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Inverse Transform

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The Laplace Transform

Inverse Laplace Transform


The Inverse Problem
If F(s) represents the Laplace transform of a function f(t); that is
ℒ{f(t)} = F(s), we then say that f(t) is the Inverse Laplace Transform of
F(s) and write that f(t) = ℒ-1{F(s)}.

For example,

ℒ-1   1
1
s
ℒ-1 s1  t
2 ℒ-1   e
1
s 3
3t
Some Inverse Transforms
=

= = =

= =

= =

When evaluating inverse transforms, it often happens that a function of s


under consideration does not match exactly the form of Laplace transform F(s)
given in the table. It may be necessary to “fix up” the function of s by multiplying
and dividing by an appropriate constant.
Example 1
Evaluate ℒ-1{1/s5}

Ans: f(t) = t4/24


Example 2
Evaluate ℒ-1{1/(s2 + 7)}

7 sin 7t
Ans: f(t) =
7
Example 3
 2s  6 
Evaluate ℒ 
-1 
 s 4 
2

Ans: f(t) = -2 cos 2t + 3 sin 2t


Partial Fractions
Partial fractions play an important role in finding inverse Laplace
Transforms. We will review some of the basic algebra in the important
cases in which the denominator of a Laplace Transform F(s) contains
distinct linear factors, repeated linear factors, and quadratic
polynomials with no real factors.

For example:

F(s) = 3/(s2 - 3s + 2); F(s) = 3/(s2 - 4s + 4); F(s) = 3/[(s2 +4)(s-3)]


Example 4
 s 2  6s  9 
Evaluate ℒ-1  ( s  1)( s  2)( s  4) 
 

16et 25e2t e4t


Ans: f(t) =   
5 6 30
Solving ODE's
REMEMBER: Transform of a Derivative

ℒ{f'(t)} = sF(s) - f(0)

Similarly, ℒ{f''(t)} = sℒ{f'(t)} - f'(0) = s2F(s) - sf(0) - f'(0)

In like manner, we can show that


ℒ{f(n)(t)} = snF(s) - sn-1f(0) - sn-2f'(0) - ... - sf(n-2)(0) - f(n-1)(0)
Example 5
Use Laplace transform to solve the first-order IVP
dy/dt + 3y = 13 sin 2t, y(0) = 6

Ans: y(t) = 8 e-3t - 2 cos 2t + 3 sin 2t


Seatwork
Use Laplace transform to solve the 2nd-order IVP
y'' - 3y' + 2y = e-4t, y(0) = 1, y'(0) = 5

16et 25e2t e4t


Ans: y(t) =   
5 6 30

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