problemset9

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 −...
View Full Document

## 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.

Ask a homework question - tutors are online