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1 Lin. Alg. & Diff. Eq., Spring 16 Student ID Circle your section: 31 MWF 8-9A 11 LATIMER LIANG 33 MWF 9-1A 11 LATIMER SHAPIRO 36 MWF 1-11A 37 CORY SHAPIRO 37 MWF 11-1P 736 EVANS WORMLEIGHTON 39 MWF -5P 1 WHEELER RABINOVICH 313 MWF -3P 115 KROEBER LIANG 31 MWF 1-P 11 WHEELER WORMLEIGHTON 315 MWF 3-P 11 LATIMER RABINOVICH If none of the above, please explain: Only this exam and a pen or pencil should be on your desk. (You can get scratch paper from me if you need it. Problem Points Possible Your Score A 1 B 1 C 1 D 1 E 1 F 1 G 1 H 1 I 1 J 1

2 Lin. Alg. & Diff. Eq., Spring 16 Problem A. Decide if the following are always true or at least sometimes false. Enter your answers as T or F in the following chart. Correct answers receive 1 points, incorrect answers receive -1 points, and blank answers receive points. No justification is necessary, although if you believe the question is ambiguous, record your interpretation below it. A is always a matrix. Statement Answer T F F F F T F T T F 1. A linear [homogeneous] constant coefficient order n differential equation in one variable always admits n linearly independent solutions.. The sum of solutions to an inhomogenous equation is again a solution. 3. The equation x + x = has no real solutions.. The determinant of the Wronskian of a collection of solutions to an equation df/dt = Af(t does not depend on t. 5. There exists a unique solution to the equation y (t = y(t with y( = 1 and y ( =. 6. Every linear ODE can be written as a first order vector valued ODE. 7. The wave equation is a nonlinear equation 8. The Fourier series of a continuous function with continuous derivative always converges pointwise. 9. The constant term in a Fourier series for an odd function is zero. 1. The function which is -1 on [ 1, ] and 1 on (, 1] does not have a Fourier series, because it is discontinous. 1

3 Lin. Alg. & Diff. Eq., Spring 16 Problem B. Give an example, or explain why none exists. 1. (3 pts A nonlinear differential equation. Solution: (y = t.. (3 pts A linear, constant coefficient, third order ODE with solutions e 3x, xe 3x, and e x. Solution: An ODE with these as solutions has characteristic polynomial divisible by (m 3 (m 1 with the (m 3 term giving xe 3x, and so expanding gives so that a suitable ODE will be (m 6m + 9(m 1 = m 3 7m + 15m 9 y y + 15y 9y =. Note that this is linear and third order with constant coefficients. 3. ( pts A second order, linear, constant coefficient ODE with solutions xe x and x e x. Solution: No such ODE exists. To see this, note that such an ODE must have characteristic polynomial divisible by (m 1 3 to include a solution of the form x e x. But then the ODE must be of order at least 3. Alternative solution: The Wronskian of xe x and x e x is x e x, which is zero somewhere (at x = but not everywhere and so, by the Wronskian lemma, xe x and x e x cannot both solve a second order linear ODE.

4 Lin. Alg. & Diff. Eq., Spring 16 Problem C. (1 pt Give the general solution to the equation y = y. Solution: y(t = αe t. ( pts Give the general solution to the equation y = y. Be sure to use only real-valued functions. Solution: y(t = α cos x + β sin x. (3 pts Give the general solution to the equation y + y + y =. Solution: y(t = αe t + βte t. ( pts Give the general solution to the equation y + y + y = e t. Solution: The general solution to the homogeneous version of this ODE is as above: y h (t = αe t + βte t. A good ansatz for a particular solution (or particular integral is y p (t = γt e t since lower order terms will be killed by the ODE. Substituting in... y p + y p + y p = γe t γte t γte t + γt e t + ( γte t γt e t + γt e t = γe t Hence γ = 1/ and y p (t = 1 t e t. The general solution is then y(t = y h (t + y p (t = αe t + βte t + 1 t e t. 3

5 Lin. Alg. & Diff. Eq., Spring 16 Problem D. ( pts Determine the Wronskian of the functions te t, t e t. Where is it zero? Where is it nonzero? Solution: Compute Wr(te t, t e t = te t e t + te t t e t te t + t e t = t e t + t 3 e t t e t t 3 e t = t e t. This is zero at t = and nonzero elsewhere. (6 pts Are the functions te t, t e t linearly independent? Justify your answer. Solution: Yes! The Wronskian is nonzero somewhere (e.g. t = 1 and so te t and t e t are independent.

