = e r x How to find the second fundamental solution? Let y = v(x)e r x be a solution of (32) Plugging y = ve r x, y = v e r x + r ve r x and y = v e r x + 2r v e r x + r 2 ve r x into (32), we get a(v e r x + 2r v e r x + r 2 ve r x ) + b(v e r x + r ve r x ) + cve r x = 0 [av + (2ar + b)v + (ar 2 + br + c)ve r x = 0 av + (2ar + b)v + (ar 2 + br + c)v = 0 (since e r x 0) av + (2ar + b)v = 0 (since ar 2 + br + c = 0) av = 0 (since r = b/2a, 2ar + b = 0) v = 0 (since a 0) Then v (x) = c 2, a constant and v(x) = v (x) dx = c 2 dx = c 2 x + c Thus y = (c + c 2 x)e r x = c e r x + c 2 xe r x We can verify that y = e r x and y 2 = xe r x are solutions of (32) This technique of getting another solution from a given solution is called reduction of order Let s see if they are linearly independent: W (e r x, xe r x ) = er x xe r x (e r x ) (xe r x ) = erx xe rx r e r x (r x + )e r x = e r x (r x + )e r x xe r x r e r x = e 2r x 0 Thus y = e r x and y 2 = xe r x are fundamental solutions of (32) and the general solution of (32) is y = c e r x + c 2 xe r x Example Find the general solution of the following ODE Solution The characteristic equation is d 2 y dx 2 6dy dx + 9y = 0 r 2 6r + 9 = 0 (r 3) 2 = 0 r = 3, 3 32

So the general solution is y = c e 3x + c 2 xe 3x Example Solve the following IVP y + 4y + 4y = 0, y(0) =, y (0) = 8 Solution The characteristic equation is r 2 + 4r + 4 = 0 (r + 2) 2 = 0 r = 2, 2 So the general solution is y = c e 2x + c 2 xe 2x Using the initial condition y(0) =, we get c = (33) Note that y = 2c e 2x + c 2 ( 2x + )e 2x Using the initial condition y (0) = 8, we get 2c + c 2 = 8 (34) Solving (33) and (34), we get c = and c 2 = 20 Thus the solution of the IVP is y = e 2x + 20xe 2x Y 4 3 2 y = e 2x + 20xe 2x 0 2 3 4 5 X 33

34 Linear Homogeneous ODEs : Complex Roots Consider the linear homogeneous second order ODE ay + by + cy = 0, a 0 (35) In the last sections we discussed the cases when the roots r and r 2 of the characteristic equation ar 2 + br + c = 0 are real Now assume b 2 4ac < 0 Then r and r 2 are complex conjugate and they can be written as r = r + iθ and r 2 = r iθ Then we get two solutions y = e (r+iθ)x and y 2 = e (r iθ)x How to find two real fundamental solutions from y and y 2? We use Euler s formula: e iθ = cos θ + i sin θ for real θ Explanation: Recall from calculus that Applying this to e iθ, we get e x = x n n! = + x! + x2 2! + x3 3! + e iθ (iθ) n = n! = + iθ! + (iθ)2 + (iθ)3 + 2! 3! = + iθ! θ2 2! iθ3 + ) 3! ( ) = ( θ2 θ 2! + + i! θ3 3! + = cos θ + i sin θ Using Eule r formula we get y = e (r+iθ)x = e rx e iθx = e rx (cos(θx) + i sin(θx)) and similarly y 2 = e (r iθ)x = e rx (cos(θx) i sin(θx)) Let s take their linear combination to get some real solutions: y = d y + d 2 y 2 = d e rx (cos(θx) + i sin(θx)) + d 2 e rx (cos(θx) i sin(θx)) = (d + d 2 )e rx cos(θx) + i(d d 2 )e rx sin(θx) = c e rx cos(θx) + c 2 e rx sin(θx), where c = d + d 2, c 2 = i(d d 2 ) We can verify that y = e rx cos(θx) and y 2 = e rx sin(θx) are solutions of (35) Let s see if they are linearly independent: W (e rx cos(θx), e rx sin(θx)) = erx cos(θx) e rx sin(θx) (e rx cos(θx)) (e rx sin(θx)) = e rx cos(θx) e rx (r cos(θx) θ sin(θx)) = e 2rx θ(cos 2 (θx) + sin 2 (θx)) = e 2rx θ 0 e rx sin(θx) e rx (r sin(θx) + θ cos(θx)) 34

Thus y = e rx cos(θx) and y 2 = e rx sin(θx) are fundamental solutions of (35) and the general solution of (35) is y = c e rx cos(θx) + c 2 e rx sin(θx) Example Find the general solution of the following ODE Solution The characteristic equation is So the general solution is d 2 y dx 2 6dy dx + 3y = 0 r 2 6r + 3 = 0 (r 3) 2 = 4 r = 3 + 2i, 3 2i y = c e 3x cos(2x) + c 2 e 3x sin(2x) Example Solve the following IVP Solution The characteristic equation is So the general solution is y + 2y + 0y = 0, y(0) = 4, y (0) = 7 r 2 + 2r + 0 = 0 (r + ) 2 = 9 Using the initial condition y(0) = 4, we get r = + 3i, 3i y = c e x cos(3x) + c 2 e x sin(3x) c = 4 (36) Note that y = c ( cos(3x) 3 sin(3x))e x + c 2 ( sin(3x) + 3 cos(3x))e x Using the initial condition y (0) = 8, we get c + 3c 2 = 7 (37) Solving (36) and (37), we get c = 4 and c 2 = Thus the solution of the IVP is y = 4e x cos(3x) e x sin(3x) 35

Y 4 3 2 y = 4e x cos(3x) e x sin(3x) 0 2 3 4 5 X 36

35 Linear Nonhomogeneous ODEs : Undetermined Coefficients Consider the linear nonhomogeneous second order ODE y + p(x)y + q(x)y = g(x) (38) Theorem The general solution of the nonhomogeneous ODE (38) is y = c y + c 2 y 2 + y p, for arbitrary constants c and c 2 where y and y 2 are fundamental solutions of the corresponding homogeneous ODE y + p(x)y + q(x)y = 0 and y p is a particular solution of (38) Proof Let y = φ(x) be a solution of (38) Then φ + pφ + qφ = g Since y p is a particular solution of (38), we have Using preceding two equations we get y p + py p + qy p = g (φ y p ) + p(φ y p ) + q(φ y p ) = g g = 0 Thus φ y p is solution of the homogeneous ODE y + p(x)y + q(x)y = 0 So φ y p can be written as a linear combination of fundamental solutions y and y 2 Thus φ y p = c y + c 2 y 2 φ = c y + c 2 y 2 + y p Theorem The general solution of the nonhomogeneous ODE is y + p(x)y + q(x)y = g (x) + g 2 (x) + + g n (x) (39) y = c y + c 2 y 2 + y p + y p2 + + y pn, for arbitrary constants c and c 2 where y and y 2 are fundamental solutions of the corresponding homogeneous ODE y + p(x)y + q(x)y = 0 and y pi is a particular solutions of y + p(x)y + q(x)y = g i (x), i =,, n Proof Since y p i + p(x)y p i + q(x)y pi = g i (x), i =,, n, we have (y p + + y pn ) + p(x)(y p + + y pn ) + q(x)(y p + + y pn ) = g (x) + + g n (x) = g(x) Thus y p + + y pn is a particular solution of (39) By the preceding theorem, the general solution of the nonhomogeneous ODE (39) is for arbitrary constants c and c 2 y = c y + c 2 y 2 + y p + y p2 + + y pn, 37

