Page 341 - Advanced engineering mathematics

10.5 Applications and Illustrations of Techniques 321

Notice that row three is the derivative of row one, and row four is the derivative of row two,
consistent with the fact that x 3 = y = x and x 4 = y = x . This serves as a partial check on the

1 1 2 2
computations.
The general solution of the system is X(t) = (t)C. To solve the initial value problem, we
need
⎛ ⎞⎛ ⎞ ⎛ ⎞
10 2 0 c 1 1
⎜ 20 −1 0 ⎟⎜ ⎟ ⎜ −1 ⎟
c 2
(0)C = ⎜ ⎟⎜ ⎟ = ⎜ ⎟ .
⎝ 02 0 6 ⎠⎝ c 3 ⎠ ⎝ 0 ⎠
04 0 −3 c 4 0
This has the unique solution
⎛ ⎞
−1/5
⎜ 0 ⎟
C = ⎜ ⎟ .
⎝ 3/5 ⎠
0
The solution of the initial value problem is
⎛ ⎞⎛ ⎞
cos(2t) sin(2t) 2cos(3t) 2sin(3t) −1/5
⎜ 2cos(2t) 2sin(2t) −cos(3t) −sin(3t) ⎟⎜ 0 ⎟
X(t) = ⎜ ⎟⎜ ⎟
⎝ −2sin(2t) 2cos(2t) −6sin(3t) 6cos(3t) ⎠⎝ 3/5 ⎠
−4sin(2t) 4cos(2t) 3sin(3t) −3cos(3t) 0
⎛ ⎞
−cos(2t) + 6cos(3t)
−2cos(2t) − 3cos(3t) ⎟
1 ⎜
= ⎜ ⎟ .
5 ⎝ 2sin(2t) − 18sin(3t) ⎠
4sin(2t) + 9sin(3t)
Since x 1 = y 1 and x 2 = y 2 , we may write, in the notation of Figure 10.2,
1 6
y 1 (t) =− cos(2t) + cos(3t)
5 5
2 3
y 2 (t) =− cos(2t) − cos(3t).
5 5
We could also have used the exponential matrix to produce a fundamental matrix. MAPLE
may produce a different fundamental matrix than that found using Theorem 10.7, but of course,
the solution of the initial value problem is the same.

EXAMPLE 10.17 An Electrical Circuit
Assume that the currents and charges in the circuit of Figure 10.3 are zero until time t = 0, at
which time the switch is closed. We want to determine the current in each loop.
Use Kirchhoff’s voltage and current laws on the left and right loops to obtain
5i 1 + 5(i − i ) = 10 (10.4)

1 2
and
q 2

5(i − i ) = 2i 2 + . (10.5)
1
2
5 × 10 −2
Using the exterior loop (around the entire circuit), we get
q 2
5i 1 + 20i 2 + = 10. (10.6)
5 × 10 −2

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October 14, 2010 20:32 THM/NEIL Page-321 27410_10_ch10_p295-342