Page 328 - Advanced engineering mathematics

308 CHAPTER 10 Systems of Linear Differential Equations

and
√ √
−t
e [sin( 3t)U + cos( 3t)V].
In terms of these solutions, a fundamental matrix is
√ √
⎛ 2t −t −t ⎞
e e cos( 3t) e sin( 3t)
√ √ √ √
−t
−t
(t) = ⎝ 0 2 3e sin( 3t) −2 3e cos( 3t) ⎠ .
√ √ √ √ √ √
−t
−t
0 e [−3cos( 3t) − 3sin( 3t)] e [ 3cos( 3t) − 3sin( 3t)]
The general solution is X(t) = (t)C.
10.2.2 Solution When A Does Not Have n Linearly Independent Eigenvectors
Two examples will give us a sense of how to proceed when A does not have n linearly
independent eigenvectors.

EXAMPLE 10.10
We will solve X = AX when

1 3
A = .
−3 7
A has eigenvalues 4,4, and all eigenvectors have the form

1
α
1
with α = 0. A does not have two linearly independent eigenvectors. One solution is

1 4t
1 (t) = e .
1
We need a second, linearly independent solution. Let

1
E 1 =
1
and attempt a second solution of the form
4t
4t
2 (t) = E 1 te + E 2 e ,
in which E 2 is a 2× 1 constant matrix to be determined. For 2 (t) to be a solution, we must have
(t) = A 2 (t). This is the equation

2
4t
4t
4t
4t
4t
E 1 [e + 4te ]+ 4E 2 e = AE 1 te + AE 2 e .
4t
Divide this by e to get
E 1 + 4tE 1 + 4E 2 = AE 1 t + AE 2 .
But AE 1 = 4E 1 , so the terms involving t cancel, leaving
AE 2 − 4E 2 = E 1
or
(A − 4I 2 )E 2 = E 1 .
If

a
E 2 = ,
b

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October 14, 2010 20:32 THM/NEIL Page-308 27410_10_ch10_p295-342