Page 127 - Advanced engineering mathematics

3.6 Solution of Systems 107

EXAMPLE 3.19
We will solve the system (with initial conditions):

x − 2x + 3y + 2y = 4,

2y − x + 3y = 0,

x(0) = x (0) = y(0) = 0.
Apply the transform to each equation of the system, making use of the initial conditions, to
obtain
4
2
s X − 2sX + 3sY + 3Y = ,
s
2sY − sX + 3Y = 0.
Solve these for X(s and Y(s):
4s + 6 2
X(s) = and Y(s) = .
2
s (s + 2)(s − 1) s(s + 2)(s − 1)
Use partial fractions to write
7 1 1 1 1 10 1
X(s) =− − 3 + +
2 s s 2 6 s + 2 3 s − 1
and
1 1 1 2 1
Y(s) =− + + .
s 3 s + 2 3 s − 1
Then
7 1 −2t 10 t
x(t) =− − 3t + e + e
2 6 3
and
1 2
t
y(t) =−1 + e −2t + e .
3 3
EXAMPLE 3.20

Figure 3.26 shows a mass/spring system. Let x 1 = x 2 = 0 at the equilibrium position, where the
weights are at rest. Choose the direction to the right as positive, and suppose the weights are at
x 1 (t) and x 2 (t) at time t.
By two applications of Hooke’s law, the restoring force on m 1 is
−k 1 x 1 + k 2 (x 2 − x 1 )
and that on m 2 is
−k 2 (x 2 − x 1 ) − k 3 x 2 .

k 1 k 2 k 3

m 1 m 2
FIGURE 3.26 Mass-spring system of Example 3.20.

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October 14, 2010 14:14 THM/NEIL Page-107 27410_03_ch03_p77-120