Old Math 330 Exams. David M. McClendon. Department of Mathematics Ferris State University

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1 Old Math 330 Exams David M. McClendon Department of Mathematics Ferris State University Last updated to include exams from Fall 07

2 Contents Contents General information about these exams 3 Exams from Fall Fall 06 Exam Fall 06 Exam Fall 06 Exam Fall 06 Final Exam Exams from Fall Fall 07 Exam Fall 07 Exam Fall 07 Final Exam

3 Chapter General information about these exams These are the exams I have given in differential equations courses. Each exam is given here, followed by what I believe are the solutions there may be some number of computational errors or typos in these answers. Typically speaking, tests labelled Exam cover Chapter in my differential equations lecture notes; tests labelled Exam cover Chapter except for the last section, and tests labelled Exam 3 cover the last section of Chapter and all of Chapter 3. 3

4 Chapter Exams from Fall 06. Fall 06 Exam. Briefly explain what is meant by existence / uniqueness in the context of ordinary differential equations.. Let y = yt be the solution of the initial value problem y = y + t y =. Suppose you wanted to estimate y3 using Euler s method with 0 steps. Find the points t, y and t, y obtained by this method. 3. Here is the picture of the slope field associated to an autonomous ODE y = φy:

5 .. Fall 06 Exam a Suppose y0 = 7. Estimate y. b Suppose y =. Find lim t yt. c On the picture above, sketch the graph of the solution satisfying the initial condition y0 = 6. Label the graph d. d Find all equilibria of this equation, and classify them as stable, semistable or unstable. 4. Find the general solution of the following ODE: dy dt y = 4e4t 5. Find the particular solution of the following initial value problem: t y = y y = Write your answer as a function y = ft. 6. Find the particular solution of the following initial value problem: ty dy dt = t y y3 = 7. Find the general solution of the following ODE: ty = y Write your answer as a function y = ft. Solutions. The Existence/Uniqueness Theorem for first-order ODEs says if the function φ is nice i.e. φ and φ are continuous, then the initial value problem y y = φt, y yt 0 = y 0 has one and only one solution, which is of the form y = ft.. First, t = n t n t 0 = 0 3 = 3. Next, we are given t 0, y 0 =,. Now φt 0, y 0 = + = 0 so t = t 0 + t = + 3 = 4 y = y 0 + φt 0, y 0 t = + 03 =. 5

6 .. Fall 06 Exam Therefore t, y = 4,. Now φt, y = + 4 = 6 so Therefore t, y = 7, 6. t = t + t = = 7 y = y + φt, y t = + 63 = a Starting at the point 0, 7 and following the vector field, we come to the point, 6 so y 6. b Starting at the point, and moving to the extreme right, we see that yt = 4. lim t c d y = 4 is stable; y = 8 is unstable. 4. Method integrating factors: The integrating factor is µt = exp [ t 0 ds] = e t. After multiplying through by µt, the equation becomes dy dt e t e t y = 4e 4t e t d ye t = 4e t dt ye t = 4e t dt ye t = 7e t + C y = e t 7e t + C y = 7e 4t + Ce t. Method undetermined coefficients: The corresponding homogeneous equation is dy y = 0 which has solution y dt h = e t exponential growth model. Now, guess y p = Ae 4t and plug into the left-hand side of the equation to get 4Ae 4t Ae 4t = 4e 4t. That means 4A A = 4, i.e. A = 7. Therefore y p = 7e 4t so y = y p + Cy h, i.e. y = 7e 4t + Ce t. 6

7 .. Fall 06 Exam 5. Start with the ODE, which is separable: t dy dt = y y dy = t dt y dy = t dt y = t + C Next, I will solve for C using the initial condition you could have solved for y first: Therefore the particular solution is = + C C = y = t +. Solve for y by first multiplying through by and then taking reciprocals to get y = this answer is fine t which, if you multiply through the numerator and denominator by t simplifies to y = t t. 6. Rewrite this equation as y t + ty dy dt. Letting M = y t and N = ty, we see that M y = y = N t so the equation is exact. Now ψt, y = M dt = y t dt = y t 3 t3 + Ay = N dy = ty dy = ty + Bt. By setting Bt = 3 t3 and Ay = 0, we reconcile these integrals to obtain ψt, y = y t 3 t3. Thus the general solution is ψt, y = C, i.e. y t 3 t3 = C. Plugging in the initial condition and solving for C, we see = C, i.e. C = 3 9 = 6. So the particular solution is y t 3 t3 = 6. 7

8 .. Fall 06 Exam 7. This equation is second-order with no y. To solve it, let v = dy so that the dt equation becomes tv = v i.e. t dv dt = v. This is separable: rewrite it as dv = dt and integrate both sides to obtain v t ln v = ln t + C. Solving for v, we get v = e ln t+c = e ln t e C = te C = Ct. Last, since v = dy, integrate to obtain y: dt y = vt dt = Ct dt = Ct + D. Renaming the first constant, this can be written as y = Ct + D. 8

9 .. Fall 06 Exam. Fall 06 Exam. Here is a picture of the phase plane of an autonomous, first-order system of ODEs y = Φy, where as usual, y = x, y: a Find the two equilibria of this system and classify each of them as a center, node, spiral or saddle. b Suppose x0 = 0 and y0 = 3. Find the following four limits: lim xt = t lim yt = t lim xt = t lim yt = t c Suppose y0 = 3, 0. Estimate the maximum value obtained by xt.. Here is the phase plane of a first-order, constant-coefficient, homogeneous linear system of ODEs y = Ay: a Find two eigenvectors of A corresponding to distinct eigenvalues. 9

10 .. Fall 06 Exam b How many positive, real eigenvalues does A have? c How many negative, real eigenvalues does A have? d How many non-real eigenvalues does A have? 3. Find the general solution of the following system of ODEs: x = x + 4y + e t y = x y + e t 4. Find the general solution of the following system of ODEs: x = 3x + y y = x 5y Write your final answer coordinate-wise. 5. Find the particular solution of the following initial value problem: y = x 4y, x + 5y y0 = 3, 0

11 .. Fall 06 Exam Solutions. a From looking at the picture, the two equilibria are 9, 3 which is a saddle and 3, 9 which is a unstable node. b By following the curve passing through 0, 3 forwards and backwards, we see that lim xt = lim t yt = 9 t lim xt = 3 lim t yt = 9. t c The maximum value obtained by xt is the right-most point on the curve passing through 3, 0, which is approximately 5.. a The eigenvectors of A go in the direction of the straight-line solutions; from the picture these are, or any multiple of, and 3, or any multiple of 3,. b Since 0 is an unstable node from the picture, both eigenvalues of A are real and positive, so the answer is two. c None since the eigenvalues of A are both positive. d None since the eigenvalues of A are both positive and real Let A =. We start by finding the solution of the homogeneous equation y = Ay. First, the eigenvalues of A: deta λi = det λ 4 λ = λ λ 4 = λ λ 6 = λ 3λ + so the eigenvalues are λ = 3 and λ =. Now for the eigenvectors let v = x, y: x + 4y = 3x λ = 3 : Av = λv x = 4y 4, x y = 3y x + 4y = x λ = : Av = λv x = y, x y = y Therefore the general solution of the homogeneous is y h = C e 3t 4 + C e t.

