Math 216 Second Midterm 28 March, 2013

Transcription

1 Math 26 Second Midterm 28 March, 23 This sample exam is provided to serve as one component of your studying for this exam in this course. Please note that it is not guaranteed to cover the material that will appear on your exam, nor to e of the same length or difficulty. In particular, the sections in the text that were covered on this exam may e slightly different from those covered y your exam. This material is (c26, University of Michigan Department of Mathematics, and released under a Creative Commons By-NC-SA 4. International License. It is explicitly not for distriution on wesites that share course materials.

2 Math 26 / Exam 2 (28 March, 23 page 2. [5 points] For this prolem note that the general solution to y + 5y + 4y is y c e t + c 2 e 4t. (Note that minimal partial credit will e given on this prolem. a. [7 points] Find a real-valued general solution to y + 5y + 4y 3e 4t. Solution: We know the general solution is y y c + y p. We use the Method of Undetermined Coefficients to find y p, guessing y p Ate 4t, after multiplying our first guess (y p Ae 4t y t ecause the forcing term is present in our homogeneous solution. Then y p2 Ae 4t 4Ate 4t and y p2 8Ae 4t + 4Ate 4t, so that on plugging in we get so that 3A 3, and A. Thus the general solution is ( 8A + 5Ae 4t 3e 4t, y c e t + c 2 e 4t te 4t. If we use Variation of Parameters, we have u e t + u 2 e 4t and u e t 4u 2 e 4t 3e 4t. Solving, we find u 2 and u e 3t, so that u 3 e 3t and u 2 t, and y p 3 e 4t te 4t.. [8 points] Find the solution to the y + 5y + 4y 6t, y( 2, y ( 2. Solution: We know the general solution is y y c + y p. We use the Method of Undetermined Coefficients to find y p, guessing y p A + Bt. Plugging in, 5B + 4A + 4Bt 6t, so that B 4 and A 5. Thus the general solution is y c e t + c 2 e 4t 5 + 4t. Applying the initial conditions, we have y( c + c 2 5 2, and y ( c 4c Thus c + c 2 7 and c 4c 2 6. Adding the two, we have 3c 2, so c 2 /3. Then the first gives c 22/3, and our solution is y 22 3 e t 3 e 4t 5 + 4t. We can, of course find y p with Variation of Parameters. Then u e t + u 2 e 4t and u e t 4u 2 e 4t 6t. Solving, we find u te4t, so that u 2 ( te4t and u 6 3 tet, so that u 6 3 ( + tet. Then y p 5 + 4t, as efore.

3 Math 26 / Exam 2 (28 March, 23 page [2 points] The eigenvalues of the matrix A are λ 4 ± 4i. Use the eigenvalue 5 7 method to find a real-valued general solution to the system x Ax. (Note that minimal partial credit will e given on this prolem. Solution: We know that a complex-valued solution to the system is x ve (4+4it, where v is the eigenvector corresponding to λ 4 + 4i, and that the real and imaginary parts of this solution will themselves e linearly independent solutions to the system. The eigenvector v T v v 2 is given y λ 5 3 4i 5 v v 5 7 λ 5 3 4i v 2. We know that the two rows of this algeraic system are equivalent, ecause we are at an eigenvalue. The first gives ( 3 4iv + 5v 2, so we may take v 5 and v i. Then a complex-valued solution to the system is ve (4+4it 5 e 4t (cos(4t + i sin(4t 3 + 4i ( 5 cos(4t + 5i sin(4t (3 cos(4t 4 sin(4t + i(4 cos(4t + 3 sin(4t e 4t. Separating the real and imaginary parts of this, we have the general solution ( ( 5 cos(4t x c e 4t 5 sin(4t + c 3 cos(4t 4 sin(4t 2 e 4t. 4 cos(4t + 3 sin(4t Similarly, if we use the second equation to find v, we have v 3 4i, and 5 3 cos(4t + 4 sin(4t x c e 4t 4 cos(4t + 3 sin(4t + c 5 cos(4t 2 e 4t. 5 sin(4t Using λ 4 4i, we get v as efore. 5 and, reversing the sign of c 3 4i 2, the same general solution

