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Solution: APPM 2360 Exam 3 Fall 2011 Problem 1: (20 points) Consider the linear differential operator 2 2 L(y) = y 00 − y 0 + 2 y t t 2 (a) Verify that y(t) = t and y(t) = t are both solutions to the homogeneous equation L(y) = 0. (b) Construct a particular solution to the nonhomogeneous equation L(y) = f , where f (t) = t3 . (c) Write down the general solution to the equation L(y) = f . Solution: (a) 2 2 2 2 L(t) = (t)00 − (t)0 + 2 (t) = 0 − (1) + 2 (t) = 0 t t t t and 2 2 2 2 L(t2 ) = (t2 )00 − (t2 )0 + 2 (t2 ) = 2 − (2t) + 2 (t2 ) = 0 t t t t . (b) Variation of parameters: Two homogeneous solutions given by y1 = t, y2 = t2 . y1 y2 t t2 = t2 . W [y1, y2] = 0 = y1 y20 1 2t −t2 t3 −y2 f = = −t3 W [y1 , y2 ] t2 y1 f t · t3 v20 = = 2 = t2 W [y1 , y2 ] t v10 = So, Z 1 −t3 dt = − t4 4 Z 1 v2 = t2 dt = t3 3 v1 = Then, yp (t) = y1 v1 + y2 v2 −t4 t3 + t2 4 3 t5 = 12 =t (c) y(t) = c1 y1 (t) + c2 y2 (t) + yp (t) = c1 t + c2 t2 + t5 . 12 Problem 2: (15 points) True/False. If the statement is always true, mark true. Otherwise, mark false. You do not need to show your work. (Any work will not be graded.) (a) There exists a real 2 × 2 matrix, A, with eigenvalues λ1 = 1, λ2 = i. (b) If λ = 2 is a repeated eigenvalue of multiplicity 2 for a matrix A, then the eigenspace associated with λ has dimension 2. (c) If an n × n matrix has n distinct eigenvalues, then each eigenvalue has a corresponding one-dimensional eigenspace. Solution: (a) False. Complex eigenvalues of real are always present in complex conjugate pairs. matrices 2 1 (b) False. Not necessarily, e.g. A = . 0 2 (c) True. If {λ1 , . . . , λn } is a set of distinct eigenvalues then a corresponding set of eigenvectors {v1 , . . . , vn } is linearly independent. This is a maximally linearly independent set in Rn so if any one eigenvalue had two (or more) linearly independent eigenvectors, then one of them would be a linear combination of eigenvectors of other eigenspaces, which would be a contradiction. Problem 3: (20 points) Consider the following differential equation, x00 + 2x0 + x = et (a) Find the general solution x(t). (b) Solve the initial value problem with x(0) = x0 (0) = 0. Solution: (a) Homogeneous solution: x00h + 2x0h + xh = 0 ⇒ xh = c1 e−t + c2 te−t Undetermined coefficients: xp = Aet ⇔ ⇔ (Aet )00 + 2(Aet )0 + (Aet ) = et 1 A= 4 General Solution: x(t) = xh (t) + xp (t) 1 = c1 e−t + c2 te−t + et . 4 (b) 1 1 ⇒ c1 = − 4 4 1 1 x0 (0) = 0 ⇒ 0 = −c1 + c2 + ⇒ c2 = − 4 2 x(0) = 0 ⇒ 0 = c1 + So, 1 1 1 x(t) = − e−t − te−t + et . 4 2 4 Problem 4: (20 points) Find the general solution to the system of differential equations 1 2 0 ~x = ~x. −2 1 Solution: p(λ) = (1 − λ)2 + 4 = λ2 − 2λ + 5 ⇒ λ = 1 ± 2i. Find the eigenvector corresponding to λ = 1 + 2i: Solve (A − λI)~v = ~0: −2i 2 0 −2i 2 0 ∼ −2 −2i 0 0 0 0 1 1 0 ⇒ ~v = = +i . i 0 1 So, ~xRe ~xIm cos 2t = e (cos βt p~ − sin βt ~q) = e − sin 2t αt t sin 2t . = e (sin βt p~ + cos βt ~q) = e cos 2t αt t Then the general solution is, ~x(t) = c1 ~xRe + c2 ~xIm c1 cos 2t + c2 sin 2t t =e −c1 sin 2t + c2 cos 2t Problem 5: (25 points) Consider the matrix 1 0 −1 A = 1 2 0 . 0 0 2 (a) Find all the eigenvalues of A. (b) For each eigenvalue that you found in part (a), find a basis for the corresponding eigenspace. (c) Find the general solution of ~x0 = A~x. (Hint: you need to find a generalized eigenvector) Solution: (a) The characteristic polynomial of A is p(λ) = (1 − λ)(2 − λ)2 . The eigenvalues are λ1 = 1, λ2 = 2 (repeated). (b) λ1 = 1: Solve (A − λ1 I)~v1 = ~0: −1 0 0 −1 0 1 1 0 0 ⇒ ~v1 = 1 . 0 0 1 0 0 Then ~v1 is a basis for the eigenspace E1 corresponding to λ1 = 1. λ2 = 2: Solve (A − λ2 I)~v2 = ~0: −1 0 −1 0 0 1 0 0 0 ⇒ ~v2 = 1 0 0 0 0 0 So ~v2 is a basis for the eigenspace E2 corresponding to λ2 = 2. (c) ~x0 = A~x is a 3 × 3 system, so we need three linearly independent solutions. Two solutions are given by ~x1 (t) = eλ1 t~v1 and ~x2 (t) = eλ2 t~v2 . For the third, we need to find a generalized eigenvector for the eigenvalue λ2 = 2, since this is the repeated eigenvalue. Solve (A − λ2 I)~u = ~v2 : −1 0 −1 0 0 1 1 0 0 1 ⇒ ~u = s 1 + 0 0 0 0 0 0 −1 for a free parameter s. Since we only need one such vector, set s = 0. Then a third linearly independent solution is given by ~x3 (t) = teλ2 t~v2 + eλ2 t ~u 0 1 = te2t 1 + e2t 0 . 0 −1 The general solution is given by ~x(t) = c1 ~x1 (t) + c2 ~x2 (t) + c3 ~x3 (t) −1 0 1 = c1 et 1 + c2 e2t 1 + c3 e2t t 0 0 −1 −c1 et + c3 e2t t = c1 e + c2 e2t + c3 te2t . −c3 e2t