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104 CHAPTER 3 The Laplace Transform

EXAMPLE 3.17
We will solve

y + 2y + 2y = δ(t − 3); y(0) = y (0) = 0.

Apply the transform to the differential equation to get
2
s Y(s) + 2sY(s) + 2Y(s) = e −3s ,

so
1
Y(s) = e −3s .
2
s + 2s + 2
The solution is the inverse transform of Y(s). To compute this, first write

1 −3s
Y(s) = e .
(s + 1) + 1
2
−1
2
Because L [1/(s + 1)]= sin(t),ashiftinthe s− variable gives us

1
−1 −t
L = e sin(t).
(s + 1) + 1
2
Now shift in the t− variable to obtain
y(t) = H(t − 3)e (t−3) sin(t − 3).

Figure 3.23 is a graph of this solution.

100

50

0
2 4 6 8
t
–50
–100

–150

FIGURE 3.23 Graph of the solution in Example
3.17.

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October 14, 2010 14:14 THM/NEIL Page-104 27410_03_ch03_p77-120