problemset9

Start again for u2 u2 c1 e 2e 1 1 0 1 1 c1 c 2 c2

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Unformatted text preview: 2 so its�eigenvalues are �he same, −1 ± i. An eigenvector for value −1 + i is given by v 1 such t � � 1−i 1 1 that v1 = 0. We can take v1 = . The corresponding normal −2 −1 − i −1 + i � � � � 1 cos t (−1+i)t −t mode is e , which has real and imaginary parts u 1 = e −1 + i − cos t − sin t � � � � � � sin t 10 10 −1 = −t . �(t) = [u1 u2 ] has �(0) = . �(0) , and u2 = e − sin t + cos t −1 1 11 � � cos t + sin t sin t . The top entries coincide with x1 so eAt = �(t)�(0)−1 = e−t −2 sin t − sin t + cos t and x2 computed above. �...
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