hw03solns - Homework 4 Solution 1 7.8 7 Given A =-4...

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Homework 4 Solution 1 7.8 7. Given A = ± 1 - 4 4 -7 , eigenvalues are λ 1 , 2 = - 3 and corresponding eigenvectors ± 1 1 , ± 1 4 0 + k ± 1 1 . x = c 1 ± 1 1 e - 3 t + c 2 •± 1 1 te - 3 t + ± 1 4 0 e - 3 t . Since x (0) = ± 3 2 c 1 = 2 ,c 2 = 4 x = e - 3 t ± 4t+3 4t+2 . -10 -8 -6 -4 -2 0 2 4 6 8 10 -10 -8 -6 -4 -2 0 2 4 6 8 10 (a) x 1 - x 2 -5 0 5 10 15 20 25 30 -25 -20 -15 -10 -5 0 5 (b) x 1 - t Figure 1: Problem 7.7 10. Given A = ± 3 9 -1 -3 , eigenvalues are λ 1 , 2 = 0 and corresponding eigenvectors ± -3 1 , ± -1 0 + k ± -3 1 . x = c 1 ± -3 1 + c 2 •± -3 1 t + ± -1 0 ¶‚ . Since x (0) = ± 2 4 c 1 = 4 ,c 2 = - 14 x = ± 42t+2 -14t+4 . -10 -8 -6 -4 -2 0 2 4 6 8 10 -10 -8 -6 -4 -2 0 2 4 6 8 10 (a) x 1 - x 2 -10 -8 -6 -4 -2 0 2 4 6 8 10 -500 -400 -300 -200 -100 0 100 200 300 400 500 (b) x 1 - t Figure 2: Problem 7.10 1
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2 Given A = 2 1 0 0 0 0 2 1 0 0 0 0 2 1 0 0 0 0 2 1 0 0 0 0 2 , eigenvalues are λ 1 , ··· , 5 = 2 and corresponding eigenvectors and generalized eigenvectors 1 0 0 0 0 , 0 1 0 0 0 , 0 0 1 0 0 , 0 0 0 1 0 , 0 0 0 0 1 . x
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This note was uploaded on 04/20/2008 for the course AME 30315 taught by Professor Goodwine during the Spring '08 term at Notre Dame.

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hw03solns - Homework 4 Solution 1 7.8 7 Given A =-4...

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