Inverse Laplace
For People Who Don’t Like Partial Fractions
Let’s look at an alternative method to finding the inverse Laplace in situations where taking partial fractions can become tedious and open us up to many
potential mistakes:
The Method, Part 1
#EX1) Find
−1
L
1
s4 (s − 2)
(t)
If we solve this using partial fractions, it is entirely possible.
However, we end up solving this equation:
1
A B
C
D
E
= + 2+ 3+ 4+
− 2)
s
s
s
s
s−2
s4 (s
Let us instead consider the following relationship:
Z t
F (s)
−1
L
(t) =
f (u)du
s
0
We know that
−1
L
1
s−2
(t) = e2t
So
−1
L
1
s(s − 2)
−1
(t) = L
e2u
=
2
1
1
·
s s−2
t
=
0
Z
(t) =
t
e2u du
0
e2t 1
−
2
2
Again we can use this same relationship
Z t 2u
1
1
1
e
1
−1
−1
L
(t) = L
·
(t) =
− du
2
s (s − 2)
s s(s − 2)
2
2
0
2u
t
e
u
e2t
t 1
−
=
− −
=
4
2 0
4
2 4
Again
−1
L
=
1
3
s (s − 2)
Z
(t) =
e2u u2 u
−
−
8
4
4
0
t
=
0
t
e2u u 1
− − du
4
2 4
t 1
e2t t2
− − −
8
4
4 8
And finally we get that
Z t 2u
e
1
u2 u 1
−1
(t)
=
−
− − du
L
s4 (s − 2)
8
4
4 8
0
2u
t
e
e2t
u3 u2 u
t3
t2
t
1
=
−
−
−
−
− − −
16
12
8
8 0
16 12
8
8 16
This solution entirely agrees with the partial fraction decomposition:
1
1 1 1 1
1 1
1 1
1
1
=− · − · 2 − · 3 − · 4 +
·
− 2)
16 s 8 s
4 s
2 s
16 s − 2
s4 (s
So we see the trade-off: either a lot of algebra with partial fractions or 4 (relatively) basic integrations.
The Method, Part 2
The previous method works well with taking the inverse Laplace
of fractions of the form:
1
, for integers a, n
n
s (s − a)
But what if the piece under the exponent is not just s?
#EX2) Find
−1
L
1
(s − 1)2 (s + 1)
(t)
For these questions, we need one more relationship:
L eat f (t) (s) = F (s − a) or L−1 {F (s − a)} (t) = eat f (t)
So for the above problem:
F (s) =
1
and L−1 {F (s)} (t) = f (t)
2
(s − 1) (s + 1)
We want to make F (s) have the same form as in part 1. So let’s
call G(s) = F (s + 1) = F (s − (−1)). Now we have
G(s) =
1
and g(t) = e−t f (t)
+ 2)
s2 (s
We can solve this using our method. After 2 integrations, we get
g(t) =
t 1
e−2t
+ −
4
2 4
Now we just use our new relationship to find what f (t) is:
g(t) = e−t f (t)
−2t
e
t 1
e−t tet et
t
⇒ f (t) = e
+ −
=
+
−
4
2 4
4
2
4
So to solve more general problems, shift the function to make s
appear under the exponent in the denominator, solve for the new
function, and then get answer by multiplying by the appropriate
exponent to ”shift” back.