M 340L HW 10 Solutions

Xk 4e2t v1 2et2 v2 v1 d u t 1 t2 v2 e 2 dt

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Unformatted text preview: λ) = λ2 − 2λ − 3 for then c1 = 2, c2 = 4 and = (λ − 3)(λ + 1) , u(0) = 2v1 + 4v2 . But v1 , v2 are eigenvectors corresponding to 1 respective eigenvalues 2, 2 so set 2t −6 , −2 −2 −5 . 1 4 1. u(t) = 6. xk = 2e2t v1 − 2et/2 v2 v2 u(0) = when A is the matrix 4. xk = 4e2t v1 + 2et/2 v2 [ v1 d u ( t) 1 t/2 v2 = e , 2 dt = 2(2e2t )v1 + 4 −1 . 1 v2 = Then u(0) is the given initial value and −2 , 6 u(0) = 3 t/2 u(t) = 2e v1 + 4e v2 . the eigenvalues of A are λ1 = 3, λ2 = −1 and corresponding eigenfunctions v1 = 1 , −1 v2 = −5 1 cruz (fmc326) – HW10 – gilbert – (56540) form a basis for R2 because λ1 = λ2 . Thus u(0) = c1 v1 + c2 v2 . To compute c1 , c2 we apply row reduction to the augmented matrix [ v1 v2 u(0) ] = ∼ 10 01 1 −1 −5 −6 1 −2 4 , 2 for then c1 = 4, c2 = 2. Now set u(t) = 4e3t v1 + 2e−t v2 . Then u(0) = 4v1 + 2v2 = −6 −2 , while Au(t) = 4e3t Av1 + 2e−t Av2 = 12e3t v1 − 2 −e−t v2 = d u ( t) , dt so u(t) = 4e3t = −5 1 + 2e−t 1 −1 4e3t − 10e−t −4e3t + 2e−t solves the differential equation. 4...
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This document was uploaded on 03/16/2014 for the course M 340L at University of Texas at Austin.

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