Solutions to Final Exam Sample Problems, Math 246, Fall 2013

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1 Solutions to Final Exam Sample Problems, Math 246, Fall 23 () Consider the differential equation dy dt = (9 y2 )y 2 (a) Identify its stationary points and classify their stability (b) Sketch its phase-line portrait in the interval 5 y 5 (c) If y() =, how does the solution y(t) behave as t? Solution (a,b) The right-hand side factors as (3 + y)(3 y)y 2 The stationary solutions are y = 3, y =, and y = 3 Therefore a sign analysis of (3+y)(3 y)y 2 shows that the phase-line portrait for this equation is + + y 3 3 unstable semistable stable Solution (c) The phase-line shows that if y() = then y(t) as t (2) Solve (possibly implicitly) each of the following initial-value problems Identify their intervals of definition (a) dy dt + 2ty +t = 2 t2, y() = (b) dy dx + ex y +2x =, y() = 2y +ex Solution (a) This equation is linear and is already in normal form An integrating factor is ( t ) 2s exp +s ds = exp ( log(+t 2 ) ) = +t 2, 2 so that the integrating factor form is d ( (+t 2 )y ) = (+t 2 )t 2 = t 2 +t 4 dt Integrate this to obtain (+t 2 )y = 3 t3 + 5 t5 +c The initial condition y() = implies that c = (+ 2 ) = Therefore y = + 3 t3 + 5 t5 +t 2 This solution exists for every t, so its interval of definition is (, ) Remark Because this equation is linear, we can see that the interval of definition of its solution is (, ) without solving it because both its coefficient and forcing are continuous over (, ) Solution (b) The initial-value problem is dy dx + ex y +2x =, y() = 2y +ex

2 2 Express this equation in the differential form This differential form is exact because (e x y +2x)dx+(2y +e x )dy = y (e x y +2x) = e x = x (2y +e x ) = e x Therefore we can find H(x,y) such that x H(x,y) = e x y +2x, y H(x,y) = 2y +e x The first equation implies H(x,y) = e x y +x 2 +h(y) Plugging this into the second equation gives e x +h (y) = 2y +e x, which yields h (y) = 2y Taking h(y) = y 2, the general solution is e x y +x 2 +y 2 = c The initial condition y() = implies that c = e = Therefore y 2 +e x y +x 2 = The quadratic formula then yields the explicit solution y = ex + e 2x 4x 2 2 Here the positive square root is taken because that solution satisfies the initial condition Its interval of definition is the largest interval (x L,x R ) containing the initial time over which e 2x > 4x 2 We cannot find the endpoints of this interval explicitly (3) Consider the following Matlab function m-file function [t,y] = solveit(ti, yi, tf, n) t = zeros(n +, ); y = zeros(n +, ); t() = ti; y() = yi; h = (tf - ti)/n; for i = :n t(i + ) = t(i) + h; y(i + ) = y(i) + h*((t(i))ˆ4 + (y(i))ˆ2); end Suppose that the input values are ti =, yi =, tf = 5, and n = 4 (a) What is the initial-value problem being approximated numerically? (b) What is the numerical method being used? (c) What is the step size? (d) What are the output values for the first two approximations, y(2) and y(3)? Solution (a) The initial-value problem being approximated numerically is dy dt = t4 +y 2, y() = Solution (b) The forward Euler (explicit Euler) method is being used Solution (c) The step size is h = tf ti n = 5 4 = 4 4 = =

3 Solution (d) By carrying out the for loop in the Matlab code for i = and i = 2 we obtain the output values t(2) = t() + h = + =, y(2) = y() + h*((t())ˆ4 + (y())ˆ2) = +( ) = + 2 = 2 t(3) = t(2) + h = + = 2, y(3) = y(2) + h*((t(2))ˆ4 + (y(2))ˆ2) = 2+ ( () 4 +(2) 2) You DO NOT have to work out the arithmetic to compute y(3)! If you did then you would obtain y(3) = 494 Remark You should be able to answer similar questions that employ the Rungetrapeziodal or Runge-midpoint method (4) Give an explicit real-valued general solution of the following equations (a) y 2y +5y = te t +cos(2t) (b) u 3u u = te 2t (c) v +9v = cos(3t) Solution (a) This is a constant coefficient, nonhomogeneous, linear equation Its characteristic polynomial is p(z) = z 2 2z +5 = (z ) 2 +4 = (z ) This has the conjugate pair of roots ± i2, which yields a general solution of the associated homogeneous problem y H (t) = c e t cos(2t)+c 2 e t sin(2t) Aparticularsolution y P (t)canbefoundby either themethodofkeyidentity Evaluations or the method of Undetermined Coefficients The characteristics of the forcing terms te t and cos(2t) are r + is = and r + is = i2 respectively Because these characteristics are different, they should be treated separately Key Indentity Evaluations The forcing term te t has degree d = and characteristic r+is =, which is a root of p(z) of multiplicity m = Because m = and m+d =, we need the Key Identity and its first derivative L(e zt ) = (z 2 2z +5)e zt, L(te zt ) = (z 2 2z +5)te zt +(2z 2)e zt Evaluate these at z = to find L(e t ) = 4e t and L(te t ) = 4te t Dividing the second of these equations by 4 yields L( 4 tet ) = te t, which implies y P (t) = 4 tet The forcing term cos(2t) has degree d = and characteristic r +is = i2, which is a root of p(z) of multiplicity m = Because m = m+d =, we only need the Key Identity, L(e zt ) = (z 2 2z +5)e zt Evaluating this at z = i2 to find L(e i2t ) = ( i4)e i2t and dividing by i4 yeilds e i2t L = e i2t i4 3