12 .. Fall 06 Exam Now, we find a particular solution y p using undetermined coefficients. Since q = e t, e t, we guess y p = Ae t, Be t. Plugging this into the system, we get Ae t = Ae t + 4Be t + e t Be t Dividing through by e t, we get = Ae t Be t + e t A = A + 4B + B = A B + In the first equation, the As cancel, so we can solve for B to get B =. From the second equation, we have A = 3B = 5 so Last, the solution is y p = Ae t Be t = 5 et et. y = y p + y h = = 5 et et + C e 3t 4 5 et + 4C e 3t + C e t et + C e 3t C e t + C e t 3 4. Let A = so that the system is y 5 = Ay. First, find the eigenvalues of A: 3 λ deta λi = det = 3 λ 5 λ + 5 λ. = λ + 8λ + 6 = λ + 4 so the only eigenvalue is λ = 4 repeated twice. Now for the eigenvectors; let v = x, y: Av = λv 3x + y = 4x 3x + y = 4y y = x v =,

13 .. Fall 06 Exam We will also need a generalized eigenvector w, which satisfies A λiw = v: A λiw = v x = y x + y = x y = Any w = x, y satisfying x + y = works in both these equations; let s use x =, y = 0 so that w =, 0. Now, applying the formula from Theorem.68 from the lecture notes which should be on your index card, we see that the solution has the form y = C e λt v + C [ e λt w + te λt v ] + C [ = C e 4t e 4t + te 4t 0 C e = 4t + C e 4t + C te 4t C e 4t C te 4t C + C = e 4t + C te 4t C e 4t C te 4t. Writing this coordinate-wise as requested, we have the solution xt = C + C e 4t + C te 4t yt = C e 4t C te 4t 4 5. Let A =. We start by finding the solution of the homogeneous 5 equation y = Ay. First, the eigenvalues of A: deta λi = det λ 4 5 λ ] = λ5 λ + 8 = λ 6λ + 3. Setting this equal to zero and solving with the quadratic formula, we get λ = 6 ± = 6 ± 6 = 6 ± 4i = 3 ± i 3

14 .. Fall 06 Exam so the eigenvalues are λ = 3 + i and λ = 3 i. Now for the eigenvector corresponding to one of the eigenvalues let v = x, y: λ = 3 + i : Av = λv x 4y = 3 + ix x + 5y = 3 + iy 4y = + ix y = + ix v =, + i =, + i0, We have α = 3, β =, a =, and b = 0,. So applying the formula from Theorem.67 of the lecture notes which should be on your index card, we obtain the solution [ y = C e αt cosβta e αt sinβtb ] [ + C e αt cosβtb + e αt sinβta ] = C [e 3t cos t = e 3t sin t 0 ] + C [ C e 3t cos t C e 3t sin t C + C e 3t cos t + C C e 3t sin t. e 3t cos t 0 + e 3t sin t ] Now we plug in the initial condition y0 = 3, to find C and C : plugging in, we see 3 = C e 0 cos 0 C e 0 sin 0 = C + C e 0 cos 0 + C C e 0 sin 0 3 = C = C + C C = 3, C =. Therefore the particular solution is y = = 3 e 3t cos t e 3t sin t 3 + e3t cos t + 3 e3t sin t 3e 3t cos t e 3t sin t e 3t cos t + e 3t. sin t 4

15 .3. Fall 06 Exam 3.3 Fall 06 Exam 3. Suppose you are given a fourth-order, linear ODE. What is meant by an initial value of this ODE?. Convert this third-order ODE to a first-order system y = Ay + q. e t y 6ty + 4y y = 7 sin 3t 3. Find the particular solution of this initial value problem: 4. Find the general solution of this ODE: 5. Find the general solution of this ODE: 6. Find the general solution of this ODE: y + 9y + 8y = 0. y0 = y 0 = 7 y 4 4y y = 0 y 4y y = 48e 6t y 0y + 34y = 0 7. A 4 kg mass is attached to a fixed point by a spring whose spring constant is 40 N/m. The mass moves back and forth along a line, subject to friction where the damping coefficient is 4 N sec/m. Suppose also that initiallly, the mass is 3 m to the right of its equilibrium position, and moving to the right at m/sec. a Suppose the mass is not subject to any external force. Find the position of the mass at time t. b Suppose that the mass is subject to an external force of 0 sin t newtons. Find the position of the mass at time t. 5

16 .3. Fall 06 Exam 3 Solutions. An initial value of a fourth-order ODE consists of values of y, y, y and y at the same value of t. In other words, y 0 = yt 0 y t 0 y t 0 y t 0. First, solve for y to get y = e t y 4e t y + 6te t y + 7e t sin 3t. Then let y = y, y, y ; then y = So by setting y = y y y y y y =, A = =. 0y + y + 0y 0y + 0y + y e t y 4e t y + 6te t y + 7 sin 3t e t 4e t 6te t e t 4e t 6te t the system becomes y = Ay + q, as desired. y + and q = 0 0 7e t sin 3t e t sin 3t 3. The characteristic equation is λ + 9λ + 8 = λ + 6λ + 3, which has roots 6 and 3, so the general solution is y = C e 6t + C e 3t. Differentiating, we get y = 6C e 6t 3C e 3t, so by plugging in the given initial conditions we get y0 = y 0 = 7 = C + C 7 = 6C 3C C = 4 3, C = 3 Therefore the particular solution is y = 4 3 e 6t + 3 e 3t. 4. The characteristic equation is l 4 4λ λ = λ λ 7, which has roots 0 and 7 both repeated twice so the general solution is y = C + C t + C 3 e 7t + C 4 te 7t. 5. The characteristic equation is λ 4λ = λ 6λ +, which has roots 6 and, so the solution of the homogeneous is y h = C e 6t + C e t. Now for a particular solution y p of the non-homogeneous equation. Since q = 48e 6t, we d ordinarily guess y p = Ae 6t, but since e 6t is part of the solution, 6

