Quick way to compute the inverse Laplace
transform (without partial fractions).
1. Rational functions with only simple roots in the
denominator.
In this case the solutions can be simply written down. It is best to illustrate by
examples.
Example 1. Find the Laplace inverse of
(Lf )(s) =
1
(s − 1)(s − 2)
The Laplace inverse is given by multiplying the above function by est and then
estimating the resulting function at s = 1, with the factor s − 1 taken out, plus, the
resulting function estimated at s = 2, with the factor s − 2 taken out. Namely
f (t) =
est
s−1
O
s=2 +
est
s−2
O
s=1 =
e2t
et
+
= e2t − et
2−1 1−2
The method still works if we have more factors in the bottom, or if the top is
multiplied by a constant (or in fact any polynomial of degree less than the bottom).
Example 2. Find the inverse Laplace transform of
(Lf )(s) =
2
(s − 1)(s − 2)(s − 3)
Solution: We simply write it down,
2 est
2 est
2 est
+
+
s=1
s=2
(s − 2)(s − 3)
(s − 1)(s − 3)
(s − 1)(s − 2)
t
2t
3t
2e
2e
2e
=
+
+
(1 − 2)(1 − 3) (2 − 1)(2 − 3) (3 − 1)(3 − 2)
2 et 2 e2t 2 e3t
=
+
+
= et − 2e2t + e3t
2
−1
2
f (t) =
O
O
Example 3. Find the inverse Laplace transform of,
(Lf )(s) =
s
s2 + a 2
1
O
s=3
Solution: You could use the table, or alternatively use the method above. First we
factor
s
s
=
s2 + a2 (s + ia)(s − ia)
There are no repeated roots in the bottom and the top is a polynomial of degree less
than the degree of the polynomial in the bottom, hence the method applies. Thus
we can just write down the solution,
est · s
est · s
s=−ia +
(s + ia)
(s − ia)
−iat
iat
e
· ( − ia) e · ia
=
+
− 2ia
2ia
iat
−iat
e +e
=
= cos (a t)
2
f (t) =
O
O
s=ia
a
As an exercice try deriving L−1 s2 + a2 = sin (a t) from the method above.
Remember the formula
sin (a x) =
eiax − e−iax
2i
for this.
2. Rational functions with multiple roots in the
denominator.
This case is more difficult, and you can either use partial fraction or the method
below.
Example 4. Find the Laplace inverse of
(Lf )(s) =
1
(s − 1)(s − 2)2
Solution: One method is to simply write down the partial fraction,
A
B
C
1
2 = s−1 + s−2 +
(s − 2)2
(s − 1)(s − 2)
and then solve for A, B , C.
Another method is to reduce to the case of rational function with only simple roots
in the denominator by using a trick. The trick is to introduce a new variable a, and
express 1/(s − 2)2 in terms of the derivative of 1/(s − a). Namely,
d
1
1
2 = da · s − a
(s − 2)
2
O
a=2
because (d/da)(1/s − a) = 1/(s − a)2 so when we estimate at s = 2 we get 1/(s − 2)2.
With this trick at hand we write,
d
1
1
1
=
·
(s − 1)(s − 2)2 da s − 1 s − a
a=2
O
Therefore,
1
f (t) = L
− 1)(s − 2)2
(s d
1
1
−1
= L
·
da s − 1 s − a
−1
O
a=2
We are allowed to interchange the L−1 (Laplace inverse) with the differentiation
operation, and so we keep on writting,
d −1
1
1
f (t) =
L
·
da
s−1 s−a
O
(1)
a=2
Now we know by the previous section that,
L
−1
1
1
·
s−1 s−a
est
est
s=1 +
s−a
s−1
t
at
e
e
=
+
1−a a−1
=
O
s=a
O
Therefore returning to (1) we have
t
e
d
eat
f (t) =
+
a=2
da 1 − a a − 1
t eat
eat
et
+
−
= (1 − a)2 a − 1 (a − 1)2
= et + t e2t − e2t
O
O
a=2
which is the desired solution.
Example 5. Find the inverse Laplace transform of
(Lf )(s) =
s
(s − 1) (s − 2)2
2
Solution: I will proceed more mechanically skipping more details as I’ve explained
the method in the previous example. We have
s
s
d d
=
da db (s − a)(s − b)
(s − 1)2(s − 2)2
3
O
a=1
b=2
Therefore,
f (t) = L
−1
s
2
(s − 1) (s − 2)2
d d −1
s
=
L
da db
(s − a)(s − b)
=
O
a=1
b=2
Now,
L
−1
s
(s − a)(s − b)
s est
s est
+
s=b
s−b
(s − a)
bt
at
ae
be
=
+
b−a a−b
O
O
s=a
Therefore returning to (2) we find,
a eat
d d b ebt
+
f (t) =
a=1
da db b − a a − b
b=2
d ebt + b t ebt
b ebt
a eat
=
−
+
a=1
da
b−a
(b − a)2 (a − b)2
b=2
bt
bt
bt
at
at
e +bte
be
e +tae
=
−2
−2·
(b − a)2
(b − a)3
(a − b)3
O
O
O
a=1
b=2
= e2t + 2 t e2t − 4 e2t + 2 et + 2 t et
as you see in fact this is comparatively short. Comparatively to trying to do the
same example via partial fractions,
s
A
B
C
D
2 = s−1 +
2 + s−2 +
(s − 1) (s − 2)
(s − 1)
(s − 2)2
2
...
4