M 340L HW 10 Solutions

E 3t e3t et ta t e 3t ta ie 1 3 and 2 1

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Unformatted text preview: xk } be a solution of the difference equation 0 P −1 , d2 d1 0 002 xk+1 = Axk , then 2 , so + e−t 2(e −3t −t correct A=P 1 1 v2 = 2e−t + e−3t = 4. e −3t e−3t − e−t = tA −t +e −3t tA i.e., λ1 = −3 and λ2 = −1. Corresponding eigenvectors are 10.0 points Using the fact that 1 1 ex = 1 + x + x2 + . . . + xn + . . . , 2! n! 1. etA = 1 4 1−λ = λ − λ + 3 = (λ + 3)(λ + 1) = 0 , Compute x1 . 1. x1 = 8 4 2. x1 = 4 correct 8 3. x1 = 7 3 4. x1 = 3 7 x0 = −1 . 7 cruz (fmc326) – HW10 – gilbert – (56540) 2 5. x1 = 3 8 1. xk = 3(2k )v1 − 1 2k v2 6. x1 = 4 7 2. xk = 3(2k )v1 + 1 2k v2 Explanation: Since v1 , v2 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R2 . Thus x0 = c1 v1 + c2 v2 . v2 x0 ] = 1 1 1 −1 ∼ 0 7 −1 1 1 2k v2 correct 4. xk = 6(2k )v1 − 2 1 2k v2 5. xk = 6(2k )v1 − To compute c1 , c2 we apply row reduction to the augmented matrix [ v1 3. xk = 3(2k )v1 − 2 03 , 14 for then c1 = 3, c2 = 4 a...
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This document was uploaded on 03/16/2014 for the course M 340L at University of Texas.

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