Example Find the general solution y 2y 3y = 3e 2x 2 y 2y 3y = 65 cos(2x) 3 y 2y 3y = 9x 2 + Solution The characteristic equation is So the homogeneous solution is r 2 2r 3 = 0 (r + )(r 3) = 0 r =, 3 y h = c e x + c 2 e 3x, for arbitrary constants c and c 2 Now we will find a particular solution of the nonhomogeneous ODE Let y p = ae 2x be a particular solution for some constant a Then y p = 2ae 2x and y p = 4ae 2x Plugging these into y p 2y p 3y p = 3e 2x, we get 4ae 2x 2 2ae 2x 3 ae 2x = 3e 2x (4a 4a 3a)e 2x = 3e 2x 3a = 3 a = So y p = e 2x is a particular solution Thus the general solution is y = y h + y p = c e x + c 2 e 3x e 2x 2 Let y p = a cos(2x) + b sin(2x) be a particular solution for some constants a, b Then y p = 2a sin(2x) + 2b cos(2x) and y p = 4a cos(2x) 4b sin(2x) Plugging these into y p 2y p 3y p = 65 cos(2x), we get ( 4a cos(2x) 4b sin(2x)) 2 ( 2a sin(2x) + 2b cos(2x)) 3 (a cos(2x) + b sin(2x)) = 65 cos(2x) ( 4a 4b 3a) cos(2x) + ( 4b + 4a 3b) sin(2x) = 65 cos(2x) ( 7a 4b) cos(2x) + (4a 7b) sin(2x) = 65 cos(2x) Comparing the coefficients of cos(2x) and sin(2x) in LHS and RHS, we get 7a 4b = 65 4a 7b = 0 Solving we get a = 7, b = 4 So the particular solution is y p = 7 cos(2x) + 4 sin(2x) Thus the general solution is y = y h + y p = c e x + c 2 e 3x + 7 cos(2x) + 4 sin(2x) 38

3 Let y p = ax 2 +bx+c be a particular solution for some constants a, b, c Then y p = 2ax+b and y p = 2a Plugging these into y p 2y p 3y p = 9x 2 +, we get 2a 2 (2ax + b) 3 (ax 2 + bx + c) = 9x 2 + 3ax 2 + ( 4a 3b)x + (2a 2b 3c) = 9x 2 + Comparing the coefficients of x 2, x and the constant term in LHS and RHS, we get 3a = 9 4a 3b = 0 2a 2b 3c = Solving we get a = 3, b = 4, c = 5 So y p = 3x 2 + 4x 5 is a particular solution Thus the general solution is y = y h + y p = c e x + c 2 e 3x 3x 2 + 4x 5 To find a particular solution of the ODE ay + by + cy = g(x) use the following table: g(x) y p c 0 x n + c x n + + c n x s (a 0 x n + a x n + + a n ) (c 0 x n + c x n + + c n )e rx x s (a 0 x n + a x n + + a n )e rx (c 0 x n + c x n + + c n ) cos(θx), x s [(a 0 x n + a x n + + a n ) cos(θx)+ (c 0 x n + c x n + + c n ) sin(θx) (b 0 x n + b x n + + b n ) sin(θx) (c 0 x n + c x n + + c n ) cos(θx)e rx, x s [(a 0 x n + a x n + + a n ) cos(θx)+ (c 0 x n + c x n + + c n ) sin(θx)e rx (b 0 x n + b x n + + b n ) sin(θx)e rx where s is the smallest nonnegative integer (s = 0,, 2) that will ensure that no term in y p is a solution of the corresponding homogeneous equation 39

Example Find the general solution y + 2y 3y = 6xe x 2 y + 2y 3y = 0x sin x 3 y + 2y 3y = 6xe x 0x sin x Solution The characteristic equation is So the homogeneous solution is r 2 + 2r 3 = 0 (r )(r + 3) = 0 r = 3, y h = c e 3x + c 2 e x, for arbitrary constants c and c 2 Now we will find a particular solution of the nonhomogeneous ODE Let y p = (ax + b)e x be a particular solution for some constants a, b Note that e x is already in the homogeneous solution So let y p = x(ax + b)e x = (ax 2 + bx)e x be a particular solution for some constants a, b Then y p = [ax 2 + (2a + b)x + be x and y p = [ax 2 + (4a + b)x + 2a + 2be x Plugging these into y p + 2y p 3y p = 6e x, we get [ax 2 + (4a + b)x + 2a + 2be x + 2 [ax 2 + (2a + b)x + be x 3 (ax 2 + bx)e x = 6xe x [8ax + 2a + 4be 2x = 6xe x 8ax + 2a + 4b = 6x Comparing the coefficient x and the constant term in LHS and RHS, we get 8a = 6 2a + 4b = 0 Solving we get a = 2, b = So y p = (2x 2 x)e x is a particular solution Thus the general solution is y = y h + y p = c e 3x + c 2 e x + (2x 2 x)e x 2 Let y p = (ax + b) cos x + (cx + d) sin x be a particular solution for some constants a, b, c, d Then y p = (cx + a + d) cos x + ( ax b + c) sin x, y p = ( ax b + 2c) cos x + ( cx 2a d) sin x 40

