BIS
AEM_3e_Chapter_10

# 24 we have det a ? i ? 8 ? 1 2 0 for ? 1 8 we obtain

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24. We have det( A λ I ) = ( λ 8)( λ + 1) 2 = 0. For λ 1 = 8 we obtain K 1 = 2 1 2 . For λ 2 = 1 we obtain K 2 = 0 2 1 and K 3 = 1 2 0 . Then X = c 1 2 1 2 e 8 t + c 2 0 2 1 e t + c 3 1 2 0 e t . 27. We have det( A λ I ) = ( λ 1) 3 = 0. For λ 1 = 1 we obtain K = 0 1 1 . 179

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10.2 Homogeneous Linear Systems Solutions of ( A λ 1 I ) P = K and ( A λ 1 I ) Q = P are P = 0 1 0 and Q = 1 2 0 0 so that X = c 1 0 1 1 e t + c 2 0 1 1 te t + 0 1 0 e t + c 3 0 1 1 t 2 2 e t + 0 1 0 te t + 1 2 0 0 e t . 30. We have det( A λ I ) = ( λ + 1)( λ 1) 2 = 0. For λ 1 = 1 we obtain K 1 = 1 0 1 . For λ 2 = 1 we obtain K 2 = 1 0 1 and K 3 = 0 1 0 so that X = c 1 1 0 1 e t + c 2 1 0 1 e t + c 3 0 1 0 e t . If X (0) = 1 2 5 then c 1 = 2, c 2 = 3, and c 3 = 2. In Problems 33-45 the form of the answer will vary according to the choice of eigenvector. For example, in Problem 33, if K 1 is chosen to be 1 2 i the solution has the form X = c 1 cos t 2cos t + sin t e 4 t + c 2 sin t 2sin t cos t e 4 t . 33. We have det( A λ I ) = λ 2 8 λ + 17 = 0. For λ 1 = 4 + i we obtain K 1 = 2 + i 5 so that X 1 = 2 + i 5 e (4+ i ) t = 2cos t sin t 5cos t e 4 t + i cos t + 2sin t 5sin t e 4 t . Then X = c 1 2cos t sin t 5cos t e 4 t + c 2 cos t + 2sin t 5sin t e 4 t . 180
10.2 Homogeneous Linear Systems 36. We have det( A λ I ) = λ 2 10 λ + 34 = 0. For λ 1 = 5 + 3 i we obtain K 1 = 1 3 i 2 so that X 1 = 1 3 i 2 e (5+3 i ) t = cos3 t + 3sin3 t 2cos3 t e 5 t + i sin3 t 3cos3 t 2sin3 t e 5 t . Then X = c 1 cos3 t + 3sin3 t 2cos3 t e 5 t + c 2 sin3 t 3cos3 t 2sin3 t e 5 t . 39. We have det( A λ I ) = λ ( λ 2 + 1 ) = 0. For λ 1 = 0 we obtain K 1 = 1 0 0 . For λ 2 = i we obtain K 2 = i i 1 so that X 2 = i i 1 e it = sin t sin t cos t + i cos t cos t sin t . Then X = c 1 1 0 0 + c 2 sin t sin t cos t + c 3 cos t cos t sin t . 42. We have det( A λ I ) = ( λ 6)( λ 2 8 λ + 20) = 0. For λ 1 = 6 we obtain K 1 = 0 1 0 . For λ 2 = 4 + 2 i we obtain K 2 = i 0 2 so that X 2 = i 0 2 e (4+2 i ) t = sin2 t 0 2cos2 t e 4 t + i cos2 t 0 2sin2 t e 4 t .

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