4 4 Because cos(2t) = Re(e i2t ), the above equation implies e i2t (+i4)e i2t y P2 (t) = Re = Re i = 7 Re( (+i4)e i2t) = 7( cos(2t) 4sin(2t) ) Combining these two particular solutions with the general solution of the associated homogeneous problem found earlier yields the general solution y = y H (t)+y P (t)+y P2 (t) = c e t cos(2t)+c 2 e t sin(2t)+ 4 tet + 7 cos(2t) 4 7 sin(2t) Undetermined Coefficients The forcing term te t has degree d = and characteristic r+is =, which is a root of p(z) of multiplicity m = Because m = and m+d =, we seek a particular solution of the form Because we see that y P (t) = A te t +A e t y P (t) = A te t +(A +A )e t, y P (t) = A te t +(2A +A )e t, Ly P (t) = y P (t) 2y P (t)+5y P(t) = ( A te t +(2A +A )e t) 2 ( A te t +(A +A )e t) +5 ( A te t +A e t) = 4A te t +4A e t Setting 4A te t + 4A e t = te t, we see that 4A = and 4A =, whereby A = 4 and A = Hence, a particular solution is y P (t) = 4 tet The forcing term cos(2t) has degree d = and characteristic r + is = i2, which is a root of p(z) of multiplicity m = Because m = and m + d =, we seek a particular solution of the form Because we see that y P2 (t) = Acos(2t)+Bsin(2t) y P2(t) = 2Asin(2t)+2Bcos(2t), y P2 (t) = 4Acos(2t) 4Bsin(2t), Ly P2 (t) = y P2 (t) 2y P2 (t)+5y P2(t) = ( 4Acos(2t) 4Bsin(2t) ) 2 ( 2Asin(2t)+2Bcos(2t) ) +5 ( Acos(2t)+Bsin(2t) ) = (A 4B)cos(2t)+(B +4A)sin(2t) Setting (A 4B)cos(2t)+(B +4A)sin(2t) = cos(2t), we see that A 4B =, B +4A =

5 This system can be solved by any method you choose to find A = and B = 4 7 whereby a particular solution is y P2 (t) = 7 cos(2t) 4 7 sin(2t) Combining these two particular solutions with the general solution of the associated homogeneous problem found earlier yields the general solution y = y H (t)+y P (t)+y P2 (t) = c e t cos(2t)+c 2 e t sin(2t)+ 4 tet + 7 cos(2t) 4 7 sin(2t) Solution (b) The equation is u 3u u = te 2t This is a constant coefficient, nonhomogeneous, linear equation Its characteristic polynomial is p(z) = z 2 3z = (z 5)(z +2) This has the two real roots5and 2, which yields a general solution of the associated homogeneous problem u H (t) = c e 5t +c 2 e 2t A particular solution u P (t) can be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients Key Indentity Evaluations The forcing term te 2t has degree d = and characteristic r + is = 2, which is a root of p(z) of multiplicity m = Because m = and m+d = 2, we will need the first and second derivative of the Key Identity, which are computed from the Key Identity as L(e zt ) = (z 2 3z )e zt, L(te zt ) = (z 2 3z )te zt +(2z 3)e zt, L(t 2 e zt ) = (z 2 3z )t 2 e zt +2(2z 3)te zt +2e zt Evaluate the last two of these at z = 2 to find By adding 2 7 L(te 2t ) = 7e 2t, L(t 2 e 2t ) = 4te 2t +2e 2t of the first to the second we get By dividing this by 4 we obtain L ( t 2 e 2t te 2t) = 4te 2t L ( 4 t2 e 2t 2 98 te 2t) = te 2t, whereby a particular solution is u P (t) = 4 t2 e 2t 2 98 te 2t Therefore a general solution is u(t) = u H (t)+u P (t) = c e 5t +c 2 e 2t 4 t2 e 2t 2 98 te 2t 5 7,

6 6 Undetermined Coefficients The forcing term te 2t has degree d = and characteristic r +is = 2, which is a root of p(z) of multiplicity m = Because m = and m+d = 2, we seek a particular solution of the form Because u P (t) = A t 2 e 2t +A te 2t u P (t) = 2A t 2 e 2t +(2A 2A )te 2t +A e 2t, u P(t) = 4A t 2 e 2t +( 8A +4A )te 2t +(2A 4A )e 2t, we see that Lu P (t) = u P (t) 3u P (t) u P (t) = ( 4A t 2 e 2t +( 8A +4A )te 2t +(2A 4A )e 2t) 3 ( 2A t 2 e 2t +(2A 2A )te 2t +A e 2t) ( A t 2 e 2t +A te 2t) = 4A te 2t +(2A 7A )e 2t By setting 4A te 2t +(2A 7A )e 2t = te 2t we see that 4A =, 2A 7A = Thissystem canbesolvedbyanymethodyouchoosetofinda = anda 4 = 2 whereby a particular solution is u P (t) = 4 t2 e 2t 2 98 te 2t Therefore a general solution is u(t) = u H (t)+u P (t) = c e 5t +c 2 e 2t 4 t2 e 2t 2 98 te 2t Solution (c) The equation is v +9v = cos(3t) This is a constant coefficient, nonhomogeneous, linear equation Its characteristic polynomial is p(z) = z 2 +9 = z This has the conjugate pair of roots ±i3, which yields a general solution of the associated homogeneous problem v H (t) = c cos(3t)+c 2 sin(3t) A particular solution v P (t) can be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients Key Indentity Evaluations The forcing term cos(3t) has degree d = and characteristic r + is = i3, which is a root of p(z) of multiplicity m = Because m = and m+d =, we need the first derivative of the Key Identity, which is found as L(e zt ) = (z 2 +9)e zt, L(te zt ) = (z 2 +9)te zt +2ze zt 98,

