m345t2S11Sol - MA 345 Dierential Equations 1. (15 points)...

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MA 345 Diferential Equations Spring 2011 Exam II Solutions Dr. E. Jacobs 1. (15 points) Let A be the following matrix: A = ± 31 13 ² a) Find all eigenvalues of A 0 = det( A - λ I )= ³ ³ ³ ³ 3 - λ 1 - λ ³ ³ ³ ³ =(3 - λ ) 2 - 1 so λ 1 = 2 and λ 2 =4 b) Find the eigenvectors corresponding to the eigenvalues that you found in part (a). We can solve ( A - λ I ± 0 by matrix reduction: for λ 1 =2 , ± 11 | 0 | 0 ² -→ ± | 0 00 | 0 ² for λ 2 , ± - | 0 1 - 1 | 0 ² ± 1 - 1 | 0 | 0 ² Therefore, for λ 1 = 2, the eigenvectors will be any nonzero scalar multiples of ´ 1 - 1 µ and for λ 2 = 4, the eigenvectors will be any nonzero scalar multiples of ( 1 1 ) c) Find a matrix P so that P - 1 AP is a diagonal matrix. Check your work by multiplying P - 1 AP out. P = ± - ² check: P - 1 AP = 1 2 ± 1 - 1 ²± - ² = 1 2 ± 2 - 2 44 - ² = ± 20 04 ² 2. (10 points) Once again, let A = ( ) . Let ± X = ´ x ( t ) y ( t ) µ . Solve the following di±erential equation: d ± X dt = A ± X where ± X (0) = ± - 2 6 ² The general solution can be constructed immediately from the eigenvectors: ± X = c 1 ± 1 - 1 ² e 2 t + c 2 ± 1 1 ² e 4 t The initial condition implies that : ± X (0) = ± - 2 6 ² = c 1 ± 1 - 1 ² + c 2 ± 1 1 ² = ± - c 1 c 2 ² = P ± c 1 c 2 ² ± c 1 c 2 ² = P - 1 ± - 2 6 ² = 1 2 ± 1 - 1 - 2 6 ² = ± - 4 2 ² Therefore, ± X = - 4 ± 1 - 1 ² e 2 t +2 ± 1 1 ² e 4 t
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3. (10 points) Let y ( t
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m345t2S11Sol - MA 345 Dierential Equations 1. (15 points)...

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