17 .3. Fall 06 Exam 3 of the homogeneous, we need to multiply by t and guess y p = Ate 6t. Thus by the Product Rule, y p = Ae 6t + 6Ate 6t and y p = Ae 6t + 36Ate 6t. Plugging in the original equation, we get y p 4y p y p = 48e 6t Ae 6t + 36Ate 6t 4Ae 6t + 6Ate 6t Ate 6t = 48e 6t 8A = 48 A = 6 Therefore y p = 6te 6t, so the general solution is y = y p + y h, i.e. y = 6te 6t + C e 6t + C e t. 6. The characteristic equation is λ 0λ + 34 = 0 which has solutions λ = 0 ± = 0 ± 36 = 0 ± 6i Therefore the general solution is y = C e 5t cos 3t + C e 5t sin 3t. = 5 ± 3i. 7. Throughout this problem, let x = xt be the position of the mass at time t. From the oscillator equation, we obtain the second-order ODE mx + bx + kx = F ext t 4x + 4x + 40x = F ext t Also, throughout the problem, we have the initial value x0 = 3, x 0 =. a In this part, assume F ext t = 0. Then the characteristic equation is 4λ + 4λ + 40λ = 4λ + 6λ + 0 which has solutions λ = 6 ± 6 40 = 6 ± 4 = 6 ± i = 3 ± i. Therefore the general solution of this ODE is x = C e 3t cos t+c e 3t sin t. Since x0 = 3, we know C = 3. Differentiating, we get x = 3C e 3t cos t C e 3t sin t 3C e 3t sin t + C e 3t cos t and since x 0 =, we get 3C + C =, i.e. C =. Therefore the particular solution of this ODE is xt = 3e 3t cos t + e 3t sin t. 7

18 .3. Fall 06 Exam 3 b In this part, assume F ext t = 0 sin t. From part a, the solution of the homogeneous is x h = C e 3t cos t + C e 3t sin t. Now we need to find the x p. Since q = 0 sin t, guess x p = A sin t + B cos t. Differentiating, we get x p = A cos t B sin t and x p = 4A sin t 4B cos t, and by plugging in to the original equation we get 4x p + 4x p + 40x p = 0 sin t 4 4A sin t 4B cos t + 4A cos t B sin t + 40A sin t + B cos t = 0 sin t 4A 48B sin t + 4B + 48A cos t = 0 sin t Therefore we get the system of equations 4A 48B = 0 4B + 48A = 0 B = A A = 6, B = 3 so the particular solution is x p = sin t cos t. That makes the solution of the ODE x = x p + x h, i.e. 6 3 xt = 6 sin t 3 cos t + C e 3t cos t + C e 3t sin t. Since x0 = 3, we have C 3 obtain = 3, i.e. C = 0. Differentiating, we 3 x t = 3 cos t+ 3 sin t 3C e 3t cos t C e 3t sin t 3C e 3t sin t+c e 3t cos t and since x 0 =, we get 3C 3 + C =, i.e. C = 35. Therefore the 3 particular solution is xt = 6 sin t 3 cos t e 3t cos t e 3t sin t. 8

19 .4. Fall 06 Final Exam.4 Fall 06 Final Exam. a Explain the difference between the terms general solution and particular solution, in the context of ODEs. b What is Euler s formula? Why is this formula important, in the context of ODEs? c An evil professor tells a student to solve a 3 3 system of second-order, linear, homogeneous ODEs by hand. After hours of work, the student produces the following answer: y = C e t 3 + C e t 3 + C 3 e 7t + C 4 e t cost + C 5 e t sint. Despite not having done the problem himself, the professor knows this answer is wrong, just by looking at it. Why?. Consider the initial value problem y = y + t, x y y0 =,. Suppose you wanted to estimate y00 using Euler s method with 0 steps. Compute the first two points other than the given initial condition obtained by this method. 3. Here is a picture of the phase plane of a first-order system y = Φy: a How many stable equilibria does this system have? b How many unstable equilibria does this system have? c Give the equation of any constant solution of the system. 9

20 .4. Fall 06 Final Exam d Let yt = xt, yt be the solution to this system satisfying y0 = 4,. i. Which statement best describes the behavior of the function xt? A. xt is increasing for all t. B. xt is decreasing for all t. C. xt is increasing for small t, but decreasing for large t. D. xt is decreasing for small t, but increasing for large t. ii. Which statement best describes the behavior of the function yt? A. yt is increasing for all t. B. yt is decreasing for all t. C. yt is increasing for small t, but decreasing for large t. D. yt is decreasing for small t, but increasing for large t. iii. Find lim t xt. iv. Find lim t yt. 4. Find and classify all equilibria of the system x = y x y = y y Consider the parameterized family of ODEs y = φy; r, where φy; r = y r. Find the locations of any bifurcations occurring in this family, classify the bifurcations, and sketch the bifurcation diagram for the family. 6. a Find the particular solution of this initial value problem: y = y+ t+ y = 3 Write your answer as a function y = ft. b Find the general solution of this ODE: dy dt = ty 7. a Find the general solution of this ODE: ty + y = 4t 0

21 .4. Fall 06 Final Exam b Find the particular solution of this initial value problem: y 9y y = 0 y0 = 9 y 0 = 8 8. a Find the general solution of this ODE: y + y 0y = 56e t b Find the general solution of this system: x = 3y y = x y 9. Find the particular solution of this initial value problem: 0. Find the general solution of this ODE: x = 6x y y = x + 4y y0 = 3, y = e y y Write your answer as a function y = ft.. Find the particular solution of this initial value problem: x = 5x + y y = 5x y y0 =,. A 40 L tank contains fresh water initially. A saline solution containing.0 kg of salt per liter is pumped into the tank at a rate of 4 L/min. At the same time, the tank drains through a pipe which removes solution from the tank at a rate of 4 L/min. Assuming the tank is kept well-stirred, how much salt is in the tank 3 minutes after this procedure starts? 3. An RLC series circuit consists of a Ω resistor, a 4 H inductor, and a F capacitor. Assume that at time 0, the charge across the resistor is 3 coulombs, 5 and the current running through the system is 3 amperes. If an external power supply of 0 cos t V is applied to the circuit, find the charge in the circuit at time t.