7 7 Evaluate the first derivative of the Key Identity at z = i3 to find that L(te i3t ) = i6e i3t Because cos(3t) = Re(e i3t ), upon dividing by i6 and taking the real part we see that a particular solution is v P (t) = Re ( i6 tei3t) = Re ( i6 t(cos(3t)+isin(3t))) = 6 t sin(3t) Therefore a general solution is v(t) = v H (t)+v P (t) = c cos(3t)+c 2 sin(3t)+ 6 t sin(3t) Undetermined Coefficients The forcing term cos(3t) has degree d = and characteristic r + is = i3, which is a root of p(z) of multiplicity m = Because m = and m+d =, we seek a particular solution of the form Because we see that v P (t) = At cos(3t)+bt sin(3t) v P(t) = 3At sin(3t)+3bt cos(3t)acos(3t)+bsin(3t), v P (t) = 9At cos(3t) 9Bt sin(3t) 6Asin(3t)+6Bcos(3t), Lv P (t) = v P(t)+9v P (t) = ( 9At cos(3t) 9Bt sin(3t) 6Asin(3t)+6Bcos(3t) ) +9 ( At cos(3t)+bt sin(3t) ) = 6Asin(3t)+6Bcos(3t) By setting 6Asin(3t)+6Bcos(3t) = cos(3t) we see that A = and B = 6, whereby a particular solution is Therefore a general solution is v P (t) = 6 t sin(3t) v(t) = v H (t)+v P (t) = c cos(3t)+c 2 sin(3t)+ 6 t sin(3t) Remark Because of the simple form of this equation, if we had tried to solve it by either the Green Function or Variation of Parameters method then integrals that arise are not too difficult However, it is not generally a good idea to use these methods for such problems because evaluating the integrals that arise often involve much more work that the methods shown above (5) Solve the following initial-value problems (a) w +4w +2w = 5e 2t, w() = 3, w () = 7 (b) y 4y +4y = e2t 3+t, y() =, y () = 5 (c) tu +4u =, u() = 2, u () = 3

8 8 Try to evaluate any definite integrals that arise Solution (a) This is a constant coefficient, nonhomogeneous, linear equation Its characteristic polynomial is p(z) = z 2 +4z +2 = (z +2) 2 +6 = (z +2) This has the conjugate pair of roots 2±i4, which yields a general solution of the associated homogeneous problem w H (t) = c e 2t cos(4t)+c 2 e 2t sin(4t) A particular solution w P (t) can be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients Key Indentity Evaluations The forcing term 5e 2t has degree d = and characteristic r+is = 2, which is a root of p(z) of multiplicity m = Because m = and m+d =, we only need the Key Identity, Evaluate this at z = 2 to find that Upon multiplying this by 5 32 L(e zt ) = (z 2 +4z +2)e zt L(e 2t ) = (4+8+2)e 2t = 32e 2t we see that a particular solution is w P (t) = 5 32 e2t Undetermined Coefficients The forcing term 5e 2t has degree d = and characteristic r+is = 2, which is a root of p(z) of multiplicity m = Therefore we seek a particular solution of the form w P (t) = Ae 2t Because we see that w P (t) = 2Ae2t, w P (t) = 4Ae2t, Lw P (t) = w P(t)+4w P(t)+2w P (t) = 4Ae 2t +4(2Ae 2t )+2Ae 2t = 32Ae 2t By setting 32Ae 2t = 5e 2t, we see that A = 5, whereby a particular solution is 32 w P (t) = 5 32 e2t Solving the Initial-Value Problem By either method we find that a general solution is Because w(t) = w H (t)+w P (t) = c e 2t cos(4t)+c 2 e 2t sin(4t) e2t w (t) = 2c e 2t cos(4t) 4c e 2t sin(4t) 2c 2 e 2t sin(4t)+4c 2 e 2t cos(4t)+ 5 6 e2t, the initial conditions yield 3 = w() = c , 7 = w () = 2c +4c

9 Upon solving this system we find that c = 9 and c 32 2 = 3, whereby the solution 32 of the initial-value problem is w(t) = 9 32 e 2t cos(4t) 3 32 e 2t sin(4t) e2t Solution (b) The initial-value problem is y 4y +4y = e2t 3+t, y() =, y () = 5 This is a constant coefficient, nonhomogeneous, linear equation in normal form Its characteristic polynomial is p(z) = z 2 4z +4 = (z 2) 2 This has the double real root 2, which yields a general solution of the associated homogeneous problem y H (t) = c e 2t +c 2 te 2t A particular solution y P (t) cannot be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients Rather, we must use either the Green Function or the Variation of Parameters method Green Function The associated Green function g(t) satisfies g 4g +4g =, g() =, g () = A general solution of this equation is g(t) = c e 2t +c 2 te 2t Because = g() = c, we see that g(t) = c 2 te 2t Then g (t) = c 2 e 2t +2c 2 te 2t Because = g () = c 2, the Green function is g(t) = te 2t The particular solution y P (t) that satisfies y P () = y P () = is given by Because t t y P (t) = we find that t = e 2t t g(t s) e2s 3+s ds = t s 3+s ds = e2t t 3+s ds = log(3+s) t s 3+s ds = t t t 2t 2s e2s (t s)e 3+s ds e2t 3 ds = t 3log 3+s 3+s ds t s 3+s ds ( 3+t = log(3+t) log(3) = log 3 3+t, y P (t) = e 2t tlog(+ 3 t) e2t( t 3log(+ 3 t)) Therefore a general solution of the equation is y(t) = c e 2t +c 2 te 2t +y P (t) 3 ), 9

10 Because y (t) = 2c e 2t +2c 2 te 2t +c 2 e 2t +y P(t), and because y P () = y P () =, the initial conditions imply = y() = c, 5 = y () = 2c +c 2 We find that c = and c 2 = 5, whereby the solution of the initial-value problem is y(t) = 5te 2t +e 2t tlog(+ 3 t) e2t( t 3log(+ 3 t)) Variation of Parameters The equation is already in normal form Therefore we seek a particular solution of the form such that y P (t) = e 2t u (t)+te 2t u 2 (t), e 2t u (t)+te2t u 2 (t) =, 2e 2t u (t)+(2te 2t +e 2t )u 2(t) = e2t 3+t This system can be solved to find that u (t) = t 3+t, u 2 (t) = 3+t These can be integrted to obtain t u (t) = 3+t dt = 3 3+t dt = t+3log(3+t)+c, u 2 (t) = 3+t dt = log(3+t)+c 2, whereby a general solution is y(t) = c e 2t e 2t( t 3log(3+t) ) +c 2 te 2t +te 2t log(3+t) Because y (t) = 2c e 2t 2e 2t( t 3log(3+t) ) ( e 2t 3 ) 3+t +2c 2 te 2t +c 2 e 2t +2te 2t log(3+t)+e 2t log(3+t)+te 2t 3+t, the initial conditions imply that = y() = c +3log(3), 5 = y () = 2c +6log(3)+c 2 +log(3) We can solve this system to find that c = 3log(3) and c 2 = 5 log(3) Therefore the solution of the initial-value problem is y(t) = 3log(3)e 2t e 2t( t 3log(3+t) ) + ( 5 log(3) ) te 2t +te 2t log(3+t) Solution (c) The initial-value problem is tu +4u =, u() = 2, u () = 3