Plugging these into y p + 2y p 3y p = 0x sin x, we get [( ax b + 2c) cos x + ( cx 2a d) sin x + 2[(cx + a + d) cos x + ( ax b + c) sin x 3[(ax + b) cos x + (cx + d) sin x = 0x sin x ( 2a 4c)x sin x + ( 2a d 2b + 2c 3d) sin x + ( ax b + 2c + 2cx + 2a + 2d 3ax 3b) cos x = 0x sin x 2cx sin x + ( 2a 2b + 2c 4d) sin x + ( 4a + 2c)x cos x + (2a 4b + 2c + 2d) cos x = 0x sin x Comparing the coefficients of x sin x, sin x, x cos x and cos x in LHS and RHS, we get 2a 4c = 0 2a 2b + 2c 4d = 0 4a + 2c = 0 2a 4b + 2c + 2d = 0 Solving we get a =, b = 7/5, c = 2, d = /5 So y p = (x + 7/5) cos x + (2x /5) sin x is a particular solution Thus the general solution is y = y h + y p = c e 3x + c 2 e x + (x + 7/5) cos x + (2x /5) sin x 3 By preceding two parts and a theorem, (2x 2 x)e x + (x + 7/5) cos x + (2x /5) sin x is a particular solution Thus the general solution is y = c e 3x + c 2 e x + (2x 2 x)e x + (x + 7/5) cos x + (2x /5) sin x 4

36 Variation of Parameters Consider the linear nonhomogeneous second order ODE y + p(x)y + q(x)y = g(x), (40) where p(x), q(x) and g(x) are continuous functions Note that if g(x) is not exponential, polynomial, sine or cosine functions, then the method of undetermined coefficients does not work We will use a general method called variation of parameters to find a particular solution of (40) Theorem Suppose y and y 2 are fundamental solutions of the corresponding homogeneous ODE of (40) Then a particular solution y p of (40) is y 2 g y p = y W (y, y 2 ) dx + y y g 2 W (y, y 2 ) dx Proof The homogeneous solution of the corresponding homogeneous ODE of (40) is y = c y + c 2 y 2 Suppose c = u(x) and c 2 = v(x) are functions of x to find a particular solution y p of (40) Then we have y p = uy + vy 2 y p = uy + u y + vy 2 + v y 2 We choose u and v such that Then u y + v y 2 = 0 (4) y p = uy + vy 2 y p = uy + u y + vy 2 + v y 2 Since y p is particular solution of (40), we have y p + py p + qy p = g (uy + u y + vy 2 + v y 2) + p(uy + vy 2) + q(uy + vy 2 ) = g u(y + py + qy ) + v(y + py + qy ) + u y + v y 2 = g u y + v y 2 = g (42) Solving for u and v from (4) and (42), we get u y 2 g = = y 2g y y 2 y 2 y W (y, y 2 ) v y g y g = = y y 2 y 2 y W (y, y 2 ) 42

Then u = y 2 g dx and v = W (y, y 2 ) y p = y y g dx Thus a particular solution is W (y, y 2 ) y 2 g W (y, y 2 ) dx + y 2 y g dx W (y, y 2 ) Example Find the general solution of xy (x + )y + y = x 2 e x You may use the fact that y = x + and y 2 = e x are fundamental solutions of the corresponding homogeneous ODE Solution Let s first put it in the standard form for variation of parameters: ( y + ) y + y x x = xex The homogeneous solution is y h = c (x + ) + c 2 e x W (y, y 2 ) = x + ex e x = (x + )e x e x = xe x By variation of parameters a particular solution is y 2 g y p = y W (y, y 2 ) dx + y y g 2 W (y, y 2 ) dx e x xe x (x + )xe x = (x + ) dx + xe x ex dx xe x = (x + ) e x dx + e x (x + ) dx ( ) x = (x + )e x + e x 2 ( ) x = e x 2 2 2 + x Thus the general solution is ( ) x y = c (x + ) + c 2 e x + e x 2 2 43

Example Find the general solution of y + y = tan x Solution The characteristic equation is r 2 + = 0 = r = ±i The homogeneous solution is y h = c cos x + c 2 sin x W (y, y 2 ) = cos x sin x sin x cos x = cos2 x + sin 2 x = By variation of parameters a particular solution is y 2 g y p = y W (y, y 2 ) dx + y y g 2 W (y, y 2 ) dx sin x tan x cos x tan x = cos x dx + sin x dx sin 2 x = cos x dx + sin x sin x dx cos x cos 2 x = cos x dx + sin x( cos x) cos x = cos x (sec x cos x) dx sin x cos x = cos x (ln sec x + tan x sin x) sin x cos x = cos x ln sec x + tan x Thus the general solution is y = c cos x + c 2 sin x cos x ln sec x + tan x 44

37 Modeling with Second Order ODEs Motion of a Simple Pendulum: Consider a simple pendulum with length L Let θ(t) be its angular displacement at time t By Newton s second law (ie, mass acceleration= net force) we get d 2 θ dt 2 = g L sin θ ie, d2 θ dt + g sin θ = 0 2 L This nonlinear ODE has no solution in terms of usual functions So it is linearized by taking sin θ θ for θ 0: d 2 θ dt 2 + g L θ = 0 The characteristic equation is The general solution is r 2 + g L = 0 = r = ±i g L ( ) ( ) g g θ = c cos L t + c 2 sin L t Using the initial conditions θ(0) = θ 0 and θ (0) = 0, we get c = θ 0 and c 2 = 0 Thus ( ) g θ = θ 0 cos L t Kirchhoff s Voltage Law: Consider an electric circuit with a resistor, an inductor, a capacitor, and a battery arranged in series (RLC circuit) Let Q(t) be the charge of the capacitor and I(t) current at time t Then Kirchhoffs voltage law states that the sum of the voltage drops across the resistor, inductor, and capacitor is equal to the supplied voltage V (t): L di dt + RI + Q C = V, where L is the constant inductance, R is the constant resistant and C is the constant capacitance Since I is the rate of change of Q with respect to time t, I = dq Then the preceding dt ODE becomes L d2 Q dt 2 + RdQ dt + Q C = V 45

Vibrating Springs: Consider a horizontal or a vertical spring with one end attached to a firm base and the other end an object of mass m If the spring is stretched or compresses x units from its equilibrium position, it exerts a restoring force F r which is proportional to x by Hooke s Law: F r = kx, where k > 0 is the spring constant Ignoring other forces like friction, by Newtons Second Law we get m d2 x dt = F 2 r = m d2 x dt 2 = kx = x md2 + kx = 0 dt2 The characteristic equation is mr 2 + k = 0 = r = ±i k/m So the general solution is x = c cos( k/mt) + c 2 sin( k/mt) = A cos(ωt + δ), where A = c 2 + c 2 2, ω = k/m, and δ = tan (c 2 /c ) are the amplitude, frequency, and phase angle respectively This is an example of simple harmonic motion Damped Vibrations: Consider a horizontal spring with friction of the surface or a vertical spring with a damping force of the medium The damping force F d is modeled to be proportional to the velocity, ie, F d = c dx dt, where c > 0 is the damping constant By Newtons Second Law we get m d2 x dt = F 2 r + F d = m d2 x dt 2 = kx cdx dt = m d2 x dt + cdx + kx = 0 2 dt The characteristic equation is mr 2 + cr + k = 0 = r, r 2 = ( c ± c 2 4mk)/2m c 2 4mk > 0 (overdamping): Roots are distinct real negative and x = c e r t + c 2 e r 2t Here c > 2 mk implies a high damping force resulting in no vibration 2 c 2 4mk = 0 (critical damping): r = r 2 = c/2m and x = (c + c 2 t)e ct/2m Here c = 2 mk implies the least damping force that suppresses too much vibrations 3 c 2 4mk < 0 (underdamping): Roots are complex conjugates and x = e ct/2m (c cos(ωt)+ c 2 sin(ωt)), ω = 4mk c 2 /2m Here c < 2 mk implies a low damping force failing to suppress too much vibrations x x x t t t overdamping critical damping 46 underdamping