4 Math 26 / Exam 2 (28 March, 23 page 4 3. [6 points] Consider the system x x + 2 x 2 3x a. [8 points] Find a real-valued general solution to this system. Solution: Letting x T x, this is equivalent to x Ax with the coefficient 2 λ 2 matrix A. Eigenvalues of A are given y det λ 3 3 λ 2 λ 6 (λ 3(λ+2, so that eigenvalues are λ 2 and λ 3. Then eigenvectors v satisfy the equation λ 2 v, 3 λ so that if λ 2 we have v ( 2 3 T (, and if λ 3, v T. Thus the general solution is ( x 2 x c e 2t + c 3 2 e 3t 2c e + c 2 e 3t 3c e 2t + c 2 e 3t. This can also e solved y elimination: x 2 3x, so x 2 3x. Then the first equation ecomes 3 x 2 3 x 2 + 2, or x 2 x 2 6. With e rt we have r 2 r 6, so r 2 and 3. Then k e 2t + k 2 e 3t. With x 3 x 2, x 2 3 k e 2t + k 2 e 3t, which is the same as we found aove with k 3c and k 2 c 2.. [4 points] Find the particular solution if x (., (.35. Solution: We have x ( 2c + c 2. and ( 3c + c Sutracting the first from the second we get c.5. Then either equation gives c 2.2, and our particular solution is ( x. e 2t +.2 e 3t.5 e 2t +.2 e 3t. c. [4 points] Briefly explain why the direction field and solution trajectory shown to the right could not match this system and your solution from (. Solution: In the long run the negative exponentials in the solution we found in ( will decay to zero and therefore we expect the solution to look like x (.2e 3t.2e 3t T. Thus the trajectory should end up on the line y x ( x. This is not shown in the figure to the right. Also note that when x the system predicts that trajectories will e horizontal, and when trajectories will have slope 3, neither of which appear to e the case here x

5 Math 26 / Exam 2 (28 March, 23 page 5 4. [2 points] Three linear constant-coefficient homogeneous systems are descried elow. Included in the description is one of the following three charateristics; for each, specify the missing two characteristics y circling the correct answers.. whether it is a node, saddle point or spiral point, and 2. whether the equilirium point (, is asymptotically stale, stale or unstale; 3. the sign of the constant a in the system. No explanation is necessary for your answers. a. [4 points] The system x Ax having direction field and representative trajectories in ( the phase plane shown 2 in the figure to the right, if A and a >. a This is a node saddle point spiral point and is asymptotically stale stale unstale x Solution: From the figure we see that this is a spiral point. We know the eigenvalues of A are given y (λ 2 + 2a, so λ ± 2ai, and trajectories increase away from the equilirium point, indicating that it is unstale.. [4 points] The system x Ax if the equilirium point is a saddle point and the eigenvalues are λ 3 and λ a. The equilirium point is asymptotically stale stale unstale and a is < > Solution: a <. Saddle points are unstale, and have real eigenvalues with opposite signs, so c. [4 points] The system x 2x + a, x 2 x 2 whose solution with the initial conditions x (, ( is shown in the figure to the right, if the equilirium point is asymptotically stale. The equilirium point is a node saddle point spiral point and a is < > x, x 5 t Solution: The solution shown has no oscillatory characteristic, so this must( e a node 2 a with oth eigenvalues real and negative. Then the coefficient matrix A, 2 so that eigenvalues are given y det(a λi (λ a. For real eigenvalues, λ 2 ± a must have a > (and a < 4, for that matter.

