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90 CHAPTER 3 The Laplace Transform

EXAMPLE 3.9
Compute L[g] where

0 for t < 2
g(t) =
2
t + 1 for t ≥ 2.
To apply the second shifting theorem, we must write g(t) as a function, or perhaps sum of
2
functions, of the form f (t − 2)H(t − 2). To do this, first write t + 1asafunctionof t − 2:
2
2
2
t + 1 = (t − 2 + 2) + 1 = (t − 2) + 4(t − 2) + 5.
Then
2
g(t) = H(t − 2)(t + 1)
2
= (t − 2) H(t − 2) + 4(t − 2)H(t − 2) + 5H(t − 2).
Now apply the second shifting theorem to each term on the right:
2
L[g]= L[(t − 2) H(t − 2)]+ 4L[(t − 2)H(t − 2)]+ 5L[H(t − 2)]
2
= e −2s L[t ]+ 4e −2s L[t]+ 5e −2s L[1]

2 4 5
−2s
= e + + .
s 3 s 2 s
−1
As usual, any formula for L can be read as a formula for L . The inverse version of the
second shifting theorem is
−1
L [e −as F(s)](t) = H(t − a) f (t − a). (3.7)
This enables us to compute the inverse transform of a known transformed function that is
multiplied by an exponential e −as .

EXAMPLE 3.10
Compute
−3s
se
−1
L .
s + 4
2
The presence of e −3s suggests the use of equation (3.7). From the table, we read that

s
−1
L = cos(2t).
2
s + 4
Then
se
−3s
−1
L (t) = H(t − 3)cos(2(t − 3)).
2
s + 4
EXAMPLE 3.11
Solve the initial value problem

y + 4y = f (t); y(0) = y (0) = 0
where

0for t < 3
f (t) =
t for t ≥ 3.

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October 14, 2010 14:14 THM/NEIL Page-90 27410_03_ch03_p77-120