Page 129 - Advanced engineering mathematics

3.6 Solution of Systems 109

and
5 1 1
x 2 (t) = − cos(2t) + cos(3t)
36 4 9
5 1 1

+ − + cos(2(t − 3)) − cos(3(t − 3)) H(t − 3).
36 4 9
The next example involves an electrical circuit and will require that we know how to take
the transform of a function defined by an integral. To see how to do this, suppose

t
f (t) = g(τ)dτ.
0
Then f (0) = 0, and assuming that g is continuous, f (t) = g(t),so

t

L[ f ]= L[g]= sL g(τ)dτ .
0
But this means that
t 1

L g(τ)dτ = L[g].
s
0
EXAMPLE 3.21
We will use this result to analyze the circuit of Figure 3.27.
Suppose the switch is closed at time zero. We want to solve for the current in each loop.
Assume that both loop currents and the charges on the capacitors are initially zero, and apply
Kirchhoff’s laws to each loop to obtain

40i 1 + 120(q 1 − q 2 ) = 10

60i 2 + 120q 2 = 120(q 1 − q 2 ).
Since i = q , we can write

t

q(t) = i(τ)dτ + q(0).
0

40 Ω 60 Ω

1/120 F
1/120 F

10 V

FIGURE 3.27 Circuit in Example 3.21.

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October 14, 2010 14:14 THM/NEIL Page-109 27410_03_ch03_p77-120