6 Math 26 / Exam 2 (28 March, 23 page 6 5. [6 points] Consider a clown on a spring in a oat, as suggested y the figure to the right. At time t we place a large ox in the oat. Then, with some not entirely unreasonale assumptions, the displacement x of the oat and of the clown are given y x 425x x 2 5x 5. Spring Boat Clown Box Water Plan View Box (Here, x and are ( measured in meters and t in seconds. Letting A, the eigenvalues and eigenvectors of A are λ 6 and λ 25 with 3 v and v. 6 Side View a. [6 points] What are the natural frequencies at which the oat and clown will oscillate? Explain. x 3 Solution: Letting x and A, we guess x ve 2 2 rt, so that r 2 v Av, and therefore r 2 λ, the eigenvalues of A. Thus r ±i 6 or r ±i 25, and solutions will look like cosines and sines of 6 t and 25 t. Thus the frequencies are ω 6( 24 and ω 2 25(. (Note that if we were recently indoctrinated y some other field we might also talk aout the ordinary frequency, f ω/2π. x Water. [6 points] Find the general solution to the homogeneous system associated with this system. Solution: Given the eigenvalues and eigenvectors provided, we have ( 3 x c (c cos( 6 t + c 2 sin( 6 t + (c 3 cos( 25 t + c 4 sin( 25 t. 6 c. [4 points] A solution to this system is shown to the right. What initial conditions were applied to x and to otain this solution? Solution: The initial conditions were x (., and x ( ( x 2 (. We see that x starts at., at, and that each solution curve starts at with zero slope. x, x.4.6 t

7 Math 26 / Exam 2 (28 March, 23 page 7 6. [5 points] Consider a mass-spring system modeled y y + cy + ky f(t, where c is the damping coefficient associated with the system and k the spring constant. For each of the following give a short explanation of your answers. a. [5 points] If f(t 3 sin(2t and the system is at resonance, are c and k positive, negative or zero? Give specific values for c and k if possile. Solution: If the system experiences resonance, we know that c. To have resonance with a forcing that has a frequency 2, we know that the natural frequency of the system, which is ω k must e 2, so that k 4.. [5 points] If f(t 3 sin(2t and the system is at resonance, sketch a qualitatively accurate graph of y p, the particular solution to the prolem. Solution: Because the system is at resonance, we know that y A t cos(2t (we may omit the t sin(2t term ecause there are only even derivatives in the prolem. Thus the solution will e something like the following graph. y p t c. [5 points] If c >, k >, and f(t 3 sin(2t, what can you say (without solving the differential equation aout the long-term ehavior of y? Solution: If c > the homogeneous solutions will e decaying. Thus the long-term response will e y p, the particular solution, which will e y p A cos(2t + B sin(2t A 2 + B 2 cos(2t α for some A and B. That is, the long-term response will e purely sinusoidal with period π.

8 Math 26 / Exam 2 (28 March, 23 page 8 7. [4 points] In this prolem we consider the system with the initial condition ( x x ( ( t/2 x 4t 3 t, a (a,. a. [4 points] Find the Euler s method approximation for the solution to the system after one step with a step size h (your answer will involve a, and h. What is the meaning of your result? Solution: After one step we have a x + h ( 2 4 This is the approximation for x at t + h. ( a a + 2 h + h( 4a +.. [4 points] Rewrite your approximation in the form P a. What is P? Solution: The matrix P is the coefficient matrix from the approximation, ( P 2 h. 4h ( + h c. [2 points] Find det(p. Solution: det(p + h + 2h 2. d. [4 points] Is it possile that the Euler step could end at x,? Explain. Solution: Note that det(p 2h 2 + h +. Therefore P is nonsingular (has an a inverse, and the only way that this approximation, P, could equal the zero vector is if a. Because we know that a,, this is therefore not possile. We could also work this out directly: we have the approximations x ( + h a + 2 h and ( + h 4ah + ( + h. Setting oth to zero requires that a + 2h, so that a 2 h, and then that 4ah + ( + h 2h2 + ( + h (2h 2 + h +. We know, and 2h 2 + h +, so this is not possile. Note that this condition is, not surprisingly, the same as the determinant.