Page 345 - Advanced engineering mathematics

10.5 Applications and Illustrations of Techniques 325

We therefore choose to use equations (10.7), (10.9), and (10.10). In rearranged order and
some manipulation, these can be rewritten as

i − i =−2i 1 + 18,
1 2

4i + 5i =−10i 1 + 10i 3 ,
1 2
and
i = 2i 2 − 2i 3 .

3
Henceforth, we refer to this as the system. We must determine the initial conditions. The
conductor current i 1 − i 2 is continuous, and i 1 (0−) − i 2 (0−) = 0. Therefore,
i 1 (0+) − i 2 (0+) = 0,
so
i 1 (0+) = i 2 (0+).
The capacitor charge q 1 − q 2 is also continuous. Since q 1 (0−) − q 2 (0−) = 0, then

q 1 (0+) = q 2 (0+).
By equation (10.9),
i 3 (0+) = 0.
Now use equation (10.11) to write
4i 1 (0+) + 5i 2 (0+) + 5i 3 (0+) = 36.
Since i 3 (0+) = 0 and i 1 (0+) = i 2 (0+), then 9i 1 (0+) = 36. So

i 1 (0+) = i 2 (0+) = 4.
In summary, we now have the system
⎞⎛ ⎞
⎛ ⎛ ⎞⎛ ⎞ ⎛ ⎞
1 −10 i 1 −2 0 0 i 1 18
⎝ 4 5 0 ⎠⎝ i 2 ⎠ = ⎝ 0 −10 10 ⎠⎝ i 2 ⎠ + ⎝ 0 ⎠ . (10.13)
0 0 1 i 3 0 2 −2 i 3 0
This has the form Bi = Di + F with B nonsingular. The initial condition is

⎛ ⎞
4
.
⎝
i(0+) = 4 ⎠
0
Multiply the system (10.13) by
⎛ ⎞
5 1 0
1
−1 ⎝ −4
B = 1 0 ⎠
9
0 0 9
to obtain a system
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞
−10/9 −10/9 10/9 10
i 1 i 1
⎝ 8/9
⎝ i 2 ⎠ = −10/9 10/9 ⎠⎝ i 2 ⎠ + −8 ⎠ .
⎝
0 2 −2 0
i 3 i 3
This is in the standard form i =Ai+G. A has eigenvalues 0,−2, and −20/9 with corresponding

eigenvectors
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 5 10
, , and .
⎝ 1 ⎠ ⎝ 0 ⎠ ⎝ 1 ⎠
1 −4 −9
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October 14, 2010 20:32 THM/NEIL Page-325 27410_10_ch10_p295-342