4 Higher Order ODEs 4 General Solutions Consider the linear homogeneous nth order ODE y (n) + p (x)y (n ) + p 2 (x)y (n 2) + + p n (x)y = 0 (43) Principle of superposition: If y, y 2,, y n are solutions of (43), then c y +c 2 y 2 + +c n y n is also a solution of (43) (verify) Theorem If y, y 2,, y n are linearly independent solutions of (43), ie, W (y, y 2,, y n ) 0, then the general solution of (43) is for arbitrary constants c,, c n y = c y + c 2 y 2 + + c n y n, Any n linearly independent solutions of (43) are called fundamental solutions of (43) because any solution of (43) is a linear combination of them Note the general formula for Wronskian: W (y, y 2,, y n ) = y y 2 y n y y 2 y n y (n ) y (n ) 2 y n (n ) Example Show that e x, cos x and sin x are linearly independent functions Solution W (e x, cos x, sin x) = = e x cos x sin x (e x ) (cos x) (sin x) (e x ) (cos x) (sin x) e x cos x sin x e x sin x cos x e x cos x sin x = e x sin x cos x cos x sin x cos x ex cos x e x sin x + sin x ex sin x e x cos x = e x (sin 2 x + cos 2 x) cos x( e x sin x e x cos x) + sin x( e x cos x + e x sin x) = e x (sin 2 x + cos 2 x) + e x (sin 2 x + cos 2 x) = 2e x 0 Since W (e x, cos x, sin x) 0, e x, cos x and sin x are linearly independent functions 47

Consider the linear nonhomogeneous nth order ODE y (n) + p (x)y (n ) + p 2 (x)y (n 2) + + p n (x)y = g(x) (44) Theorem If y, y 2,, y n are fundamental solutions of the corresponding homogeneous ODE of (44) and y p is a particular solution of (44), then the general solution of (44) is for arbitrary constants c,, c n Consider the IVP y = c y + c 2 y 2 + + c n y n + y p, y (n) + p (x)y (n ) + p 2 (x)y (n 2) + + p n (x)y = g(x) y(x 0 ) = a 0, y (x 0 ) = a,, y (n ) (x 0 ) = a n (45) Theorem If p, p 2,, p n and g are continuous functions on an interval containing x 0, then the ODE (45) has a unique solution y = φ(x) on interval containing x 0 Consider the linear homogeneous nth order ODE with constant coefficients y (n) + a y (n ) + a 2 y (n 2) + + a n y = 0 (46) The characteristic equation of (46) is r n + a r n + + a n = 0 If r, r,, r are k characteristic roots, then e rx, xe rx,, x k e r x are k linearly independent solutions of (46) If r ± iθ, r ± iθ,, r ± iθ are k pairs of characteristic roots, then e rx cos(θx), e rx sin(θx); xe rx cos(θx), xe rx sin(θx); ; x k e rx cos(θx), x k e rx sin(θx) are k pairs of linearly independent solutions of (46) 48

Example Solve the IVP y 2y + y 2y = 0 y(0) = 3, y (0) = 3, y (0) = 7 Solution The characteristic equation is r 3 2r 2 + r 2 = 0 r 2 (r 2) + (r 2) = 0 (r 2)(r 2 + ) = 0 r = 2, ±i So the general solution is Taking derivatives we get y = c e 2x + c 2 cos x + c 3 sin x y = 2c e 2x c 2 sin x + c 3 cos x y = 4c e 2x c 2 cos x c 3 sin x Using the initial conditions y(0) = 3, y (0) = 3, y (0) = 7, we get c + c 2 = 3 2c + c 3 = 3 4c c 2 = 7 Solving we get c = 2, c 2 =, c 3 = Thus the solution is y = 2e 2x + cos x sin x Example Find the general solution of (D 2) 3 (D 2 2D + 2) 2 y = 0 Here the differential operator D is defined as D k y = dk y dx k Solution The characteristic equation is So the general solution is (r 2) 3 (r 2 2r + 2) 2 = 0 r = 2, 2, 2, ± i, ± i y = c e 2x + c 2 xe 2x + c 3 x 2 e 2x + c 4 e x cos x + c 5 e x sin x + c 6 xe x cos x + c 7 xe x sin x 49

42 Higher Order ODEs : Undetermined Coefficients Consider the linear nonhomogeneous second order ODE y (n) + p (x)y (n ) + p 2 (x)y (n 2) + + p n (x)y = g(x) (47) To find a particular solution of the ODE (47) use the following table: g(x) y p c 0 x n + c x n + + c n x s (a 0 x n + a x n + + a n ) (c 0 x n + c x n + + c n )e rx x s (a 0 x n + a x n + + a n )e rx (c 0 x n + c x n + + c n ) cos(θx), x s [(a 0 x n + a x n + + a n ) cos(θx)+ (c 0 x n + c x n + + c n ) sin(θx) (b 0 x n + b x n + + b n ) sin(θx) (c 0 x n + c x n + + c n ) cos(θx)e rx, x s [(a 0 x n + a x n + + a n ) cos(θx)+ (c 0 x n + c x n + + c n ) sin(θx)e rx (b 0 x n + b x n + + b n ) sin(θx)e rx where s is the smallest nonnegative integer (s = 0,, 2,, n) that will ensure that no term in y p is a solution of the corresponding homogeneous equation Example Find the general solution (D 3 + 2D 2 3D)y = 6xe x 2 (D 3 + 2D 2 3D)y = 0x sin x 3 (D 3 + 2D 2 3D)y = 6xe x 0x sin x Solution The characteristic equation is So the homogeneous solution is r 3 + 2r 2 3r = 0 r(r )(r + 3) = 0 r = 0, 3, y h = c + c 2 e 3x + c 3 e x, for arbitrary constants c, c 2 and c 3 Now we will find a particular solution of the nonhomogeneous ODE Let y p = (ax + b)e x be a particular solution for some constants a, b Note that e x is already in the homogeneous solution So let y p = x(ax + b)e x = (ax 2 + bx)e x be a particular solution for some constants a, b Then y p = [ax 2 + (2a + b)x + be x, y p = [ax 2 + (4a + b)x + 2a + 2be x, y p = [ax 2 + (6a + b)x + 6a + 3be x 50

