Page 342 - Advanced engineering mathematics

322 CHAPTER 10 Systems of Linear Differential Equations

5 Ω 20 Ω

–2
5 × 10 F
i 1 5 H
i 2
10 V

FIGURE 10.3 Circuit of Example 10.17.

Any two of these equations contain all of the information needed to solve the problem. We will
use equations (10.4) and (10.6). The reason for this choice is that we would have to differentiate
the second equation to eliminate the q 2 term in equation (10.5), producing second derivative
terms in the currents. This is avoided by using the first and third equations.
Divide equations (10.4) and (10.6) by 5 and differentiate the new equation (10.6), using the
fact that q = i 2 , to obtain

2

i − i =−i 1 + 2

1 2
i + 4i =−4i 2 .

1 2
We must determine the initial conditions. We know that
i 1 (0−) = i 2 (0−) = q 2 (0−) = 0.
Then (i 1 − i 2 )(0−) = 0. Since the current i 1 − i 2 through the inductor is continuous, then
(i 1 − i 2 )(0+) = 0 also. Therefore,
i 1 (0+) = i 2 (0+) = 0.
Put this into equation (10.6) and use the fact that the charge on the capacitor is continuous to
obtain
5i 1 (0+) + 20i 2 (0+) + 20q 2 (0+) = 10
or

25i 1 (0+) = 10.
Then
10 2
i 1 (0+) = i 2 (0+) = =
25 5
amperes. Finally, the initial value problem for the currents is

i − i =−i 1 + 2

1
2

i + 4i =−4i 2
1
2
2
i 1 (0+) = i 2 (0+) = .
5
In matrix form,

1 −1 i 1 −1 0 i 1 2
= + .
1 4 i 2 0 −4 i 2 0
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October 14, 2010 20:32 THM/NEIL Page-322 27410_10_ch10_p295-342