APPM 36: Midterm exam 3 April 19, 17 On the front of your Bluebook write: (1) your name, () your instructor s name, (3) your lecture section number and (4) a grading table. Text books, class notes, cell phones and calculators are NOT permitted. A one page (letter sized 1 side only) handwritten crib sheet is allowed. Problem 1: (3 points) We will consider the following two 3 3 matrices A 3 1 1, B 1 3. 5 3 3 Show all of your work in this problem. (a) (1 points) Find all eigenvalues of A. (b) (1 points) Given that the eigenvalues of B are λ 1 λ 3, λ 3, find the corresponding eigenvectors. (c) (4 points) What are the dimensions of the eigenspaces from part (b)? (d) (4 points) Using the definition of eigenvalue/eigenvector, verify directly that the vector you found in part (b) corresponding to λ 3 is in fact an eigenvector of B with eigenvalue λ 3. (a) Solve det(a λi) Find such that det 3 λ 1 λ 1 5 3 λ, p 3 (λ) (3 λ)[(1 λ)( 3 λ) + 5] (3 λ)[(λ 1)(λ + 3) + 5] (3 λ)[λ + λ + ] λ 1 3, λ,3 ± 4 8 (b) For λ 1, 3 solve (B 3I)v, i.e., 1 1 1 1 ± i. RREF Set v (v 1, v, v 3 ) T, then v 1 + v 3 v 1 v 3, v 3 is free. Also v is free. Set v s and v 3 t so that v t s s 1 + t t 1
This gives two linearly independent eigenvectors v 1 1, v 1 that span a two-dimensional space. For λ 3 solve (B I)v, i.e., 1 1 1 1 1 1 RREF Hence letting v (v 1, v, v 3 ) T, v 1 + v v 1 v, v is free, and v 3. Setting v t we have Thus v t t v 3 which spans a one-dimensional space. t 1 1 1 1 (c) The dimension of the eigenspace associated with λ 1, is. The dimension of the eigenspace associated with λ 3 1 is 1. (d) Note that Bv 3 1 3 3 1 1 v 3. Problem : (3 points) Consider the second order differential equation: (1 + t)y + ty y (1 + t) e t. (1) Note that y 1 (t) t and y (t) e t are two linearly independent solution to (1 + t)y + ty y. Show all of your work in this problem. (a) (4 points) Compute the Wronskian of y 1 and y. (b) (6 points) Write down (1) in the form required to use the method of variation of parameters (the normalized equation). In this form, what is the nonhomogeneous term f(t)? (c) (16 points) Use the method of variation of parameters to find a particular solution y p to equation (1). Do not use any other method. (d) (4 points) Write down the general solution of (1). (a) W (t) y 1 y y 1 y t e t 1 e t (t + 1)e t. (b) The normalized ODE is y + t 1 + t y 1 1 + t y (1 + t)e t. Therefore, f(t) (1 + t)e t. (c) Assume that y p (t) v 1 (t)y 1 (t) + v (t)y (t).