Plugging these into y p + 2y p 3y p = 6e x, we get [ax 2 + (6a + b)x + 6a + 3be x + 2 [ax 2 + (4a + b)x + 2a + 2be x 3 [ax 2 + (2a + b)x + be x = 6xe x [8ax + 0a + 4be 2x = 6xe x 8ax + 0a + 4b = 6x Comparing the coefficient x and the constant term in LHS and RHS, we get 8a = 6 0a + 4b = 0 Solving we get a = 2, b = 5 So y p = (2x 2 5x)e x is a particular solution Thus the general solution is y = y h + y p = c + c 2 e 3x + c 3 e x + (2x 2 5x)e x 2 Let y p = (ax + b) cos x + (cx + d) sin x be a particular solution for some constants a, b, c, d Then y p = (cx + a + d) cos x + ( ax b + c) sin x, y p = ( ax b + 2c) cos x + ( cx 2a d) sin x, y p = ( cx 3a d) cos x + (ax + b 3c) sin x Plugging these into y p + 2y p 3y p = 0x sin x, we get [( cx 3a d) cos x + (ax + b 3c) sin x + 2[( ax b + 2c) cos x + ( cx 2a d) sin x 3[(cx + a + d) cos x + ( ax b + c) sin x = 0x sin x ( 2a 4c)x cos x + ( 6a 2b + 4c 4d) cos x + (4a 2c)x sin x + ( 4a + 4b 6c 2d) sin x = 0x sin x Comparing the coefficients of x cos x, cos x, x sin x and sin x in LHS and RHS, we get 2a 4c = 0, 6a 2b + 4c 4d = 0, 4a 2c = 0, 4a + 4b 6c 2d = 0 Solving we get a = 2, b = 6/5, c =, d = 7/5 So y p = ( 2x + 6/5) cos x + (x + 7/5) sin x is a particular solution Thus the general solution is y = y h + y p = c + c 2 e 3x + c 3 e x + ( 2x + 6/5) cos x + (x + 7/5) sin x 3 By preceding two parts and a theorem, (2x 2 5x)e x +( 2x+6/5) cos x+(x+7/5) sin x is a particular solution Thus the general solution is y = c + c 2 e 3x + c 3 e x + (2x 2 5x)e x + ( 2x + 6/5) cos x + (x + 7/5) sin x 5

5 Systems of Differential Equations 5 Introduction to Linear Algebra Matrix: An m n matrix A is an m-by-n array of scalars of the form a a 2 a n a 2 a 22 a 2n A = a m a m2 a mn The order (or size) of A is m n (read as m by n) It means A has m rows and n columns The (i, j)-entry of A = [a i,j is a i,j [ 2 0 Example A = is a 2 3 real matrix The (2, 3)-entry of A is 3 0 The identity matrix of order n, denoted by I n, is the n n diagonal matrix whose diagonal 0 0 entries are Example I 3 = 0 0 is the 3 3 identity matrix 0 0 An n matrix is called a column matrix or n-dimensional (column) vector It is denoted by x Example x = 0 is a 3-dimensional vector which represents the position vector 2 of the point (, 0, 2) in R 3 Matrix Operations: Transpose: The transpose of an m n matrix A, denoted by A T, is an n m matrix whose columns are corresponding rows of A, ie, (A T ) ij = A ji [ 3 2 0 Example If A =, then A 3 0 T = 2 0 0 Scalar Multiplication: Let A be a matrix and c be a scalar The scalar multiple, denoted by ca, is[ the matrix whose entries are c[ times the corresponding entries of A 2 0 2 4 0 Example If A =, then 2A = 3 0 6 0 2 Sum: If A and B are m n matrices, then the sum A + B is the m n matrix whose entries are the sum [ of the corresponding entries [ of A and B, ie, (A + B) ij [ = A ij + B ij 2 0 0 2 0 0 0 Example If A = and B =, then A + B = 3 0 3 0 2 0 0 Exercise Find 2A B 52

Multiplication: Matrix-vector multiplication: If A is an m n matrix and x is an n-dimensional vector, then their product A x is an n-dimensional vector whose (i, )-entry is a i x + a i2 x 2 + + a im x n, the dot product of the row i of A and x Note that a x + a 2 x 2 + + a n x n A a 2 x + a 22 x 2 + + a 2n x n x = = x a m x + a m2 x 2 + + a mn x n [ Example If A = 2 0 3 0 and x = 0 a a 2 a m +x 2 a 2 a 22 a m2, then A x = + +x n [ 3 which is a linear combination of first and second columns of A with weights and respectively a n a 2n a mn 2 Matrix-matrix multiplication: If A is an m n matrix and B is an n p matrix, then their product AB is an m p matrix whose (i, j)-entry is the dot product the row i of A and the column j of B (AB) ij = a i b j + a i2 b 2j + + a im b mj Note that AB BA in general Also note column i of AB is A(column i of B) [ 2 2 [ 2 2 Example If A = and B = 4 0 0 0, then AB = 0 0 2 2 2 [ 2 2 4 0 Note that AB = = A 0 A 0 2 2 Determinant: The determinant of an n n matrix A is denoted by det A and A It is defined recursively We will only use determinant of order 2 and 3 a a 2 a 2 a 22 = a a 22 a 2 a 2 Example a a 2 a 3 a 2 a 22 a 23 = a a 22 a 23 a a 3 a 32 a 32 a 33 a 2 a 2 a 23 a 3 a 33 + a 3 a 2 a 22 a 3 a 32 33 2 7 3 0 8 0 3 = 2 0 8 3 3 8 0 3 + 7 3 0 0 = 4 53