11 This is a variable coefficient, homogeneous, second-order linear equation Because it does not depend explictly on u, we can reduce its order by setting w = u Then w satisfies the initial-value problem tw +4w =, w() = 3 This is a variable coefficient, homogeneous, first-order linear equation Its normal form is By setting A(t) = t w + 4 w =, w() = 3 t 4 s ds = 4log(s) t the solution of the initial-value problem for w is = 4log(t) 4log() = 4log(t), w(t) = w()e A(t) = 3e 4log(t) = 3t 4 Then because u = w and u() = 2 we see that u(t) = u()+ t w(s)ds = 2 3 t s 4 ds = 2+ [ s 3] t = 2+t 3 = +t 3 The interval of definition for this solution is (, ), which can be seen by putting the original initial-value problem into normal form (6) Give an explicit general solution of the equation h +2h +5h = Sketch a typical solution for t If this equation governs a spring-mass system, is the system undamped, under damped, critically damped, or over damped? Solution This is a constant coefficient, homogeneous, linear equation Its characteristic polynomial is p(z) = z 2 +2z +5 = (z +) This has the conjugate pair of roots ±i2, which yields a general solution h(t) = c e t cos(2t)+c 2 e t sin(2t) When c 2 +c 2 2 > this can be put into the amplitute-phase form h(t) = Ae t cos(2t δ), where A > and δ < 2π are determined from c and c 2 by A = c 2 +c2 2, cos(δ) = c A, sin(δ) = c 2 A In other words, (A,δ) are the polar coordinates for the point in the plane whose Cartesian coordinates are (c,c 2 ) The sketch should show a decaying oscillation with amplitude Ae t and quasiperiod 2π = π A sketch might be given during the 2 review session The equation governs an under damped spring-mass system because its characteristic polynomial has a conjugate pair of roots with negative real part

12 2 (7) When a mass of 2 kilograms is hung vertically from a spring, it stretches the spring 5 meters (Gravitational acceleration is 98 m/sec 2 ) At t = the mass is set in motion from 3 meters below its equilibrium (rest) position with a upward velocity of 2 m/sec It is acted upon by an external force of 2cos(5t) Neglect drag and assume that the spring force is proportional to its displacement Formulate an initial-value problem that governs the motion of the mass for t > (DO NOT solve this initialvalue problem; just write it down!) Solution Let h(t) be the displacement (in meters) of the mass from its equilibrium (rest) position at time t (in seconds), with upward displacements being positive The governing initial-value problem then has the form m d2 h dt +kh = 2cos(5t), h() = 3, 2 h () = 2, where m is the mass and k is the spring constant The problem says that m = 2 kilograms The spring constant is obtained by balancing the weight of the mass (mg = 2 98 Newtons) with the force applied by the spring when it is stetched 5 m This gives k5 = 2 98, or k = 2 98 = 4 98 Newtons/m 5 Therefore the governing initial-value problem is 2 d2 h dt h = 2cos(5t), h() = 3, h () = 2 Had you chosen downward displacements to be positive then the sign of the initial data would change! You should make your convention clear! (8) Find the Laplace transform Y(s) of the solution y(t) to the initial-value problem where y +4y +8y = f(t), y() = 2, y () = 4 f(t) = { 4 for t < 2, t 2 for 2 t You may refer to the table of Laplace transforms on the last page (DO NOT take the inverse Laplace transform to find y(t); just solve for Y(s)!) Solution The Laplace transform of the initial-value problem is where L[y ](s)+4l[y ](s)+8l[y](s) = L[f](s), L[y](s) = Y(s), L[y ](s) = sy(s) y() = sy(s) 2, L[y ](s) = s 2 Y(s) sy() y () = s 2 Y(s) 2s 4 To compute L[f](s), first write f as f(t) = ( u(t 2) ) 4+u(t 2)t 2 = 4 u(t 2)4+u(t 2)t 2 = 4+u(t 2)(t 2 4) = 4+u(t 2)j(t 2),

13 where j(t) = (t+2) 2 4 = t 2 +4t Referring to the table of Laplace transforms, line, line 3 with c = 2, and line 3 with n = and n = 2 then show that L[f](s) = 4L[](s)+L [ u(t 2)j(t 2) ] (s) = 4L[](s)+e 2s L [ j(t) ] (s) = 4L[](s)+e 2s L[4t+t 2 ](s) = 4L[](s)+4e 2s L[t](s)+e 2s L[t 2 ](s) = 4 s +4e 2s s 2 +e 2s 2 s 3 The Laplace transform of the initial-value problem then becomes ( s 2 Y(s) 2s 4 ) +4 ( sy(s) 2 ) +8Y(s) = 4 s +e 2s 4 s 2 +e 2s 2 s 3, 3 which becomes (s 2 +4s+8)Y(s) 2s 2 = 4 s +e 2s 4 s 2 +e 2s 2 s 3 Hence, Y(s) is given by Y(s) = (2s+2+ 4s 4s 2s ) s 2 +4s+8 +e 2s 2 +e 2s 3 (9) Find the function y(t) whose Laplace transform Y(s) is given by e 3s 4 (a) Y(s) = s 2 6s+5, (b) Y(s) = e 2s s s 2 +4s+8 You may refer to the table of Laplace transforms on the last page Solution (a) The denominator factors as (s 5)(s ), so we have the partial fraction identity 4 s 2 6s+5 = 4 (s 5)(s ) = s 5 s Referring to the table of Laplace transforms, line 2 with a = 5 and with a = gives [ ] [ ] L (t) = e 5t, L (t) = e t, s 5 s whereby [ ] [ L 4 (t) = L s 2 6s+5 s 5 ] = e 5t e t s It follows from line 3 with c = 3 and j(t) = e 5t e t that [ ] y(t) = L [Y(s)](t) = L e 3s 4 (t) s 2 6s+5 [ ] = u(t 3)L 4 (t 3) = u(t 3) ( e 5(t 3) e t 3) s 2 6s+5