Then we get and Therefore, y (t)f(t) v 1 (t) dt W (t) v (t) y1 (t)f(t) dt W (t) e t (1 + t)e t (t + 1)e t dt t(1 + t)e t dt (t + 1)e t e t dt e t. tdt t. y p (t) v 1 (t)y 1 (t) + v (t)y (t) e t t 1 t e t (t + 1 t )e t. (d) The general solution to (1) is y c 1 y 1 + c y + y p c 1 t + c e t (t + 1 t )e t, c 1, c R. Problem 3: (3 points) Consider the differential equation y 6y + 9y () whose solution is y c 1 e 3t + c te 3t. Using the Method of Undetermined Coefficients, construct a guess for the particular solutions of the following differential equations (DO NOT SOLVE FOR THE COEFFICIENTS IN PARTS (a) (c)): (a) (5 points) y 6y + 9y e 6t (b) (5 points) y 6y + 9y cos(3t) sin(4t) (c) (5 points) y 6y + 9y te 3t (d) (15 points) FIND y p BY CONSTRUCTING A GUESS AND SOLVING FOR THE CO- EFFICIENTS: y 6y + 9y t (a) y p Ae 6t (b) y p A cos(3t) + B sin(3t) + C cos(4t) + D sin(4t) (c) y p (At + B)t e 3t (d) y p At + B y p A y p Plug-in and collect terms to get: 9A 1 and 9B 6A so A 1/9 and B 6/81 /7 so y p (t) (1/9)t + /7 OVER FOR MORE PROBLEMS Problem 4: (3 points) Show all of your work in this problem. (a) (6 points) Use linearity and the provided table to find the Laplace transform F (s) of the function f(t) (e t + e t ) + 4 sin(3t)
(b) (1 points) Given the Laplace transform F (s) 4(s + 3) s 3 + 3s + 4s, find its inverse f(t). (c) (1 points) Using the definition of the Laplace transform only, find L{g(t)} given { 1 t if t < 1 g(t) if t 1 (a) Write So that f(t) e 4t + + e 4t + 4 sin(3t) (b) Partial fraction decomposition Now and completing the square L{f(t)} L{e 4t } + L{1} + L{e 4t } + 4L{sin(3t)} 1 s 4 + s + 1 s + 4 + 4 3 s + 9 4(s + 3) s(s + 3s + 4) A s + Bs + C s + 3s + 4 3 s + s 9 s + 3s + 4 3 s L{3} s 9 s + 3s + 4 s 9 s + 3s + 9/4 9/4 + 4 s 9 (s + 3/) + 7/4 s + 3/ 3/ 9 (s + 3/) + ( 7/) s + 3/ (s + 3/) + ( 7/) 1 1 (s 3/) + ( 7/) 7/ 7 s 3/ (s + 3/) + ( 7/) 1 { L e 3t/ cos( 7t/) 1 e 3t/ sin( 7t/) 7 (s + 3/) + ( 7/) } Thus f(t) L 1 {F (s)} 3 + e 3t/ cos( 7t/) 1 7 e 3t/ sin( 7t/)
(c) By definition L{g(t)} 1 e st g(t) dt e st (1 t) dt g(t) for t 1 (1 t) e st 1 1 e st s ( ) dt s ] [(1 1) e s (1 )e s s s e s s + 1 [ ] 1 s + s e st e s s + 1 s + s [e s e ] e s s + 1 s + e s s s 1 e st dt u 1 t, dv e st dt Problem 5: (3 points, 6 points each) True/False questions: Write the full word TRUE if the given statement is always true, otherwise write the full word FALSE. You don t have to show any work. (a) The space of solutions to the differential equation: is spanned by {e t, e t }. (b) All the solutions to: y + 4y + 5y + y y + 5y + 4y are strictly decreasing functions of time. (c) The solutions to: y + 4y cos(4t) exhibit growing oscillations due to a pure resonance. (d) If you know two different particular solutions, y p1 and y p, to the nonhomogeneous second order linear differential equation y + a 1 (t)y + a (t)y f(t), then you can find a solution to the associated homogeneous differential equation without any additional information. (e) If λ is an eigenvalue of the matrix A, then the vector equation A x λ x has exactly one solution. (a) False: The dimension of the space of solutions S to an homogeneous, linear differential equation of order 3 has dimension 3. This set of functions cannot span S. The characteristic equation r 3 + 4r + 5r + has a simple root r and a double root r 1. Hence: S Span ( e t, te t, e t) (b) False: The characteristic equation has two real, negative roots 1 and 4, both associated with decaying exponentials e t and e 4t. Their combination, however, is not guaranteed to decrease. Counterexample: y e t e 4t has a positive slope y 3 at t. (c) False: The natural frequency is ω 4, which is different from the forcing frequency ω f 4. No resonance is observed. (Solutions have beats.) (d) True: y(t) y p1 (t) y p (t) will be a solution to the associated homogeneous diffeq.
(e) False: This equation has infinitely many solutions (any nonzero multiple of an eigenvector is also an eigenvector of A with the same λ). Table 8.1.1: Short table of Laplace Transforms L{1} 1 s L{cos(bt)} L{t n } n! s n+1 L{e at } 1 s a s s + b L{e at sin(bt)} L{sinh(bt)} L{t n e at } n! (s a) n+1 L{sin(bt)} b (s a) + b L{e at cos(bt)} b s b L{cosh(bt)} s s b s a (s a) + b b s + b