Page 343 - Advanced engineering mathematics

10.5 Applications and Illustrations of Techniques 323

This can be written as
Bi = Ai + K.

This is not quite in the form we have studied. However, |B|= 5, so B is nonsingular. We find that
1 4 1
−1
B = .
5 −1 1
Multiply the system by B −1 to obtain

−4/5 −4/5 8/5
i 1 i 1
= + .
i 2 1/5 −4/5 i 2 −2/5
This is in the standard form
i = Ai + G

in which

−4/5 −4/5
A = .
1/5 −4/5
We will use the exponential matrix and variation of parameters to solve for the currents, assuming
availability of software to compute e At and carry out integrations and matrix products that are
needed. First, compute the fundamental matrix
−4t/5 −4t/5
e cos(2t/5) −2e sin(2t/5)
At
(t) = e = 1 −4t/5 −4t/5 .
2 e sin(2t/5) e cos(2t/5)
Then

e 4t/5 cos(2t/5) 2e 4t/5 sin(2t/5)
−1
(t) = .
1 4t/5
− e sin(2t/5) e 4t/5 cos(2t/5)
2
The general solution of the associated homogeneous system i = Ai is C.

For a particular solution p of the nonhomogeneous system, first compute
8 4t/5 4 4t/5
8/5 e cos(2t/5) − e sin(2t/5)
−1
G = = 5 4 4t/5 5 2 4t/5 .
−2/5 − e sin(2t/5) − e cos(2t/5)
5 5
Form
4t/5
2e cos(2t/5)
−1
U(t) = (t)G(t)dt = 4t/5 .
−e sin(2t/5)
A particular solution of the nonhomogeneous system is
2 2
2cos (2t/5) + 2sin (2t/5) 2
(t) = (t)U(t) = = .
cos(2t/5)sin(2t/5) − cos(2t/5)sin(2t/5) 0
The general solution of i = Ai + G is

−4t/5 −4t/5
e cos(2t/5) −2e sin(2t/5) c 1 2
i = 1 −4t/5 sin(2t/5) e −4t/5 cos(2t/5) + 0 .
e
2 c 2
Since i 1 (0) = i 2 (0) = 2/5, then we must choose c 1 and c 2 so that

2/5 10 c 1 2
= + .
2/5 01 c 2 0
Then c 1 =−8/5 and c 2 = 2/5. The solution for the currents is
−4t/5 −4t/5
e cos(2t/5) −2e sin(2t/5) −8/5 2
i 1
= 1 −4t/5 −4t/5 + .
i 2 e sin(2t/5) e cos(2t/5) 2/5 0
2
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October 14, 2010 20:32 THM/NEIL Page-323 27410_10_ch10_p295-342