14 4 Solution (b) The denominator does not have real factors The partial fraction identity is s s 2 +4s+8 = s (s+2) 2 +4 = s+2 (s+2) (s+2) Referring to the table of Laplace transforms, lines 8 and 7 with a = 2 and b = 2 give [ [ ] L s+2 ](t) = e 2t cos(2t), L 2 (t) = e 2t sin(2t), (s+2) (s+2) whereby [ [ [ L s ](t) = L s+2 ](t) L s 2 +4s+8 (s+2) = e 2t( cos(2t) sin(2t) ) 2 (s+2) ] (t) It follows from line 3 with c = 2 and j(t) = e 2t( cos(2t) sin(2t) ) that [ [ ] y(t) = L [Y(s)](t) = L e 2s s ](t) = u(t 2)L s (t 2) s 2 +4s+8 s 2 +4s+8 = u(t 2)e 2(t 2)( cos(2(t 2)) sin(2(t 2)) ) () Consider two interconnected tanks filled with brine (salt water) The first tank contains 8 liters and the second contains 3 liters Brine flows with a concentration of 3 grams of salt per liter flows into the first tank at a rate of 2 liters per hour Well stirred brine flows from the first tank to the second at a rate of 6 liters per hour, from the second to the first at a rate of 4 liters per hour, and from the second into a drain at a rate of 2 liters per hour At t = there are 7 grams of salt in the first tank and 25 grams in the second Give an initial-value problem that governs the amount of salt in each tank as a function of time Solution: The rates work out so there will always be 8 liters of brine in the first tank and 3 liters in the second Let S (t) be the grams of salt in the first tank and S 2 (t) be the grams of salt in the second tank These are governed by the initial-value problem ds dt = 3 2+ S S 8 6, S () = 7, ds 2 dt = S 8 6 S S 2 3 2, S 2() = 25 You could leave the answer in the above form It can however be simplified to ds dt = 6+ 2 S S 4, S () = 7, ds 2 dt = 3 S 4 S 5 2, S 2 () = 25 () Consider the real vector-valued functions x (t) = (a) Compute the Wronskian W[x,x 2 ](t) ( t 3, x t) 2 (t) = 3+t 4

15 (b) Suppose that x and x 2 comprise a fundamental set of solutions to the linear system x = A(t)x Give a general solution to this system Solution (a) The Wronskian is given by t 3 W[x,x 2 ](t) = det t 3+t 4 = (3+t 4 ) t t 3 = 3+t 4 t 4 = 3 Solution (b) Because x (t), x 2 (t) is a fundamental set of solutions for the linear system whose coefficient matrix is A(t), a general solution is given by t 3 x(t) = c x (t)+c 2 x 2 (t) = c +c t 2 3+t 4 (2) Give a general real vector-valued solution of the linear planar system x = Ax for (a) A =, (b) A = 4 2 Solution (a) The characteristic polynomial of A is p(z) = z 2 tr(a)z +det(a) = z 2 6z 6 = (z 3) 2 25 = (z 3) The eigenvalues of A are the roots of this polynomial, which are 3±5, or simply 2 and 8 Therefore we have [ e ta = e 3t cosh(5t)i+ sinh(5t) ] (A 3I) 5 = e [cosh(5t) 3t + sinh(5t) ] ( cosh(5t)+ = e 3t 3 sinh(5t) 4 sinh(5t) ) sinh(5t) cosh(5t) 3 sinh(5t) 5 5 Therefore a general solution is given by ( x(t) = e ta c cosh(5t)+ = c c e 3t 3 sinh(5t) ( 5 4 )+c 2 sinh(5t) 2 e 3t 4 sinh(5t) ) 5 cosh(5t) 3 sinh(5t) 5 5 Alternative Solution (a) The characteristic polynomial of A is p(z) = z 2 tr(a)z +det(a) = z 2 6z 6 = (z 3) 2 25 = (z 3) The eigenvalues of A are the roots of this polynomial, which are 3±5, or simply 2 and 8 Because A+2I =, A 8I =, we see that A has the eigenpairs ( ) 2, 2, ( ( 2 8, )) 5

16 6 Form these eigenpairs we construct the solutions 2 x (t) = e 2t, x 2 2 (t) = e 8t, Therefore a general solution is x(t) = c x (t)+c 2 x 2 (t) = c e 2t 2 +c 2 e 8t ( 2 ) Solution (b) The characteristic polynomial of A is p(z) = z 2 tr(a)z +det(a) = z 2 2z +5 = (z ) 2 +4 = (z ) The eigenvalues of A are the roots of this polynomial, which are ±i2 Therefore we have [ e ta = e t cos(2t)i+ sin(2t) ] (A I) 2 = e [cos(2t) t + sin(2t) ] cos(2t) sin(2t) = e t sin(2t) cos(2t) Therefore a general solution is given by x(t) = e ta c cos(2t) sin(2t) = c c e t +c 2 sin(2t) 2 e t cos(2t) Alternative Solution (b) The characteristic polynomial of A is p(z) = z 2 tr(a)z +det(a) = z 2 2z +5 = (z ) 2 +4 = (z ) The eigenvalues of A are the roots of this polynomial, which are ±i2 Because i2 2 i2 2 A (+i2)i =, A ( i2)i =, 2 i2 2 i2 we see that A has the eigenpairs ( ( +i2,, i)) ( i2, ) i Because cos(2t)+isin(2t) e (+i2)t = e i t, sin(2t)+icos(2t) two real solutions of the system are cos(2t) sin(2t) x (t) = e t, x sin(2t) 2 (t) = e t cos(2t) Therefore a general solution is cos(2t) x(t) = c x (t)+c 2 x 2 (t) = c e t sin(2t) sin(2t) +c 2 e t cos(2t)

