problemset9

T cos t sin t so x1 y1 y2 and x2 y2 form a normalized

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Unformatted text preview: 0 So a fundamental matrix is given by �(t) = . �(0) = , sin(bt) − cos(bt) 0 −1 � � � � 10 cos(bt) − sin(bt) �(0)−1 = , so eAt = �(t)�(0)−1 = eat . 0 −1 sin(bt) cos(bt) � � cos(bt) − sin(bt) (a+bi)t ) = A(eat (cos(bt) + i sin(bt))) = eat A(e = eA(a+bi)t . sin(bt) cos(bt) e at � (b) [9] s2 + 2s + 2 = (s + 1)2 + 1 so the roots of the characteristic p olynomial are −1 ± i. Basic solutions are given by y1 = e−t cos(t) and y2 = e−t sin(t). (I write y instead of x b ecause the ˙ ˙ problem wrote x for the normalized solutions.) y 1 (0) = 1, y1 (0) = −1, y2 (0) = 0, y2 (0) = 1. −t (cos t + sin t), So x1 = y1 + y2 and x2 = y2 form a normalized pair of solutions: x 1 (t) = e x2 (t) = e−t sin t. � � 0 1 . Its characteristic p olynomial is the same, � 2 +2�+2, The companion matrix is A = −2 −...
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This note was uploaded on 01/22/2014 for the course MATH 18.03 taught by Professor Vogan during the Spring '09 term at MIT.

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