6 Lin. Alg. & Diff. Eq., Spring 16 Problem E. Consider the equation y (t + 3y (t + y(t =. (3 pts Write this second order equation as a system of first order equations Solution: Let x 1 = y, x = y, then x 1 = x. Let [ 1 A = 3 ] and x = [x 1, x ] T, then the equation is equivalent to x = Ax. (3 pts Write a fundamental matrix for this system of equations Solution: The eigenvalues and eigenvectors for A are: [ ] [ 1 1 λ 1 = 1, u 1 =, λ 1 =, u = ]. So a fundamental solution matrix could be: [ X = e t e t e t e t ] (pts Find y(t satisfying the above equation, such that y( = 1 and y ( = 3. [ ] [ ] 1 Solution: Let x =, y =, then we are trying to find constant vector c such that 3 Xc(x = y, this can be done by solving the linear system X(x c = y. The results are given by: [ ] [ ] 5 5e c =,, Xc = t e t 5e t + 8e t. y(t is the top entry by construction, and so the solution to the IVP is y(t = 5e t e t. 5

7 Lin. Alg. & Diff. Eq., Spring 16 Problem F. Consider the equation y (t = y(t. (5pts Find a basis of real solutions Solution: The auxiliary equation is r = 1, which has roots ±1, ±i. Thus, a basis of real solutions to the differential equation is {e t, e t, cos t, sin t}. (5pts Solve the initial value problem y( =, y ( = 1, y ( =, y ( = 3. Solution: The general solution to the differential equation is c 1 e t + c e t + c 3 cos t + c sin t, and so y( = c 1 + c + c 3 y ( = c 1 c + c y ( = c 1 + c c 3 y ( = c 1 c c. We need to solve for the c i s that give us the desired initial values for y and its derivatives. This amounts to row reducing the following augmented matrix: , which has reduced row echelon form so the solution to the initial value problem is 1 3/ 1 1/ , 3 et 1 e t cos t sin t 6

8 Lin. Alg. & Diff. Eq., Spring 16 Problem G. (7pts Compute the eigenvalues and eigenvectors of the matrix ( 1 3 Solution: The eigenvalues and eigenvectors of the matrix (denoted as A are λ 1 = , u 1 = [ ], λ = 5 33, u = [ ] ( 1 1 (3pts Compute 3 [ ] λ1 Solution: Let Q = [u 1 u ], and Λ =., then A λ 1 = QΛ 1 Q 1. (I didn t calculate the actual answer since the numbers are complicated. Remark: If the eigenvalues are complex, writing them in polar form is going to simplify the computation. 7

9 Lin. Alg. & Diff. Eq., Spring 16 Problem H. (5pts Determine the dimension of the null space (aka kernel of the matrix 1 3 M = You can row reduce M to The number of pivot columns is, so the dimension of the null space is 1. (5pts For which vectors b does the equation Mx = b have a solution? Solution: This is just the image of the linear map v Mv or the column space of M. Such vectors b then are exactly those in Col(M = Span((1,, 7, (, 5, 8, (3, 6, 9 = Span((1,, 7, (, 5, 8.. 8

