M 340L HW 10 Solutions

Xk 2e2t v1 4et2 v2 correct 2 xk 4e2t v1 4et2 v2 2t t2

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Unformatted text preview: – HW10 – gilbert – (56540) Find the solution of the differential equation du = A u ( t) , dt when A is a 2 × 2 matrix with eigenvalues 2, and corresponding eigenvectors 1 , 1 v1 = 1 2 1. xk = 2e2t v1 + 4et/2 v2 correct 2. xk = 4e2t v1 − 4et/2 v2 2t t/2 3. xk = 4e v1 + 4e Au(t) = 2e2t Av1 + 4et/2 Av2 so u(t) = 2e2t v1 + 4et/2 v2 solves the differential equation. 005 10.0 points Find the solution of the differential equation du = A u ( t) , dt v2 A= 5. xk = 2e2t v1 + 2et/2 v2 Explanation: Since v1 , v2 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R2 . Thus 4e3t − 10e−t −4e3t + 2e−t 2. u(t) = 4e−3t − 10et −4e−3t + 2et 3. u(t) = −4e−3t + 10et 4e3t − 2et 4. u(t) = −4e3t + 10e−t 4e3t − 2e−t u(0) = c1 v1 + c2 v2 . To compute c1 , c2 we apply row reduction to the augmented matrix u(0) ] = ∼ 1 0 1 1 −1 1 −2 6 02 , 14 correct Explanation: Since det[A − λI ] = −2 − λ 1 −5 4−λ = 5 − (2 + λ)(4...
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