22 .4. Fall 06 Final Exam Solutions. a The general solution of an ODE or system of ODEs is a description of all its solutions this description has arbitrary constants in it. Given an initial value problem, you can plug in the initial conditions to the general solution and solve for the constants, obtaining a particular solution of the IVP which has no constants in it. b Euler s formula says that for any complex number t, e it = cos t + i sin t. This formula is important in solving systems of ODEs and higher-order ODEs because it tells you how to rewrite solutions obtained from complex eigenvalues and eigenvectors in terms of cosines and sines. c If you reduce the order of a 3 3 second-order system, you will get a 6 6 first-order linear, homogeneous system because 6 = 3 The solution of any 6 6 first-order linear, homogeneous system is a 6-dimensional subspace, so it has to have six arbitrary constants in it. The student s answer only has five arbitrary constants, so it has to be wrong.. First, let t = tn t 0 = 00 0 = 5. Next, to establish notation, let the system n 0 be y = Φt, y = φ t, x, y, φ t, x, y. We are given t 0, y 0 = 0,,. Therefore t = t 0 + t = = 5 and x y = x 0 + φ 0,, t = = + 0 = = y 0 + φ 0,, t = + 5 = 5 = 3 Next, t = t + t = = 0 and x = x + φ 0,, t = = + 35 = 46 y = y + φ 0,, t = = = 67 t, y = 5,, 3 t, y = 0, 46, a The system has one stable equilibrium at 7, 5. b The system has two unstable equilibria at 8, 5 and 0, 6. c The equilibria are constant solutions, so any of these three answers are valid: x = 7 y = 5 x = 8 y = 5 x = 0 y = 6 There are other ways to write this; for example, y = 7, 5, etc.

23 .4. Fall 06 Final Exam d Let yt = xt, yt be the solution to this system satisfying y0 = 4,. i. From the phase plane, xt decreases then increases. The answer is D. ii. From the phase plane, yt is always decreasing. The answer is B. iii. From the phase plane, lim xt = 8. t iv. From the phase plane, lim yt = 5. t 4. Thinking of the system as y = Φy, we set Φy = 0 and solve for x and y: 0 = y x x = 0 = y y 80 y 0 = y 0y + 8 From the second equation, y = 0 or y = 8. From the first equation, the corresponding x-values are 5 and 4, so the two equilibria are 5, 0 and 4, 8. To classify the equilibria, compute the total derivative: DΦ = φ x φ x φ y φ y = 0 y Since this matrix is upper triangular, its eigenvalues are and y. That means that for the equilibrium 5, 0, the eigenvalues of DΦ5, 0 are and 8. Since there is one positive and one negative eigenvalue, 5, 0 is an unstable saddle. For the equilibrium 4, 8, the eigenvalues of DΦ 4, 8 are and 8. Since both eigenvalues are negative and real, 4, 8 is a stable node. 5. We start by finding the equilibria of the system in terms of r: set φy; r = 0 and solve for y to get 0 = y r 0 = y ry + r y = r, y = r. Next, classify these equilibria using the sign of φ. φ y = y, so we conclude φ r = r y = r is φ r = r y = r is stable if r < 0 semistable if r = 0 unstable if r > 0 unstable if r < 0 semistable if r = 0 stable if r > 0 3

24 .4. Fall 06 Final Exam We can now sketch the bifurcation diagram: That means there is a transcritical bifurcation at r = 0 because the two equilibria cross and change behavior. 6. a Separate the variables and integrate both sides: dy dt = y + t + y + dy = t + dt y + dy = t + dt lny + = lnt + + C y + = e lnt++c = Ct + y = Ct + Now plug in the initial condition y = 3 to get 3 = C + and solve for C to get C = 4. Thus the particular solution is y = 4 t +, 3 3 i.e. y = 4 3 t

25 .4. Fall 06 Final Exam b Separate the variables and integrate both sides: dy dt = ty dy = t dt y dy = t dt y y = 3 t3/ + C. If you solved for y, you d get y = 3 t3/ + C = 9 t3 + 3 Ct t + C. 7. a First, divide through by t to write the equation as y + y = 4t. Then, t compute the integrating factor: µt = e t 0 p 0s ds = e Multiply through by µ to obtain the equation t d dt yµ = 4tt d yt = 4t 3 dt yt = t 4 + C 0 s ds = e ln t = t. y = t + Ct. b The characteristic equation is 0 = λ 9λ = λ λ + which has roots λ = and λ =. Therefore the general solution is y = C e t + C e t. To find the particular solution, differentiate to get y = C e t C e t and plug in the initial values to get 9 = C + C 8 = C C C =, C = 7 Therefore the particular solution is y = e t + 7e t. 8. a The characteristic equation is λ 3 + λ 0λ = λλ + 5λ 4 which has roots λ = 0, λ = 4 and λ = 5. Therefore the general solution of the homogeneous is y = C + C e 4t + C 3 e 5t. Since q = 56e t, guess y p = Ae t and plug in the original equation to get 8Ae t + 4Ae t 40Ae t = 56e t 8A = 56 A =. Therefore y p = e t so the general solution is y = y p + y h, i.e. y = e t + C + C e 4t + C 3 e 5t. 5

26 .4. Fall 06 Final Exam b Thinking of the system as y = Ay, start with the eigenvalues of A: deta λi = 0 λ λ 3 = λ +λ+6 = λ+3λ λ = 3, λ = Next, eigenvectors. Let v = x, y and solve Av = λv for each eigenvalue: 3y = 3x λ = 3 : y = x v =, x y = 3y 3y = x λ = : 3y = x v = 3, x y = y Thus the general solution is y = C e 3t + C e t 3 Written coordinate-wise not required, this is xt = C e 3t + 3C e t yt = C e 3t + C e t 9. Thinking of the system as y = Ay, start by finding eigenvalues of A: deta λi = 6 λ4 λ + = λ 0λ + 5 = λ 5 so the only eigenvalue of A is λ = 5. Next, find eigenvectors: let v = x, y and set Av = λv to get 6x y = 5x x = y v =, x + 4y = 5y Now, find a generalized eigenvector w = x, y by solving A λiw = v: x y = w =, 0 x y = Now, our theorem on repeated eigenvalues tells us that the general solution is [ ] + C y = C e 5t e 5t 0 Coordinate-wise, this is x = C + C e 5t + C te 5t y = C e 5t + C te 5t + te 5t To find the particular solution, plug in the initial condition to get 3 = C + C and = C. Therefore C = 4 so the particular solution is x = 3e 5t + 4te 5t y = e 5t + 4te 5t... 6

27 .4. Fall 06 Final Exam 0. This is a second-order, non-linear equation with no t in it. Think of y as the independent variable and let v = y = dy. Then dt y = v dv so the equation dy becomes v dv dy = e y v. This equation can be solved by separating variables and integrating both sides: dv = e y dy v = e y + C. Now back-substitute for v to get the equation dy dt = e y + C = e + C = Cey. y e y This equation is separable and can be rewritten as e y dy = dt; Ce y integrate both sides you need the u-substitution u = Ce y on the left-hand side to get C lncey = t + D. Solve for y to get [ ] y = ln C ect+d +, i.e. y = lne Ct+D + ln C.. Think of the system y = Ay + q and start by finding the eigenvalues of A: deta λi = 5 λ λ+0 = λ 4λ+5 λ = 4 ± 4 45 Solve for the eigenvector v = x, y corresponding to + i: 5x + y = + ix y = 3+ix v = = 5x y = + iy 3 + i 3 = 4 ± 4 We have α =, β =, a =, 3 and b = 0,, so by the theorem governing solutions with complex eigenvalues we have [ ] 0 0 y = C [e t cos t e t sin t ]+C 3 e t cos t + e t sin t. 3 To find the particular solution, plug in the initial condition to get = C = 3C + C C =, C = 5 +i 0 7. = ±i

