# solutions_6 - MATH 286 Sections D1 X1 Assignment 6...

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MATH 286 Sections D1 & X1: Assignment 6 Solutions for the graded problems. Section 5.1, #28 [2 points] To verify that x 1 = 1 6 - 13 = e λ 1 t v 1 , x 2 = e 3 t 2 3 - 2 = e λ 2 t v 2 , x 3 = e - 4 t - 1 2 1 = e λ 3 t v 3 are solutions to x 0 = 1 2 1 6 - 1 2 - 1 - 2 - 1 | {z } = A x , we just need to check that v i is an eigenvector of A with eigenvalue λ i for each i = 1, 2, 3. That is, we must show that ( A - λ i I ) v i = 0 ( i = 1, 2, 3 ) . We leave this calculation to the reader. Next, to verify linear independence, we calculate the Wronskian W ( t ) = det [ x 1 ( t ) x 2 ( t ) x 3 ( t )] = det 1 2 e 3 t - 1 e - 4 t 6 3 e 3 t 2 e - 4 t - 13 - 2 e 3 t 1 e - 4 t = e 3 t e - 4 t det 1 2 - 1 6 3 2 - 13 - 2 1 = - 84 e - t . Since W ( t ) 6 = 0 for all t , we see that these solutions are linearly independent. The general solution is then given by a general linear combination x ( t ) = 3 i = 1 c i x i ( t ) . Section 5.2, #38 [2 points] Given A = 1 0 0 0 2 0 0 0 0 3 3 0 0 0 4 4 , the eigenvalues λ are given by det ( A - λ I ) = 0 ⇐⇒ det 1 - λ 0 0 0 2 2 - λ 0 0 0 3 3 - λ 0 0 0 4 4 - λ = 0 ⇐⇒ ( 1 - λ )( 2 - λ )( 3 - λ )( 4 - λ ) = 0. 1
So the eigenvalues are λ = 1, 2, 3, 4. The corresponding solutions to the ODE x 0 = A x are x 1 ( t ) = e t v 1 , x 2 ( t ) = e 2 t v 2 , x 3 ( t ) = e 3 t v 3 , x 4 ( t ) = e 4 t v 4 , where v λ = a λ b λ c λ d λ 6 = 0 is an eigenvextor associated to the eigenvalue λ . I.e., v λ solves the equation ( A - λ I ) v λ = 0 ( λ = 1, 2, 3, 4 ) .
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