10 Lin. Alg. & Diff. Eq., Spring 16 Problem I. (1pts Consider the function f(x defined on [, ] by the formula f(x = x. Determine the Fourier series of this function. Solution: Note that f is even (as x = x and so there are no sine terms (b m =. Thus, one needs only compute the cosine terms. Let s do that... a m = 1 f(x cos mx dx = For clarity, I ll move the constants onto the left hand side... a m = as f(x = x on (,. Note that f(x cos mx dx = f(x cos mx dx. ( x cos mx dx cos mx dx = and so this reduces to x cos mx dx = 1 m x sin mx + 1 sin mx dx m after applying integration by parts. The boundary term 1 sin m mx = as m is an integer, leaving 1 sin mx dx = 1 m m cos mx = 1 m (1 ( 1m observing that cos m = ( 1 m. This is { m m is odd m is even and so { m is odd m a m = m is even Finally, compute the constant term a = 1 f(x dx = x dx = ( 1 =. After reindexing to just include the odd terms, the Fourier series is... + cos (m 1x (m 1 m=1 9

11 Lin. Alg. & Diff. Eq., Spring 16 Problem J. (1 pts Consider a wire of length, which is stretched from x = to x =. Assume the ends are kept at temperature zero, and the initial temperature is given by the function u(x, = x. Using these initial and boundary conditions, solve the heat equation u(x, t = u(x, t t x (If you did not succeed in finding the Fourier series in the previous question, you may leave the Fourier coefficients undetermined; however note you should ensure your answer is compatible with the boundary conditions. Solution: We shall use separation of variables, that is look for the solution in the form Then, the initial heat equation takes form or equivalently u(x, t = X(xT (t. X(xT (t = X (xt (t T (t T (t = X (x X(x. Since the left and right hand sides of the latter equation depend on different variables (t and x respectively, the only way for equation can hold is if for some constant λ. Therefore, we get T (t T (t = λ = X (x X(x X (x + λx(x =. (1 At the same time, since the temperature at both ends of the wire is kept zero, we have the following boundary conditions: u(, t = u(, t = which implies X( = X( =. ( Equation (1 allows for three different forms of solution depending on the sign of λ, however boundary conditions ( yield only trivial solution if λ < or λ =. Thus, the only nontrivial case is λ > which we shall restrict ourselves to from now on. Given λ > the geberal solution to the equation (1 takes form X(x = A cos( λx + B sin( λx. 1

12 Lin. Alg. & Diff. Eq., Spring 16 Then, the boundary conditions ( read and we conclude A cos( λ + B sin( λ =, A cos( λ B sin( λ = A cos( λ = B sin( λ =. Note that sin( λ = implies λ = n for n Z in which case cos( λ. Similarly, cos( λ = implies λ = n 1 for n Z in which case sin( λ. Therefore, we obtain ( n 1 X(x = A n cos x + B n sin(nx, n = 1,, 3,... Now, let us return to the equation on T (t. The general solution to the equation is pf the form Therefore, u(x, t = n=1 T (t + λt (t = T (t = Ce λt. C n e nt sin(nx + D n e (n 1 t cos It is now left to find the coefficients C n and D n from the equation u(x, = x. ( n 1 x. Clearly, where C n = sin(nx, x and D n = f, g = 1 cos f(xg(xdx. ( n 1 x, x Note that sin(x is an odd function while x is even, and thus C n =. As for D n we have D n = 1 ( n 1 ( x cos x dx = Splitting the latter integral in two we get cos ( n 1 x dx = ( n 1 n 1 sin ( n 1 ( x cos x dx. x= x = x= n 1 ( 1n+1. 11

13 Lin. Alg. & Diff. Eq., Spring 16 Integrating the second summand by parts we obtain x cos ( n 1 x dx = ( ( n 1 n 1 x sin = ( n 1 ( 1n+1 + = n 1 ( 1n+1 x= x sin x= n 1 ( n 1 (n 1 cos x 8 (n 1. ( n 1 x= x= x dx Finally, and we get D n = ( n 1 ( 1n+1 n 1 ( 1n+1 u(x, t = n=1 8 (n 1 (n 1 e t cos 8 = (n 1 ( n 1 x. 8 (n 1 1