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Partial Fractions

−1
although we could equally have written L (e −as /s) = ν (t) for
a
{
1, t > a,
ν (t) =
a
0, t ≤ a,

which is another variant of the unit step function.
Another interesting function along these lines is the following. For 0 ≤ a < b,
let

0, t < a,
1 
u (t) = (u (t) − u (t)) = 1 , a ≤ t < b,
b
ab
a
b − a  b−a

0, t ≥ b.
Then
e −as − e −bs
L(u (t)) = .
ab
s(b − a)
The process of computing the inverse Laplace transform of a function turns out

to be more challenging than most students like. It involves lots of algebra and
using a table of Laplace transforms backwards. For example, if we were asked to
[ 3 ]
−1
find L we would write
s 3
[ 3 ] 3 [ 2 ] 3
−1 −1 2
L = L = t ,
s 3 2 s 3 2
2
2
since we know that L[t ] = and we can adjust the constants to work out.
. s 3

Example 2.23, Find the inverse Laplace transform of F(s) =
s 2 11
+ + .
2
2
s − 9 s 4 s + 5
.
. . . . .
Solution. By the linearity property of the inverse Laplace transform

[ s 2 11 ] [ s ] [ 2 ]
−1 −1 −1 −1
L [F(s)] = L + + = L + L +
2
2
2
s − 9 s 4 s + 5 s − 9 s 4
√
[ 11 ] [ s ] 2 [ 3! ] 11 [ 5 ]
−1 −1 −1 −1
+L == L + L + √ L √ =
2
2
s + 5 s − 3 2 3! s 4 5 s + ( 5) 2
2
1 11 √
3
= cosh (3t) + t + √ sin ( 5t).
3 5
. . . .
Partial Fractions

In many applications of the Laplace transform it becomes necessary to find the
inverse of a particular transform, F(s). Typically it is a function that is not

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