17 (3) What answer will be produced by the following Matlab command? >> A = [ 4; 3 2]; [vect, val] = eig(sym(a)) You do not have to give the answer in Matlab format Solution The Matlab command will produce the eigenpairs of A = characteristic polynomial of A is p(z) = z 2 tr(a)z +det(a) = z 2 3z = (z 5)(z +2), so its eigenvalues are 5 and 2 Because 4 4 A 5I =, A+2I = 3 3 we can read off that the eigenpairs are ( ( 5,, )) ( 2, ) , The 3 2 (4) A real 2 2 matrix A has eigenvalues 2 and with associated eigenvectors ( 3 and ) 2 (a) Give a general solution to the linear planar system x = Ax (b) Compute e ta (c) Sketch a phase-plane portrait for this system and identify its type Classify the stability of the origin Carefully mark all sketched orbits with arrows! Solution (a) Use the given eigenpairs to construct the solutions 3 x (t) = e 2t, x 2 (t) = e t 2 Therefore a general solution is x(t) = c x (t)+c 2 x 2 (t) = c e 2t ( 3 ) +c 2 e t 2 Solution (b) The matrix A can be diagonalized as A = VDV where 3 2 V =, D =, V 2 = Then e ta = Ve td V = 7 = 7 ( 3 e 2t 2 ) ( 3e 2t e t 2 e 2t 2e t 3 ) 2 e t 3 = 7 ( 6e 2t +e t 3e 2t 3e t 2e 2t 2e t e 2t +6e t )

18 8 Alternative Solution (b) By part (a) a fundamental matrix is Ψ(t) = ( x (t) x 2 (t) ) 3e 2t e = t e 2t 2e t Then ) ( 3e e ta = Ψ(t)Ψ() 2t e = t 3 e 2t 2e t 2 = 3e 2t e t 2 7 e 2t 2e t = 6e 2t +e t 3e 2t 3e t 3 7 2e 2t 2e t e 2t +6e t Solution (c) The matrix A has two real eigenvalues of opposite sign Therefore the origin is a saddle and is thereby unstable There is one orbit moves away from (,) along each half of the line x = 3y, and one orbit moves towards (,) along each half of the line y = 2x (These are the lines of eigenvectors) Every other orbit sweeps away from the line y = 2x and towards the line x = 3y A phase-plane portrait might be sketched during the review session (5) Solve the initial-value problem x = Ax, x() = x I and describe how its solution behaves as t for the following A and x I 3 3 (a) A =, x 5 7 I = (b) A =, x 5 2 I = 3 Solution (a) The characteristic polynomial of A = is 5 7 p(z) = z 2 tr(a)z +det(a) = z 2 +4z +29 = (z +2) Therefore the eigenvlues of A are 2±i5 Then [ e ta = e 2t cos(5t)i+ sin(5t) ] (A+2I) 5 = e [cos(5t) 2t + sin(5t) ] cos(5t)+sin(5t) 2sin(5t) = e 2t sin(5t) cos(5t) sin(5t) Therefore the solution of the initial-value problem is cos(5t)+sin(5t) 2sin(5t) 3 x(t) = e ta x I = e 2t sin(5t) cos(5t) sin(5t) 2 3cos(5t)+sin(5t) = e 2t 2 cos(5t) + sin(5t) This solution decays to zero as t

19 Alternative Solution (a) The characteristic polynomial of A = p(z) = z 2 tr(a)z +det(a) = z 2 +4z +29 = (z +2) Therefore the eigenvlues of A are 2±i5 Because 5 i5 A ( 2+i5)I =, 5 5 i5 we can read off that A has the conjugate eigenpairs +i i 2+i5,, 2 i5, Because +i e 2t+i5t = e 2t( cos(5t)+isin(5t) ) +i a fundamental set of real solutions is cos(5t) sin(5t) x (t) = e 2t cos(5t) Then a fundamental matrix Ψ(t) is given by Because = e 2t ( cos(5t) sin(5t)+icos(5t)+isin(5t) cos(5t) isin(5t) 3 is 5 7 ), cos(5t)+sin(5t), x 2 (t) = e 2t sin(5t) Ψ(t) = ( x (t) x 2 (t) ) ( cos(5t) sin(5t) cos(5t)+sin(5t) = e 2t cos(5t) sin(5t) Ψ() = = =, we see that cos(5t) sin(5t) cos(5t)+sin(5t) e ta = Ψ(t)Ψ() = e 2t cos(5t) sin(5t) cos(5t)+sin(5t) 2sin(5t) = e 2t sin(5t) cos(5t) sin(5t) Therefore the solution of the initial-value problem is cos(5t)+sin(5t) 2sin(5t) 3 x(t) = e ta x I = e 2t sin(5t) cos(5t) sin(5t) 2 3cos(5t)+sin(5t) = e 2t 2 cos(5t) + sin(5t) This solution decays to zero as t Remark After we have constructed the fundmental set of solutions x (t) and x 2 (t), we could also have solved the initial-value problem by finding constants c and c 2 such that x(t) = c x (t) + c 2 x 2 (t) satisfies the initial condition Had we done this using the x (t) and x 2 (t) constructed above, we would have found that c = 2 and c 2 = ) 9

20 2 Solution (b) The characteristic polynomial of A = 8 5 is 5 2 p(z) = z 2 tr(a)z +det(a) = z 2 6z +9 = (z 3) 2 The ony eigenvalue of A is 3 Then e ta = e 3t[ I+t(A 3I) ] [ = e 3t ( +5t 5t = e 3t 5t 5t ] 5 5 +t 5 5 ) Therefore the solution of the initial-value problem is +5t 3 3+2t x(t) = e ta x I = e 3t = e 5t 5t)( 3t +2t This solution grows like 2te 3t as t Alternative Solution (b) The characteristic polynomial of A = p(z) = z 2 tr(a)z +det(a) = z 2 6z +9 = (z 3) 2 The only eigenvalue of A is 3 Because A 3I = 5 5, 5 5 we can read off that A has the eigenpair ( ( 3, )) 8 5 is 5 2 We can use this eigenpair to construct the solution x (t) = e 3t A second solution can be constructed by x 2 (t) = e 3t w+te 3t (A 3I)w, where w is any nonzero vector that is not an eigenvector associated with 3 For example, taking w = ( )T yields 5 +5t x 2 (t) = e 3t +te 3t = e 5 5)( 3t 5t Then a fundamental matrix Ψ(t) is given by Ψ(t) = ( x (t) x 2 (t) ) +5t = e 3t 5t Because Ψ() = = =,

