Page 329 - Advanced engineering mathematics

10.2 Solution of X = AX for Constant A 309

then we have the linear system of two equations in two unknowns:

a 1
(A − 4I 2 ) = .
b 1
This is the system

−3 3 a 1
=
−3 3 b 1
with general solution

α
E 2 = .
(1 + 3α)/2
Since we need only one E 2 ,let α = 1 to get

1
E 2 = .
4/3
Therefore, a second solution is

1 1 1 + t
4t 4t 4t 4t 4t
2 (t) = E 1 te + E 2 e = te + e = e .
1 4/3 4/3 + t
1 and 2 are linearly independent solutions and can be used as columns of the fundamental
matrix
4t 4t
e (1 + t)e
(t) = 4t 4t .
e (4/3 + t)e
The general solution is X(t) = (t)C.

EXAMPLE 10.11
We will solve X = AX when

−2 −1 −5
⎛ ⎞
⎝ 25
A = −7 0 ⎠ .
0 1 3
A has eigenvalues −2,−2,−2, and all eigenvectors are scalar multiples of
⎛ ⎞
−1
.
⎝
E 1 = −5 ⎠
1
Onesolutionis 1 (t)=E 1 e −2t . We will try a second solution, linearly independent from the first,
of the form
2 (t) = E 1 te −2t + E 2 e −2t
in which E 2 must be determined. Substitute this proposed solution into the differential equation
to get
E 1 [e −2t − 2te −2t ]+ E 2 [−2e −2t ]= AE 1 te −2t + AE 2 e −2t .
Divide out e −2t and recall that AE 1 =−2E 1 to cancel terms in the last equation, leaving
AE 2 + 2E 2 = E 1 .

This is the system
(A + 2I 3 )E 2 = E 1 .

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October 14, 2010 20:32 THM/NEIL Page-309 27410_10_ch10_p295-342