28 .4. Fall 06 Final Exam Therefore the particular solution, written coordinate-wise, is xt = e t cos t + 5e t sin t yt = e t cos t 8e t. sin t. Let yt be the amount of salt in the tank at time t. The rate at which salt enters the tank is.0 kg/l 4 L/min =.08 = kg/min, and the rate at 5 which salt leaves the tank is y/40 kg/l 4 L/min = y kg/min. This leads 0 to the initial value problem y = y0 = 0 y 5 0 y0 = 0 because the water is initially fresh. To solve the ODE, one can use integrating factors or undetermined coefficients. Using undetermined coefficients, the corresponding homogeneous equation is y = y which has 0 solution y h = Ce t/0. Since q =, guess y 5 p = A and plug in to obtain 0 = A. Solve for A to get A = 4, so y p = 4 and the general solution is 5 therefore y = y p + y h = Ce t/ Solve for C using the initial condition y0 = 0 to get the particular solution y = 4 5 e t/ Therefore at time 3, the amount of salt in the tank is 5 3. The RLC series circuit equation is y3 = 4 5 e 3/ Lq t + Rq t + C qt = E St where q is the charge. In this problem, the above equation leads to the initial value problem 4q + q + 5q = 0 cos t q0 = 3 q 0 = 3 To solve the IVP, start with the characteristic equation 0 = 4λ + λ + 5 = 4λ + 3λ + 5 which has roots 4 λ = 3 ± 9 5 = 3 ± i Therefore the general solution of the homogeneous is q h = C e 3t/ cos t + C e 3t/ sin t. 8

29 .4. Fall 06 Final Exam To find q p, guess q p = A cos t + B sin t. Then q p = A sin t + B cos t and q p = A cos t B sin t. Plugging in the original equation, we get 4 4 A 4 4 cos t B 4 sin t A + sin t + B cos t + 5 A cos t + B sin t = 0 cos t 4A + 6B cos t + 4B 6A sin t = 0 cos t Therefore 4A + 6B = 0 and 4B 6A = 0; solving for A and B we get B = and A = 4. Therefore y p = 4 cos t + sin t so the general solution is q = q p + q h = 4 cos t + sin t + C e 3t/ cos t + C e 3t/ sin t. To find C and C, differentiate to get q = sin t + cos t 3 C e 3t/ cos t C e 3t/ sin t 3 C e 3t/ sin t+c e 3t/ cos t; then plug in the initial conditions to get 3 = 4 + C 3 = 3 C + C C = 9, C = 8 Therefore the particular solution is qt = q p + q h, i.e. qt = 4 cos t + sin t + 9e 3t/ cos t + 8e 3t/ sin t. 9

30 Chapter 3 Exams from Fall Fall 07 Exam. Briefly explain the difference between the terms general solution and particular solution in the context of ordinary differential equations.. Sketch the phase line for the autonomous ODE y = 5y y Here is the picture of the slope field associated to some first-order ODE y = φt, y: a Write the equation of any one solution of this ODE. b Suppose y0 = 5. Find lim t yt. c Suppose y =. Estimate y. d On the picture above, sketch the graph of the solution satisfying the initial condition y3 = 4. 30

31 3.. Fall 07 Exam e Let y = ht be the solution of the initial value problem y = φt, y y0 = Suppose you used Euler s method to estimate h using 3 steps. What are the coordinates of the point you would you obtain as t, y? 4. Find the general solution of the following ODE: dy dt = 5y + 0e5t Write your answer as a function y = ft. 5. Find the particular solution of the following initial value problem: dy = dt ty t + y0 = 4 Write your answer as a function y = ft. 6. Find the particular solution of the following initial value problem: y = y t te t y = 0 7. Find the general solution of the following ODE: y = cos t y cos y t t sin y 3

32 3.. Fall 07 Exam Solutions. Given an ODE, the set of all solutions of that ODE is called the general solution of the ODE. This solution will have one or more arbitrary constants in it. If you are given an initial value, you can plug that initial value into the general solution, solving for the constant in the general solution. This produces a solution of the ODE with no arbitrary constants, which is called a particular solution of the ODE.. φy = 5y y 3 ; set φy = 0 and factor to solve for y: this gives 0 = y 5 y so the two equilibria are y = 0 and y = 5. To classify them, compute derivatives of φ and plug in the equilibria. When y = 5, φ 5 = < 0 so 5 is stable; when y = 0; φ 0 = 0 but φ 0 = 0 0 so 0 is semistable. Therefore the phase line looks like this: Here is the picture of the slope field associated to some first-order ODE y = φt, y: a y = 3 and y = are solutions. b If y0 = 5, then lim t yt =. c If y =, then y. d Here is the graph: e Notice that φ0, is the slope of the vector field at the point 0,, which is 0. So if you used Euler s method with 3 steps, t = tn t 0 = 0 = 4 n 3 so t = t 0 + t = = 3 y = y 0 + φt 0, y 0 t = + φ0, 4 = = and therefore t, y = 4,. 4. Solution # : Rewrite in standard form as y 5y = 0e 5t ; then the integrating factor is µ = exp 5 dt = e 5t. 3

33 3.. Fall 07 Exam After multiplying through by the integrating factor, the equation becomes Integrate both sides to get solve for y to get y = 0te 5t + Ce 5t. d ye 5t = 0e 5t e 5t = 0. dt ye 5t = 0t + C; Solution # : Rewrite in standard form as y 5y = 0e 5t and use undetermined coefficients. The corresponding homogeneous equation is y 5y = 0 which has solution y h = e 5t. Since the right-hand side of the ODE is 0e 5t, a normal guess for the particular solution would be y p = Ae 5t but since this is the same as y h up to a constant, you need to multiply the guess by t, i.e. y p = Ate 5t. Plugging in the equation, we get y 5y = 0e 5t Ate 5t 5Ate 5t = 0e 5t Ae 5t + 5Ate 5t 5Ate 5t = 0e 5t Ae 5t = 0e 5t A = 0 Therefore y p = 0te 5t, so the solution is y = y p + Cy h = 0te 5t + Ce 5t. 5. This equation is separable; divide both sides by y and multiply both sides by dt to obtain y dy = t t + dt. Then integrate both sides you need the u-sub u = t + on the right to get ln y = lnt + + C; solving for y by exponentiating both sides gives y = e lnt ++C = C t +. Now, plug in the initial condition t = 0, y = 4 to get 4 = C 0 +, i.e. C = 4. Thus the particular solution is y = 4 t This equation is first-order linear; rewrite it as y t y = te t. 33

