Page 344 - Advanced engineering mathematics

324 CHAPTER 10 Systems of Linear Differential Equations

This gives us
4 −4t/5
i 1 (t) = 2 − e (2cos(2t/5) + sin(2t/5))
5
2 −4t/5
i 2 (t) = e (cos(2t/5) − 2sin(2t/5)).
5
The currents can also be obtained by diagonalizing A and changing variables by setting
i = PZ, where P diagonalizes A. This is a straightforward computation but is a little tedious
because the eigenvalues of A are complex.

EXAMPLE 10.18 Another Electrical Circuit
The circuit of Figure 10.4 has three connected loops (and of course, the external loop). The
currents in these three loops are zero prior to t = 0, at which time the switch is closed. The
capacitor is in a discharged state at time zero. We want to determine the current in each loop at
all later times.
Apply Kirchhoff’s current and voltage laws to obtain

4i 1 + 2i − 2i = 36, (10.7)

1 2

2i − 2i = 5i 2 + 10q 2 − 10q 3 , (10.8)
1 2
10q 2 − 10q 3 = 5i 3 , (10.9)
4i 1 + 5i 2 + 10q 2 − 10q 3 = 36, (10.10)
4i 1 + 5i 2 + 5i 3 = 36, (10.11)

2i − 2i = 5i 2 + 5i 3 . (10.12)
1 2
Any three of these equations are enough to determine the currents. Because equation
(10.8) involves both charge and current terms, we would have to differentiate to put every-

thing in terms of currents (recall that q = i j ). This would introduce second derivatives, which
j
we want to avoid. Cross out this equation. We could use equation (10.11) to eliminate one
variable and reduce the problem to a two by two system, but this would involve a lot of
algebra.
Equation (10.9) is a likely candidate to retain. If we then use equations (10.7) and (10.12),

we obtain a system of the form Bi = Di + F with B as singular, so we would not be able to
multiply by B −1 to obtain a system in standard form.

4 Ω 5 Ω 5 Ω

10 –1 F

i 1 2 H i 2
i 3
36 V

FIGURE 10.4 Circuit of Example 10.18.

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October 14, 2010 20:32 THM/NEIL Page-324 27410_10_ch10_p295-342