? 2 t 2 1 ? n t 1 2 ? n t 2 e ? 1 t e ? 2 t e ? n t

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( λ 2 t ) 2 + · · · · · · 0 . . . . . . . . . . . . 0 0 · · · 1 + λ n t + 1 2! ( λ n t ) 2 + · · · = e λ 1 t 0 · · · 0 0 e λ 2 t · · · 0 . . . . . . . . . . . . 0 0 · · · e λ n t CHAPTER 10 REVIEW EXERCISES 3. Since 4 6 6 1 3 2 1 4 3 3 1 1 = 12 4 4 = 4 3 1 1 , we see that λ = 4 is an eigenvalue with eigenvector K 3 . The corresponding solution is X 3 = K 3 e 4 t . 6. We have det( A λ I ) = ( λ + 6)( λ + 2) = 0 so that X = c 1 1 1 e 6 t + c 2 1 1 e 2 t . 9. We have det( A λ I ) = ( λ 2)( λ 4)( λ + 3) = 0 so that X = c 1 2 3 1 e 2 t + c 2 0 1 1 e 4 t + c 3 7 12 16 e 3 t . 191
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CHAPTER 10 REVIEW EXERCISES 12. We have X c = c 1 2cos t sin t e t + c 2 2sin t cos t e t . Then Φ = 2cos t 2sin t sin t cos t e t , Φ 1 = 1 2 cos t sin t 1 2 sin t cos t e t , and U = Φ 1 F dt = cos t sec t sin t dt = sin t ln | sec t + tan t | cos t , so that X p = ΦU = 2cos t ln | sec t + tan t | 1 + sin t ln | sec t + tan t | e t . 15. (a) Letting K = k 1 k 2 k 3 we note that ( A 2 I ) K = 0 implies that 3 k 1 +3 k 2 +3 k 3 = 0, so k 1 = ( k 2 + k 3 ). Choosing k 2 = 0, k 3 = 1 and then k 2 = 1, k 3 = 0 we get K 1 = 1 0 1 and K 2 = 1 1 0 , respectively. Thus, X 1 = 1 0 1 e 2 t and X 2 = 1 1 0 e 2 t are two solutions. (b) From det( A λ I ) = λ 2 (3 λ ) = 0 we see that λ 1 = 3, and 0 is an eigenvalue of multiplicity two. Letting K = k 1 k 2 k 3 , as in part ( a ), we note that ( A 0 I ) K = AK = 0 implies that k 1 + k 2 + k 3 = 0, so k 1 = ( k 2 + k 3 ). Choosing k 2 = 0, k 3 = 1, and then k 2 = 1, k 3 = 0 we get K 2 = 1 0 1 and K 3 = 1 1 0 , respectively. Since the eigenvector corresponding to λ 1 = 3 is K 1 = 1 1 1 , 192
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CHAPTER 10 REVIEW EXERCISES the general solution of the system is X = c 1 1 1 1 e 3 t + c 2 1 0 1 + c 3 1 1 0 . 193
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