34 3.. Fall 07 Exam From this point there are two methods of solution: Solution # : The integrating factor is µ = exp dt = exp ln t = t = t t ; after multiplying through the equation by the integrating factor we get d y = e t. dt t Integrate both sides to get y t = e t + C. Now plug in the initial condition t =, y = 0 to get 0 = e + C, i.e. C = e. Thus the particular solution is y t = e t e. If you solved for y not required, this can be rewritten as y = te t et. Solution # : The corresponding homogeneous equation is y y = 0 which t has solution y h = exp t dt = e ln t = t. For the particular solution, guess y p = Ate t +Be t ; plugging in the equation we get y t y = e t Ate t + Be t t Ate t + Be t = e t Ae t Ate t Be t Ae t + B t e t = e t Ate t Be t + B t e t = e t Therefore A = so A = and B = 0, so y p = te t. That makes the general solution y = y p + Cy h = te t + Ct. Plugging in the initial condition as in Solution # gives the same particular solution: C = e and therefore y = te t et. 34

35 3.. Fall 07 Exam 7. Rewrite the equation as y + cos y cos t + t t sin yy = 0. Now let M = y + cos y cos t and let N = t t sin y. We see that M y = N t = sin y so the equation is exact. Now find ψ by integrating: ψ = M dt = y + cos y cos t dt = yt + t cos y sin t + Ay ψ = N dy = t t sin y dy = ty + t cos y + Bt To reconcile these answers, set Bt = sin t and Ay = 0 so that ψt, y = ty+t cos y sin t. The general solution is therefore ψt, y = C, i.e. ty+t cos y sin t = C. 35

36 3.. Fall 07 Exam 3. Fall 07 Exam. Let yt = xt, yt be the solution of the initial value problem y = x y + t, y 3x y0 =,. Suppose you wanted to estimate y80 using Euler s method with 40 steps. Compute the points y and y that would be obtained by this method.. Here is the phase plane of a first-order, autonomous system of ODEs y = Φy: Use this picture to answer the following questions: a Find all saddles of this system if there are no saddles, say so. b Find all nodes of this system if there are no nodes, say so. c Find all centers of this system if there are no centers, say so. d Is the trace of DΦ3, positive, negative or zero? e Is the determinant of DΦ3, positive, negative or zero? f Suppose y0 = 5, 6. In this situation, which statement best describes the behavior of xt? A. xt increases for all t B. xt decreases for all t C. initially, xt is increasing, but then it becomes decreasing D. initially, xt is decreasing, but then it becomes increasing g Suppose y0 = 5, 6. In this situation, which statement best describes the behavior of yt? A. yt increases for all t 36

37 3.. Fall 07 Exam B. yt decreases for all t C. initially, yt is increasing, but then it becomes decreasing D. initially, yt is decreasing, but then it becomes increasing 3. Find the particular solution of the following initial value problem: x = 7x + 5y y = x 3y x0 = y0 = 4 4. Find the general solution of the following system of ODEs: x = 5x 4y + 3e t y = 4x 5y + 4e t 5. Find the particular solution of the following initial value problem: y = 7x + y, 5x + 3y y0 =, 5 Write your final answer coordinate-wise. 37

38 3.. Fall 07 Exam Solutions. First, t = tn t 0 = 80 0 =. Now, t n 40 0, y 0 = 0,, and Φy 0 = + 0, 3 = 0, 5 so by the Euler s method formula, t = t 0 + t = 0 + = y = y 0 + Φt 0, y 0 t =, + 0, 5 =, 9. Now, Φt, y = 9 +, 9 3 =, 5 so t = t + t = + = 4 y = y + Φt, y t =, 9 +, 5 = 46, 39.. a The only saddle is at 3, 3. b This system has no nodes. c The only center is at 3,. d Since 3, is a center, trdφ3, = 0. e Since 3, is a center, det DΦ3, is positive. f B. Since the graph of the solution always moves to the left in the direction of increasing t, xt decreases for all t. g C. In the direction of increasing t, the graph of the solution initially goes upwards, then downwards Let A = so that the system is y 3 = Ay. First, find the eigenvalues of A: 7 λ 5 deta λi = det = 7 λ 3 λ 5 3 λ = λ + 0λ + 6 = λ + 8λ + so the eigenvalues are λ = 8 and λ =. Now for the eigenvectors; let v = x, y: 7x + 5y = 8x λ = 8 : Av = λv x 3y = 8y 7x + 5y = x λ = : Av = λv x 3y = y x = 5y 5, x = y, Therefore the general solution is y = C e 8t 5 + C e t. 38

39 3.. Fall 07 Exam To find the particular solution, plug in the initial value y0 =, 4 to get = 5C + C 6C 4 = C + C = C = 3, C = 3. Therefore the particular solution is y = 5 3 e 8t + 5 = 3 e 8t + 3 e t 3 e 8t +. 3 e t 3 e t Let A = so that the corresponding homogeneous system is y 4 5 = Ay. First, find the eigenvalues of A: 5 λ 4 deta λi = det = 5 λ 5 λ λ = λ 9 = λ + 3λ 3 so the eigenvalues are λ = 3 and λ = 3. Now for the eigenvectors; let v = x, y: 5x 4y = 3x λ = 3 : Av = λv y = x, 4x 5y = 3y 5x 4y = 3x λ = 3 : Av = λv x = y, 4x 5y = 3y Therefore the general solution of the homogeneous is y h = C e 3t + C e 3t. Now, we find a particular solution y p using undetermined coefficients. Since q = 3e t, 4e t, we guess y p = Ae t, Be t. Plugging this into the system, we get Ae t = 5Ae t 4Be t + 3e t A = 5A 4B + 3 Be t = 4Ae t 5Be t + 4e t B = 4A 5B + 4 From the second equation 4B = 4A + 4, i.e. B = A +. Plugging this into the first equation gives A = 5A 4A + + 3, i.e. A = A so A = and B = 3. Therefore Ae t yp = = e t. Be t 3 e t 39

