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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.
 Spring '09
 vogan
 Differential Equations, Equations

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