Inverse of a matrix: An n n matrix A is called invertible if there is an n n matrix B such that AB = BA = I n Here B is called the inverse of A which is denoted by A So [ a b Example = c d [ ad bc AA = A A = I n d b c a Theorem An n n matrix A is invertible iff det A 0 A can be found by the Gauss-Jordan row reductions and adjoint formula: A = adj A det A A system of linear equations with n variables x,, x n and m equations can be written as follows: A = a x + a 2 x 2 + + a n x n = b a 2 x + a 22 x 2 + + a 2n x n = b 2 a m x + a m2 x 2 + + a mn x n = b m This linear system is equivalent to the matrix equation A x = b, where a a 2 a n x b a 2 a 22 a 2n x 2 b 2, x = a m a m2 a mn x n and b = b m A is called the coefficient matrix A x = b There are multiple ways to solve the matrix equation Linear independent vectors: v, v 2,, v k are linearly independent if c v + c 2 v2 + + c k vk = 0 = c = c 2 = = c k = 0 In other words v, v 2,, v k are linearly independent if [ v v2 zero solution x = 0 v k x = 0 has only the Theorem Consider n vectors v, v 2,, v n in R n Then the following are equivalent v, v 2,, v n are linearly independent 2 det[ v v2 v n 0 3 any vector x in R n is a linear combination of v, v 2,, v n, ie, x = c v + c 2 v2 + + c n vn for some scalar c, c 2,, c n 54

Example Show that the following vectors are linearly independent: Solution det 2 7 3 0 8 0 3 2 3 0, 0, 7 8 3 = 4 0 So the vectors are linearly independent Also note that any 3-dimensional vector can be written as a linear combination of these three vectors Span of vectors: The span of v, v 2,, v k, denoted by Span{ v, v 2,, v k }, is the set of all linear combinations {[ of v, v[ 2, }, v k 0 Example R 2 = Span, 0 Eigenvalues and eigenvectors: Let A be an n n matrix If A x = λ x for some nonzero vector x and some scalar λ, then λ is an eigenvalue of A and x is an eigenvector of A corresponding to λ [ 2 Example Consider A =, λ = 3, [ v =, [ 2 u = 0 3 Since A [ [ [ [ 2 3 v = = = 3 = λ v, 3 is an eigenvalue of A and v is an 0 3 3 eigenvector of A corresponding to the eigenvalue 3 Since A [ [ [ [ 2 2 0 2 u = = λ = λ u for all scalars λ, u is not an 0 3 3 eigenvector of A Note that an eigenvalue can be[ a complex number [ and an eigenvector [ [ can be a[ complex 0 0 i vector Example Consider A = Since = = i, [ 0 0 i i i is an eigenvalue of A and is an eigenvector of A corresponding to the eigenvalue i i The characteristic polynomial of A is det(a λi), a polynomial of λ The characteristic equation of A is det(a λi) = 0 Since the roots of the characteristic polynomial are the eigenvalues of the n n matrix A, A has n eigenvalues, not necessarily distinct The multiplicity of a root λ in det(a λi) is the algebraic multiplicity of the eigenvalue λ of A Suppose λ is an eigenvalue of the matrix A Then Nul(A λi) = { x (A λi) x = O } is the eigenspace of A corresponding to the eigenvalue λ 55

Example Let A = [ 2 0 3 (a) Find the eigenvalues of A with their algebraic multiplicities (b) Find the eigenvectors and eigenspaces of A Solution (a) The characteristic polynomial of A is det(a λi) = λ 2 0 3 λ = ( λ)(3 λ) det(λi A) = ( λ)(3 λ) = 0 = λ =, 3 So and 3 are eigenvalues of A with algebraic multiplicities and respectively (b) The eigenspace of A corresponding to the eigenvalue is (A I) x = O = Nul(A I) = { x (A I) x = O } [ 0 2 0 2 [ x x 2 = So we get x 2 = 0 where x is a free variable Thus [ [ [ x x x = = = x x 2 0 0 [ 0 0 = {[ Span 0 Thus the eigenvector of A corresponding to the eigenvalue is A corresponding to the eigenvalue is Nul(A I) = Span {[ 0 The eigenspace of A corresponding to the eigenvalue 3 is (A 3I) x = O = } Nul(A 3I) = { x (A 3I) x = O } [ 2 2 0 0 [ x x 2 = [ 0 0 So we get x = x 2 where x 2 is a free variable Thus [ [ [ x x2 x = = = x x 2 2 x 2 { 2x2 = 0 2x 2 = 0 [ 0 } and the eigenspace of = { 2x +2x 2 = 0 {[ Span } 56

Thus the eigenvector of A corresponding to the eigenvalue 3 is A corresponding to the eigenvalue 3 is Nul(A 3I) = Span {[ } [ and the eigenspace of Matrix of functions: Entries of an m n matrix A may be functions of x: a (x) a 2 (x) a n (x) a 2 (x) a 22 (x) a 2n (x) A(x) = a m (x) a m2 (x) a mn (x) Calculus on matrix of functions: We take limit, derivative and integration of a matrix A(x) entry-wise: and lim A(x) = x a lim a (x) x a lim a 2(x) x a lim a m(x) x a lim a 2 (x) lim a n (x) x a x a lim a 22 (x) lim a 2n (x) x a x a, lim a m2 (x) lim a mn (x) x a x a a (x) a 2(x) a n(x) a 2(x) a 22(x) a 2n(x) A (x) =, a m(x) a m2(x) a mn(x) a (x) dx a 2 (x) dx an (x) dx a2 (x) dx a 22 (x) dx a2n (x) dx A(x) dx = am (x) dx a m2 (x) dx amn (x) dx [ e x cos x Example Consider A(x) = sin x 2e 2x Then lim A(x) = [ x 0 e A x sin x (x) = cos x 4e 2x and [ e A(x) dx = x sin x cos x e 2x [ 0 2, 57

52 Introduction to Systems of Linear ODEs Suppose x, x 2,, x n are n functions of t Consider the following system of n linear ODEs: where A = x = a (t)x + a 2 (t)x 2 + + a n (t)x n + g (t) x 2 = a 2 (t)x + a 22 (t)x 2 + + a 2n (t)x n + g 2 (t) x n = a m (t)x + a m2 (t)x 2 + + a mn (t)x n + g n (t) It can be simply written in the following matrix form: x = A x + g, (48) a a 2 a n x g a 2 a 22 a 2n x 2 g 2, x = a n a n2 a nn x n and g = g n A is called the coeffi- cient matrix of (48) When g = 0, (48) is a homogeneous system Similarly when g 0, (48) is a nonhomogeneous system Example A multiple mass-spring system can have the following equations of motion: m x = k x + k 2 (x 2 x ) + F (t) m 2 x 2 = k 2 (x 2 x ) + F 2 (t) Example Write the following system in matrix form Solution [ x x 2 where A = [ 4 2 x = x + x 2 + e t x 2 = 4x 2x 2 + t 2 [ x + x = 2 + e t 4x 2x 2 + t 2 =, [ x x = x 2 [ 4 2 ie, x = A x + g, and [ e t g = Example Write the following IVP in matrix form t 2 [ [ x e t + x 2 t 2 Solution where x (0) = [ 5 6 and x = [ x x 2 x = 4x 3x 2 x 2 = 8x 6x 2 x (0) = 5, x 2 (0) = 6 x = [ 4 3 8 6 x, 58

