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Unformatted text preview: P −1 AP = α1
0α . 8. The system of diﬀerential equations is equivalent to the single matrix
4 −1
˙
equation X = AX , where A =
.
4
8
The characteristic polynomial of A is λ2 − 12λ + 36 = (λ − 6)2 , so we
can use the previous question with α = 6. Let
−2 −1
4
2 B = A − 6I2 =
−2
=
4
P = [X1 X2 ], we have
Then BX2 = 1
. Also let X1 = BX2 . Then if
0 0
, if X2 =
0
P −1 AP = . 61
06 . Now make the change of variables X = P Y , where Y =
˙
Y = (P −1 AP )Y = 61
06 x1
y1 . Then Y, or equivalently x1 = 6x1 + y1 and y˙1 = 6y1 .
˙
Solving for y1 gives y1 = y1 (0)e6t . Consequently
x1 = 6x1 + y1 (0)e6t .
˙
Multiplying both side of this equation by e−6t gives
d −6t
(e x1 ) = e−6t x1 − 6e−6t x1 = y1 (0)
˙
dt
e−6t x1 = y1 (0)t + c,
where c is a constant. Substituting t = 0 gives c = x1 (0). Hence
e−6t x1 = y1 (0)t + x1 (0)
and hence
x1 = e6t (y1 (0)t + x1 (0)).
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.
 Fall '10
 Dreibelbis

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