21 we see that e ta = Ψ(t)Ψ() = +5t +5t 5t = 5t 5t 5t Therefore the solution of the initial-value problem is +5t 5t 3 3+2t x(t) = e ta x I = e 3t = e 5t 5t 3t +2t This solution grows like 2te 3t as t Remark After we have constructed the fundmental set of solutions x (t) and x 2 (t), we could also have solved the initial-value problem by finding constants c and c 2 such that x(t) = c x (t) + c 2 x 2 (t) satisfies the initial condition Had we done this using the x (t) and x 2 (t) constructed above, we would have found that c = and c 2 = 4 (6) Consider the nonlinear planar system x = 2xy, y = 9 9x y 2 (a) Find all of its stationary points (b) Find a nonconstant function H(x,y) such that every orbit of the system satisfies H(x,y) = c for some constant c (c) Classify the type and stability of each stationary point (d) Sketch the level sets (contour lines) H(x,y) = c for values of c corresponding to each saddle point Use arrows to indicate the direction of the orbit along each curve that you sketch! Solution (a) Stationary points satisfy = 2xy, = 9 9x y 2 The top equation shows that x = or y = If x = then the bottom equation becomes = 9 y 2 = (3 y)(3+y), which shows that either y = 3 or y = 3 If y = then the bottom equation becomes = 9 9x = 9( x), which shows that x = Therefore the stationary points of the system are (,3), (, 3), (,) Solution (b) The associated first-order orbit equation is dy dx = 9 9x y2 2xy This equation is not linear or separable It has the differential form which is exact because (y 2 +9x+ 9)dx+2xydy =, y (y 2 +9x 9) = 2y = x (2xy) = 2y 2

22 22 Therefore there exists H(x, y) such that x H(x,y) = y 2 +9x 9, y H(x,y) = 2xy By integrating the second equation we see that H(x,y) = xy 2 +h(x) When this is substituted into the first equation we find x H(x,y) = y 2 +h (x) = y 2 +9x 9, which implies that h (x) = 9x 9 By taking h(x) = 9 2 x2 9x we obtain H(x,y) = xy x2 9x Alternative Solution (b) Notice that x f(x,y)+ y g(x,y) = x (2xy)+ y (9 9x y 2 ) = 2y 2y = Therefore the system has Hamiltonian form with Hamiltonian H(x, y) that satisfies y H(x,y) = 2xy, x H(x,y) = 9 9x y 2 By integrating the first equation we see that H(x,y) = xy 2 +h(x) When this is substituted into the second equation we find x H(x,y) = y 2 h (x) = 9 9x y 2, which implies that h (x) = 9x 9 By taking h(x) = 9 2 x2 9x we obtain H(x,y) = xy x2 9x Solution (c) Because f(x,y) = f(x,y) 2xy = g(x, y) 9 9x y 2, the Jacobian matrix f(x, y) of partial derivatives is x f(x,y) f(x,y) = y f(x,y) 2y 2x = x g(x,y) y g(x,y) 9 2y Evaluating this matrix at each stationary point yields f(,3) =, f(, 3) =, f(,) = Because the matrix f(,3) is lower triangular, we can read off that its eigenvalues are 6 and 6 Because these are real with opposite signs, the stationary point (,3) is a saddle and thereby is unstable Because the matrix f(, 3) is lower triangular, we can read off that its eigenvalues are 6 and 6 Because these are real with opposite signs, the stationary point (, 3) is a saddle and thereby is unstable

23 The characteristic polynomial of the matrix f(, ) is p(z) = z 2 +8, so the matrix f(,) has eigenvalues ±i 8 Because these are immaginary and the system has an integral while the lower left entry of f(,) is negative, the stationary point (,) is a clockwise center and thereby is stable Alternative Solution (c) If you saw that the system has Hamiltonian form with Hamiltonian H(x,y) from part (b) then you can take this approach The Hessian matrix 2 H(x,y) of second partial derivatives of the Hamiltonian H(x,y) is 2 H(x,y) = ( xx H(x,y) xy H(x,y) yx H(x,y) yy H(x,y) ) = ( 9 2y 2y 2x ) Evaluating this at the stationary points yields H(,3) =, H(, 3) =, 2 H(,) = 6 6 The characteristic polynomial of the matrix 2 H(,3) is p(z) = z 2 9z 36 = (z 2)(z +3) 9 2 Therefore the matrix 2 H(,3) has eigenvalues 2 and 3 Because these have different signs, the stationary point (, 3) is a saddle and thereby is unstable The characteristic polynomial of the matrix 2 H(, 3) is p(z) = z 2 9z 36 = (z 2)(z +3) Therefore the matrix 2 H(, 3)has eigenvalues 2 and 3 Because these have different signs, the stationary point (, 3) is a saddle and thereby is unstable Because the matrix 2 H(,) is diagonal, we can read off that its eigenvalues are 9 and 2 Because these are both positive, the stationary point (,) is a clockwise center and thereby is stable Solution (d) The saddle points are (,3) and (, 3) Because H(,3) = H(, 3) = (±3) = Hence, the level set corresponding to these saddle points is = xy x2 9x = (y x 9)x The points on this set must satisfy either y x 9 = or x = Therefore the level set is the union of the parabola x = y2 and the y-axis Along the y-axis (x = ) the y equation reduces to y = 9 y 2 = (3 y)(3+y), whereby the arrows point towards (,3) and away from (, 3) Along the parabola x = y2 the arrows point away from (,3) and towards (, 3) because they are saddle points 23