We can transform nth order linear ODE y (n) = F (t, y, y,, y (n ) ) to the following system of n variables x = y, x 2 = y,, x n = y (n ) : x = x 2 x 2 = x 3 x n = x n x n = F (t, x, x 2,, x n ) Example Convert the following third order linear ODE to a system of linear ODEs and write its matrix form y = e t y ty + cos ty 5t Solution Let x = y, x 2 = y, x 3 = y Then x = x 2 x 2 = x 3 x 3 = e t x 3 tx 2 + cos tx 5t Its matrix form is where x = [x, x 2, x 3 T x = 0 0 0 0 cos t t e t x + 0 0 5t, 59

53 Solving Linear Systems with constant coefficients Suppose x, x 2,, x n are n functions of t Consider the following system of n linear ODEs: x = A(t) x + g (t), (49) where x = [x, x 2,, x n T Theorem The linear system (49) has a solution x = φ (t) on an interval where A(t) and g (t) are continuous Now consider the corresponding homogeneous system: x = A(t) x (50) Theorem If x,, x n are solutions of (50), then so is x = c x + + c n xn, for all scalars c,, c n (Principle of superposition) Theorem x,, x n are linearly independent solutions of (50) iff for some (actually all) value of t W ( x,, x n ) = det[ x,, x n 0 Theorem If x,, x n are linearly independent solutions of (50), then any solution x of (50) is a linear combination of x,, x n, ie, x = c x + + c n xn, for some scalars c,, c n Any n linearly independent solutions of (50) are called fundamental solutions of (50) and x = c x + + c n xn is called the general solution of (50) How to solve (50) when entries of A are constants? Recall that x = λx has the general solution x = ce λt So a solution of (50) may have a similar form x = e λt v for some constant vector v Let s verify that x = A x = λe λt v = Ae λt v = λ v = A v = A v = λ v = v is an eigenvector of A corresponding to eigenvalue λ 60

Theorem If v,, v n are n linearly independent eigenvectors of A corresponding to eigenvalues λ,, λ n respectively, then e λ t v,, e λnt v n are n linearly independent solutions of and the general solution is for arbitrary scalars c,, c n Verify: Thus x = A x x = A x x = c e λ t v + + c n e λnt v n, x = c e λ t λ v + + c n e λnt λ n vn A x = c e λ t A v + + c n e λnt A v n = c e λ t λ v + + c n e λnt λ n vn Example Find the general solution of [ x = x 4 2 Solution [ The eigenvalues [ of the coefficient matrix are 2 and 3 with corresponding eigenvectors and respectively (show all the steps) So the general solution is 4 x (t) = c e 2t [ + c 2 e 3t [ 4 Example x = 3 3 2 2 3 2 4 5 Find the general solution Solution The eigenvalues of the coefficient matrix are 2, and with corresponding 0 eigenvectors, and respectively (show all the steps) So the general 2 solution is x (t) = c e 2t 2 + c 2 e t x 0 + c 3 e t 6

Example Solve the IVP [ [ x = x, 3 x (0) = Solution [ The eigenvalues [ of the coefficient matrix are 0 and 2 with corresponding eigenvectors and respectively (show all the steps) So the general solution is x (t) = c [ [ + c 2 e 2t x (0) = [ 3 [ [ = c + c 2 = = c c 2 = 3, c + c 2 = = c = 2, c 2 = [ 3 So the solution is x (t) = 2 [ [ e 2t Geometric view: [ [ x (t) = 2 e 2t = [ x x 2 [ 2 + e 2t = 2 e 2t = x = 2+e 2t, x 2 = 2 e 2t Eliminating t, we get x + x 2 = 4, the trajectory of the particle whose planar motion is described by the given IVP x 2 4 3 x + x 2 = 4 2 (3, ) 0 2 3 4 x 62

54 Equilibrium Solutions and Phase Portraits Consider the following system of n linear ODEs: x = A x, (5) where x = [x, x 2,, x n T x is called an equilibrium solution of (5) if A x = 0 Obviously x = 0 is always an equilibrium solution of (5) Note that if A is invertible (ie, det A 0 or, all eigenvalues of A are nonzero), then A x = 0 = x = A 0 = 0 and x = 0 is the only equilibrium solution of (5) Direction field/ tangent field: A solution curve or a trajectory of (5) can be visualized by plotting direction field of tangent vectors to solutions of (5) x = A x is the tangent vector of a solution curve at x R n We mostly focus on the case n = 2 and plot trajectories on the x x 2 -plane which is called the phase plane A sketch of trajectories on the phase plane is called a phase portrait Example x = [ 2 2 x (a) Draw direction field Use the points: (0, 0), (±, 0), (0, ±), (±, ±) (b) Draw the phase portrait Solution (a) At [ 0 x =, [ [ 2 0 x 0 = = 2 0 At [ ± x =, [ [ [ 2 ± x 0 = = ± 2 0 2 At [ 0 x =, [ [ [ 2 x ± 0 2 = = ± 2 ± At [ x = ±, [ [ [ 2 3 x = ± = ± 2 3 At [ x = ±, [ [ 2 x = ± = ± 2 Now draw the tangent vector A x at each x [ 0 0 [ [(b) The eigenvalues [ of the coefficient matrix are and 3 with corresponding eigenvectors and respectively (show all the steps) So the general solution is x (t) = c e t [ [ + c 2 e 3t 63

When c 2 = 0, [ x = c x = c e t This gives the trajectory x = c e t, x 2 = c e t, ie, the line x 2 = x (the line through the origin parallel to [ v = ) This trajectory is in the quadrant 4 when c > 0 and in the quadrant 2 when c < 0 Note that x = c [ x = c e t is far from the origin for large negative t and [ [ x = c x = c e t 0 0 as t When c = 0, [ x = c 2x2 = c 2 e 3t This gives the trajectory x = c 2 e 3t, x 2 = c 2 e 3t, ie, the line x 2 = x (the line through the origin parallel to [ v 2 = ) This trajectory is in the quadrant when c > 0 and in the quadrant 3 when c < 0 Note that x = c [ 2x2 = c 2 e 3t is far from the origin for large positive t and [ [ x = c 0 2x2 = c 2 e 3t as 0 t Note: For these kind of trajectories (when eigenvalues are real of opposite signs) in the preceding example, the origin is called a saddle point which is unstable as trajectories move away from the origin as t increases 2 When eigenvalues are distinct real of the same sign, the origin is called a node which is asymptotically stable when eigenvalues are negative (trajectories move toward the origin as t increases) and unstable when eigenvalues are positive (trajectories move away from the origin as t increases) 64