40 3.. Fall 07 Exam Last, the solution is 5. Let A = of A: y = y p + y h = = 7 0 e t 3 0 e t + C e 3t e t + C e 3t + C e 3t 3 e t + C e 3t + C e 3t + C e 3t. so that the system is y = Ay. First, find the eigenvalues 7 λ deta λi = det = 7 λ3 λ λ By the quadratic formula, the eigenvalues are λ = 4 ± 6 49 = 4 ± 00 = λ + 4λ + 9. = 4 ± 0i so α = and β = 5. Next, find the eigenvectors; let v = x, y: λ = + 5i : Av = λv 7x + y = + 5ix 5x + 3y = + 5iy y = 5 + 5ix = ± 5i, 5 + 5i =, 5 + i0, 5. Therefore a =, 5 and b = 0, 5. By the theorem from the lecture notes, the general solution is therefore ] [ y = C [e αt cosβta e αt sinβtb + C = C [e t cos 5t = 5 e t sin 5t 0 5 C e t cos 5t + C e t sin 5t 5C + 5C e t cos 5t + 5C 5C e t sin 5t ] e αt cosβtb + e αt sinβta ] [ + C e t 0 cos 5t 5 + e t sin 5t Now for the particular solution. Plugging in t = 0, y =, 5, we get. 5 ] = C 5 = 5C + 5C C =, C = 3. 40

41 3.. Fall 07 Exam Therefore the particular solution is y = Written coordinate-wise, this is e t cos 5t 3e t sin 5t 5e t cos 5t 0e t sin 5t xt = e t cos 5t 3e t sin 5t yt = 5e t cos 5t 0e t sin 5t.. 4

42 3.3. Fall 07 Final Exam 3.3 Fall 07 Final Exam. a Write down an example of a matrix A such that the constantcoefficient system y = Ay has a stable node at the origin. b Consider the third-order differential equation y + y 5y = e t. Convert this equation to a first-order system of the form y = Ay + q, clearly defining what y, A and q are. c Let y = φt, y be a first-order, linear ODE and suppose that y t = t + t and y t = t + t 3 are particular solutions of y = φt, y. Find the general solution of the ODE y = φt, y. d Find lim t ht, where y = ht is the solution of the IVP y = y y 9 y0 = 5.. Consider the initial value problem dy dt = t y+t y =. Estimate y0 by using Euler s method with 3 steps. 3. Here is a picture of the phase plane of some first-order system y = Φy: Use this picture to answer the following questions: a Give the location of all equilibria of the system. Classify each equilibrium. b Let y = xt, yt be the solution of this system satisfying y0 = 6,. 4

43 3.3. Fall 07 Final Exam i. Find lim t xt. ii. Find lim t yt. iii. Find lim t xt. c Let y = ft, gt be the solution of this system satisfying f0 = and g0 = 6. i. Is f 0 positive, negative or zero? ii. Is g 0 positive, negative or zero? 4. Find and classify all equilibria of this system: x = x y y = y 3y 5. Find the particular solution of this initial value problem: y = 0 6y + y4 = 3 6. Find the general solution of this ODE writing your answer as a function y = ft. dy + 6ty = t dt 7. Find the particular solution of this initial value problem: 8. Find the general solution of this ODE: y = 6y y0 = 3 y 0 = 5 y 7y 8y = 54t Find the particular solution of this initial value problem: y = 3x + 4y, x 7y y0 = 7, 0. Find the general solution of this system, and write your answer coordinatewise: x = 7x + 5y y = 5x + 3y 43

44 3.3. Fall 07 Final Exam. A 4 kg mass is attached to the end of a spring with spring constant N/cm. Assume that the damping coefficient is 6 N sec/cm and the entire system is subject at time t to an external force of cos t + 6 sin t. If at time t = 0, the mass is moving with initial velocity 4 cm/sec and has position, find the position of the mass at time t = π.. Two large tanks each hold some volume of liquid. Tank X holds 00 L of sulfuric acid solution which is initially 4% hydrochloric acid; tank Y holds 50 L of solution which is initially % hydrochloric acid. Pure water flows into tank Y at a rate of 3 L/min and pure water flows into tank X at a rate of L/min. Tank Y drains into tank X at a rate of L/min, and drains out of the system at a rate of L/min. Tank X drains out of the system at a rate of 3 L/min, and does not drain into tank Y. Assuming that at all times the solution in each tank is kept mixed, find the concentration of hydrochloric acid in tank X at time 5. 44

45 3.3. Fall 07 Final Exam Solutions. a Any matrix with two negative, real eigenvalues works: the easiest thing is to write down any triangular or diagonal matrix with negative 0 numbers along the diagonal, like for example A =. 0 y b Rewrite the equation as y = y +5y +e t. Then, let y = y, let A = y as desired. and let q = 0 0 e t. The equation becomes y = Ay + q c From the theory of first-order linear equations, we know that the difference of any two solutions of a linear ODE solves the corresponding homogeneous equation. So y t y t = t + t t 3 is a solution of the corresponding homogeneous equation. Furthermore, we know that for a homogeneous first-order linear ODE, the solution set is the span i.e. set of multiples of any one nonzero solution, so the homogeneous equation has as its solution set y h = Ct + t t 3. The solution set of the original equation is therefore any one particular solution of the equation like either y or y plus the solution set of the homogeneous, i.e. the general solution is y = t + t + Ct + t t 3. d Let φy = y y 9. Clearly φy = 0 when y = or y = 9. Then φ y = y, so φ = 7 < 0 and φ 9 = 7 > 0. Thus is a stable equilibrium and 9 is an unstable equilibrium of this equation. As t, ht will approach the stable equilibrium, so lim t ht =.. Let φt, y = t so that the equation becomes y+t y = φt, y. Next, t = tn t 0 0 = 3. We are given t 3 0, y 0 =,. Therefore φ3, 0 = = 0 so + t = t 0 + t = + 3 = 4 y = y 0 + φt 0, y 0 t = + 03 = t, y = 4,. Next, φt, y = 4 4+ = so n = t = t + t = = 7 y = y + φt, y t = + 3 = 7 t, y = 7, 7. 45

46 3.3. Fall 07 Final Exam Last, φt, y = 7 = = 4 7 so t3 = t + t = = 0 y 3 = y + φt, y t = = = 73 4 t 3, y 3 = 0, So y a, 4 is a stable node; 7, 8 is a saddle. b i. lim xt =. t ii. lim yt = 4. t iii. lim xt =. t c i. f 0 = dx > 0 since the graph of y is moving to the right at dt x=,y=6 t = 0. ii. g 0 = dy < 0 since the graph of y is moving downward at dt x=,y=6 t = Let Φx, y = x y, y 3y. Setting Φx, y = 0, we obtain 0 = x y 0 = y 3y From the second equation which factors as 0 = yy 3, either y = 0 or y = 3. When y = 0, from the first equation x =, so x = ±. When y = 3, from the first equation x = 4, so x = ±. We therefore have four equilibria, which we classify by looking at the eigenvalues of the total derivative DΦx, y = x : 0 y 3, 0: DΦ, 0 = saddle., 0: DΦ, 0 =, 0 a stable node., 3: DΦ, 3 = unstable node., 3: DΦ, 3 =, 3 a saddle has eigenvalues and 3, making, 0 a has eigenvalues and 3, making has eigenvalues 4 and 3, making, 3 an has eigenvalues 4 and 3, making 46