24 24 (7) Consider the nonlinear planar system x = 5y, y = x 4y x 2 (a) Find all of its stationary points (b) Compute the Jacobian matrix at each stationary point (c) Classify the type and stability of each stationary point (d) Sketch a plausible global phase-plane portrait Use arrows to indicate the direction of the orbit along each curve that you sketch! Solution (a) Stationary points satisfy = 5y, = x 4y x 2 The first equation implies y =, whereby the second equation becomes = x x 2 = x( x), which implies either x = or x = Therefore all the stationary points of the system are (,), (,) Solution (b) Because f(x,y) = f(x,y) 5y = g(x, y) x 4y x 2, the Jacobian matrix of partial derivatives is x f(x,y) f(x,y) = y f(x,y) = x g(x,y) y g(x,y) 5 2x 4 Evaluating this matrix at each stationary point yields the coefficient matrices 5 5 A = at (,), A = at (,) 4 4 Solution (c) The coefficient matrix A at (, ) has eigenvalues that satisfy = det(zi A) = z 2 tr(a)z +det(a) = z 2 +4z +5 = (z +2) The eigenvalues are thereby 2±i Because a 2 = >, the stationary point (,) is a counterclockwise spiral sink, which is attracting This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, ) The coefficient matrix A at (, ) has eigenvalues that satisfy = det(zi A) = z 2 tr(a)z +det(a) = z 2 +4z 5 = (z +2) The eigenvalues are thereby 2±3, or simply 5 and The stationary point (,) is thereby a saddle, which is unstable This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, ) Solution (d) The nullcline for x is the line y = This line partitions the plane into regions where x is increasing or decreasing as t increases The nullcline for y is the parabola y = 4 (x x2 ) This curve partitions the plane into regions where y is increasing or decreasing as t increases Neither of these nullclines is invariant The stationary point (, ) is a counterclockwise spiral sink

25 The stationary point (, ) is a saddle The coefficient matrix A has eigenvalues 5 and Because A+5I =, A I =, 5 it has the eigenpairs ( ( 5,, )) (, ) 5 Near (,) there is one orbit that emerges from (,) tangent to each side of the line x = 5y There is also one orbit that approaches (,) tangent to each side of the line y = x These orbits are separatrices A global phase-plane portrait might be sketched during the review session Remark The global phase-plane portrait becomes clearer if we had seen that H(x,y) = 2 x y2 3 x3 satisfies d dt H(x,y) = xh(x,y)x + y H(x,y)y = (x x 2 )( 5y)+5y(x 4y x 2 ) = 2y 2 The orbits of the system are thereby seen to cross the contour lines of H(x,y) so as to decrease H(x,y) You would not be expected to see this on the Final Exam (8) Consider the nonlinear planar system x = x(3 3x+2y), y = y(6 x y) Do parts (a-d) as for the previous problem (e) Why do solutions that start in the first quadrant stay in the first quadrant? Solution (a) Stationary points satisfy = x(3 3x+2y), = y(6 x y) The first equation implies either x = or 3 3x+2y =, while the second equation implies either y = or 6 x y = If x = and y = then (,) is a stationary point If x = and 6 x y = then (,6) is a stationary point If 3 3x+2y = and y = then (,) is a stationary point If 3 3x+2y = and 6 x y = then upon solving these equations we find that (3, 3) is a stationary point Therefore all the stationary points of the system are Solution (b) Because (,), (,6), (,), (3,3) f(x,y) = f(x,y) 3x 3x = 2 +2xy g(x, y) 6y xy y 2, the Jacobian matrix of partial derivatives is x f(x,y) f(x,y) = y f(x,y) 3 6x+2y 2x = x g(x,y) y g(x,y) y 6 x 2y 25

26 26 Evaluating this matrix at each stationary point yields the coefficient matrices 3 5 A = at (,), A = at (,6), A = at (,), A = at (3,3) Solution (c) The coefficient matrix A at (,) is diagonal, so we can read-off its eigenvalues as 3 and 6 The stationary point (,) is thereby a nodal source, which is repelling This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (,) The coefficient matrix A at (,6) is triangular, so we can read-off its eigenvalues as 6 and 5 The stationary point (,6) is thereby a saddle, which is unstable This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, 6) The coefficient matrix A at (,) is triangular, so we can read-off its eigenvalues as 3 and 5 The stationary point (,) is thereby a saddle, which is unstable This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, ) The coefficient matrix A at (3, 3) has eigenvalues that satisfy = det(zi A) = z 2 tr(a)z +det(a) = z 2 +2z +45 = (z +6) Its eigenvalues are thereby 6 ± i3 Because a 2 = 3 <, the stationary point (3,3) is a clockwise spiral sink, which is attracting This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (3, 3) Solution (d) The nullclines for x are the lines x = and 3 3x+2y = These lines partition the plane into regions where x is increasing or decreasing as t increases The nullclines for y are the lines y = and 6 x y = These lines partition the plane into regions where y is increasing or decreasing as t increases Next, observe that the lines x = and y = are invariant A orbit that starts on one of these lines must stay on that line Along the line x = the system reduces to y = y(6 y) Along the line y = the system reduces to x = 3x( x) The arrows along these invariant lines can be determined from a phase-line portrait of these reduced systems The stationary point (, ) is a nodal source The coefficient matrix A has eigenvalues 3 and 6 Because it has the eigenpairs A 3I =, A 6I = 3 ( ( 3,, )) ( ( 6, )) 3,

27 Near (,) there is one orbit that emerges from (,) along each side of the invariant lines y = and x = Every other orbit emerges from (,) tangent to the line y =, which is the line corresponding to the eigenvalue with the smaller absolute value The stationary point (, 6) is a saddle The coefficient matrix A has eigenvalues 6 and 5 Because 2 A+6I =, A 5I =, it has the eigenpairs ( ( 6,, )) ( ) 7 5, 2 Near (,6) there is one orbit that approaches (,6) along each side of the invariant line x = There is also one orbit that emerges from (,6) tangent to each side of the line y = 6 2 x These orbits are separatrices 7 The stationary point (, ) is a saddle The coefficient matrix A has eigenvalues 3 and 5 Because A+3I =, A 5I =, 8 it has the eigenpairs ( ( 3,, )) ( ( 5, 4)) Near (,) there is one orbit that emerges from (,) along each side of the invariant line y = There is also one orbit that approaches (,) tangent to each side of the line y = 4(x ) These orbits are also separatrices Finally, the stationary point (3, 3) is a clockwise spiral sink All orbits in the positive quadrant will spiral into it A phase-plane global portrait might be sketched during the review session Be sure your portrait be correct near each stationary point! Solution (e) Because the lines x = and y = are invariant, the uniqueness theorem implies that no other orbits can cross them 27