Example [ 5 4 x = x 2 Solution The general solution is [ x (t) = c e t + c 2 e 6t [ 4 c 2 x2 will dominate c x for large negative t (ie, when far from the origin) c x will dominate c 2 x2 for large positive t (ie, when near the origin) The origin is an asymptotically stable node Example [ 3 x = x 2 4 Solution The general solution is [ x (t) = c e 2t [ + c 2 e 5t 2 The origin is an unstable node 65

Example [ x = x Solution The general solution is [ x (t) = c [ + c 2 e 2t Each point on the line x 2 = x is an equilibrium The origin is a degenerate node Note The origin is asymptotically stable when both eigenvalues are negative (trajectories move toward the origin as t increases) and unstable when at least one eigenvalue is positive (trajectories move away from the origin as t increases) 66

55 Complex Eigenvalues for Linear Systems Consider the following linear system: A(t) is a real-valued continuous function of t x = A(t) x, (52) Theorem If x = u (t) + i v (t) is a solution of (52) for real-valued functions u (t) and v (t), then so are u (t) and v (t) Proof ( u + i v ) = A( u + i v ) u + i v = A u + ia v Comparing real and imaginary parts of both sides, we get u = A u and v = A v Theorem If A has complex eigenvalue λ = r + iθ with corresponding eigenvector v, then two linearly independent solutions of (52) are x = e rt [cos(θt)re v sin(θt)im v and x2 = e rt [cos(θt)im v + sin(θt)re v Proof We have verified before that x = e (r+iθ)t v is a solution x = e rt [cos(θt) + i sin(θt)[re v + iim v = e rt [cos(θt)re v sin(θt)im v + ie rt [cos(θt)im v + sin(θt)re v The rest follows from the preceding theorem and Wronskian Example x = [ 2 5 3 x (a) Find the general solution (b) Draw the phase portrait Solution (a) λ = 2 ± 3i and [ + 3i v = 5 ([ x = e 2t [cos(3t) + i sin(3t) 5 [ [ 3 = e (cos(3t) 2t sin(3t) 5 0 [ 3 + i 0 A solution is ) ) + ie 2t ( cos(3t) [ 3 0 [ + sin(3t) 5 ) 67

So the general solution is x = c x + c 2 x2 = c e 2t (cos(3t) [ 5 = c e 2t [ cos(3t) 3 sin(3t) 5 cos(3t) [ 3 sin(3t) 0 ) ( [ 3 + c 2 e 2t cos(3t) 0 + c 2 e 2t [ 3 cos(3t) sin(3t) 5 sin(3t) [ + sin(3t) 5 ) (b) To draw the spiral for x, get a point on it by plugging t = 0 and then find the tangent vector x Do the same for x 2 The origin is an unstable spiral point Note that the origin would have been an asymptotically stable spiral point if eigenvalues had negative real part Note that there are 3 types of trajectories for the eigenvalue λ = r ± iθ in a system of 2 linear ODEs: r < 0, the origin is an asymptotically stable spiral point 2 r > 0, the origin is an unstable spiral point 3 r = 0, the origin is a stable center (trajectories are circle or ellipse) Example x = [ 2 x (a) Find the general solution (b) Draw the phase portrait 68

Solution (a) λ = ±i and [ + i v = A solution is x = [cos(t) + i sin(t) ([ [ + i 0 ) So the general solution is x = c x + c 2x2 [ [ ) ( = c (cos(t) sin(t) + c 0 2 cos(t) [ [ cos t sin t cos t sin t = c + c cos t 2 sin t (b) Let s draw the trajectory for x = [ cos t sin t x = cos t x = cos t sin t, x 2 = cos t [ 0 [ + sin(t) ) It gives the trajectory Solving for cos t and sin t, we get sin t = (x + x 2 ), cos t = x 2 Using sin 2 t + cos 2 t =, we get the trajectory which is the ellipse (x + x 2 ) 2 + x 2 2 = whose one axis is x + x 2 = 0 To determine the direction of a trajectory, plot tangent vector at a point, say [, 0 T : [ [ [ 2 x = = 0 So trajectories move counterclockwise Note that if A is not given, then plug t = π/2 to get a point x (π/2) and the tangent vector x (π/2) The origin is a stable center, but not asymptotically stable as trajectories do not move toward the origin as t increases 69

56 Repeated Eigenvalues for Linear Systems Consider the following linear system: x = A x, (53) where A is a 2 2 real matrix If A has two distinct eigenvalues, then there are two linearly independent eigenvectors giving two linearly independent solution of (53) But if A has only one distinct eigenvalue λ of multiplicity 2, then there is only one linearly independent eigenvector v giving only one linearly independent solution x = e λt v of (53) It can be verified that x = te λt v is not a nonzero solution of (53) Theorem Suppose λ is an eigenvalue of a 2 2 real matrix A of multiplicity 2 with an eigenvector v Then the general solution of x = A x is x = c e λt v + c 2 (te λt v + e λt w ), where (A λi) w = v or, equivalently (A λi) 2 w = 0 ( w is called a generalized eigenvector of A corresponding the eigenvalue λ) Proof First we show that x = te λt v + e λt w is a solution of x = A x LHS = (te λt v + e λt w ) = e λt v + tλe λt v + λe λt w RHS = A(te λt v + e λt w ) = te λt A v + e λt A w = te λt λ v + e λt (λ w + v ), since (A λi) w = v, ie, A w = λ w + v Thus LHS = RHS Using Wronskian it can be shown that these two solutions are linearly independent Example x = [ 4 3 x (a) Find the general solution (b) Draw the phase portrait Solution (a) The eigenvalues of the coefficient matrix are and with corresponding eigenvector [ v = (show all the steps) Now let s find a vector w that satisfies 2 (A λi) w = v [ [ [ 2 w = = 4 2 w 2 2 [ [ 2w w = 2 = 4w 2w 2 2 = 2w w 2 = = w 2 = 2w = [ w w = = w 2 [ w 2w [ [ = w + 2 0 70

Assigning a particular value of w, say w = 0, we get [ 0 w = is [ ( [ [ x = c e t + c 2 2 te t + e t 0 2 So the general solution (b) To draw the curve for x 2, get three points on it by plugging t = 0,, The origin is called an improper node which is asymptotically stable when eigenvalues are negative (trajectories move toward the origin as t increases) and unstable when eigenvalues are positive (trajectories move away from the origin as t increases) ) Note for Chapter 7: If x = A x has a solution [ [ x 2 = te t + e 2 t 0 where A is not given Since [ 0 x 2 (0) =, the solution curve passes through (0, ) To find its direction at (0, ), find the tangent vector [ [ [ x 2 = ( e t te t ) e 2 t 0 = at t = 0 3 7