47 3.3. Fall 07 Final Exam 5. This equation is separable: rewrite it as 6y + dy = 0 dt. Then integrate both sides to get y 3 + y = 0t + C. Plug in the initial condition 4, 3 to get = 04 + C, i.e. C = 7. Thus the particular solution is y 3 + y = 0t Multiply through by the integrating factor µ = exp 6t dt = exp3t to obtain the equation d dt ye 3t = te 3t. Now integrate both sides you need the u-substitution u = 3t, du = 6t dt on the right-hand side to get ye 3t = e 3t + C. Solving for y, we obtain y = + Ce 3t. 7. Rewrite the equation as y + 6y = 0 and solve the characteristic equation λ + 6 = 0 to obtain λ = ±4i. Thus the general solution is y = C cos 4t + C sin 4t. To find the particular solution, differentiate to obtain y = 4C sin 4t + 4C cos 4t. Plugging in y0 = 3 and y 0 = 5 yields 3 = C 5 = 4C C = 3, C = 5 4. Thus the particular solution is y = 3 cos 4t + 5 sin 4t First, solve the homogeneous equation y 7y 8y = 0 by considering the characteristic equation 0 = λ 7λ 8 = λ 9λ +. This equation has solutions λ = 9, λ = so the solution of the homogeneous is y h = C e 9t + C e t. Now find a particular solution using undetermined coefficients; guess y p = At + B and plug into the equation to obtain 0 7A 8At + B = 54t + 5, i.e. 8At + 7A 8B = 54t + 5. Thus 8A = 54, so A = 3. Then 73 8B = 5 so 8B = 36 so B =. Thus y p = 3t so the general solution of the equation is 9. First, find the eigenvalues: y = y p + y h = 3t + C e 9t + C e t. 3 λ 4 det = 3 λ 7 λ + 4 = λ + 0λ + 5 = λ λ so the only eigenvalue is λ = 5 repeated twice. Next, the eigenvectors: 3x + 4y = 5x x 7y = 5y x = y v =,. 47

48 3.3. Fall 07 Final Exam Since there is only one linearly independent eigenvector, we find a generalized eigenvector w = x, y: x + 4y = w = v w =, x y = Now by the formula from the lecture notes, the general solution is [ ] y = C e 5t + C e 5t + te 5t 0 C + C = e 5t + C te 5t C e 5t C te 5t. Now for the particular solution. Plugging in the initial condition y0 = 7,, we get 7 = C + C = C which leads to C = and C = 3. Thus the particular solution is 7e y = 5t + 6te 5t e 5t 3te 5t. 0. First, find the eigenvalues: 7 λ 5 det = 7 λ3 λ + 5 = λ 6λ λ so by the quadratic formula, the eigenvalues are λ = 6 ± = 3 ± 00 = 3 ± 5i. Next, the eigenvector for 3 + 5i: 7x + 5y = 3 + 5ix 5y = 0+5ix 5y = +ix v = 5, +i = 5, +i0,. 5x + 3y = 3 + 5iy Now by the formula in the lecture notes, the general solution is ] [ y = C [e 3t 5 cos 5t e 3t 0 sin 5t + C e 3t 0 cos 5t = 5C e 3t cos 5t + 5C e 3t sin 5t C + C e 3t cos 5t + C + C e 3t sin 5t + e 3t sin 5t Written coordinate-wise, this is xt = 5C e 3t cos 5t + 5C e 3t sin 5t yt = C + C e 3t cos 5t + C + C e 3t sin 5t.. 5 ] 48

49 3.3. Fall 07 Final Exam. From the spring equation mx t + bx t + kxt = F ext t, we obtain, by plugging in the constants given in the problem, the following IVP: 4x + 6x + x = cos t + 6 sin t x0 = x 0 = 4 To solve this, start with the corresponding homogeneous equation 4x +6x + x = 0 which has characteristic equation 4λ + 6λ + = 0. This equation factors as λ + λ + = 0 so it has solutions λ = and λ =. Thus the general solution of the homogeneous is y h = C e t + C e t/. Next, find a particular solution with undetermined coefficients. Guess x p = A cos t + B sin t ; then x p = A sin t + B cos t and x p = A cos t B sin t. 4 4 Plugging in the original equation, we get 4 A 4 cos t B 4 sin t + 6 A sin t + B cos t + A cos t + B sin t = cos t + 6 sin t Therefore x p = 8 5 cos t sin t so A + 3B cos t + 3A + B sin t = cos t + 6 sin t A + 3B = 3A + B = 6 A = 8 5, B = 6 5. x = x p + x h = 8 5 cos t sin t + C e t + C e t/. Now find C and C by plugging in the initial conditions: plugging in x0 =, we get = 8 + C 5 + C so C + C = 8. Plugging in 5 x 0 = 4, we get 4 = 3 C 5 C, i.e. C C = 3 ; solving the two equations together 5 gives C =, C = 8 so the particular solution of the spring equation is 5 xt = 8 5 cos t sin t e t e t/. Finally, answer the question by plugging in x = π to obtain xπ = 8 5 cos π sin π e π e π/ = e π e π/.. Let xt and yt be the amount of hydrochloric acid in tanks X and Y, respectively, at time t. Let y = xt, yt; from the given information, we obtain the following IVP: y = y/50 3x/00, 3y/50 y0 = 4, 6 49

50 3.3. Fall 07 Final Exam Written in matrix form, this equation is y = Ay where A = Since this matrix is triangular, its eigenvalues are its diagonal entries: λ = 3, λ = 3. Now find eigenvectors: λ = 3 : x + y = y = 3 y y = 0, x 3 λ = 3: x + y = y = 3y y = 3 x 4y = 3x 4, x Therefore the general solution of this system is y = C e 3t/00 0 Plugging in the initial conditions, we get. + C e 3t/ = C + 4C 6 = 3C C =, C =. Thus the particular solution is y = e 3t/00 0 e 3t/50 4 3, which coordinate-wise is xt = e 3t/00 8e 3t/50 yt = 6e 3t/50. So the amount of hydrochloric acid in tank X at time 5 is x5 = e 3/4 8e 3/ ; the concentration is given by this amount divided by the volume of tank X, which gives e 3/4 8e 3/ 00 = 7e 3/4 e 3/. 5 50