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10.4 Exponential Matrix Solutions 317
EXAMPLE 10.14
Let
2 −5
A = .
1 4
MAPLE returns the exponential matrix
5
1 − sin(2t)
3t cos(2t) − sin(2t)
At
2
e = e 1 sin(2t) cos(2t) − sin(2t).
2
1
2 2
This is a fundamental matrix for the system X = AX. We could also solve this system by
diagonalizing A, which has eigenvalues 3 ± 2i.
EXAMPLE 10.15
Let
⎛ ⎞
2 1 0
A = 0 3 −2 ⎠ .
⎝
0 1 1
Then
⎛ ⎞
1sin(t) − cos(t) + 1 2(cos(t) − 1)
At
e = e 2t ⎝ 0 sin(t) + cos(t) −2sin(t) ⎠ .
0 sin(t) cos(t) + sin(t)
This is a fundamental matrix for X = AX.
The fundamental matrix (t) = e At is sometimes called a transition matrix for the system
X = AX. This is a fundamental matrix satisfying (0) = I n .
Variation of Parameters and the Laplace Transform
We will briefly mention a connection between the Laplace transform, the exponential matrix and
the variation of parameters method for finding a particular solution p (t) of X = AX + G,in
which A is an n × n real, constant matrix.
The variation of parameters method is to write p (t) = (t)U(t), where
−1
U(t) = (t)G(t)dt
and (t) is a fundamental matrix for X = AX.
Write U(t) as a definite integral with s as the variable of integration:
t
−1
U(t) = (s)G(s)ds.
0
Then
t
−1
p (t) = (t) (s)G(s)ds
0
t
−1
= (t) (s)G(s)ds.
0
At
In this, (t) can be any fundamental matrix for the system. If we choose (t) = e , then
−1
(t) = e −At and
−1
At −As
(t) (s) = e e = e A(t−s) = (t − s).
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October 14, 2010 20:32 THM/NEIL Page-317 27410_10_ch10_p295-342