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Mathematical Expression Editor

Given , the Laplace transform of some function , we study techniques for recovering
the function .

The Inverse Laplace Transform

Definition of the Inverse Laplace Transform

In Trench 8.1 we defined the Laplace transform of by
We’ll also say that is an inverse Laplace Transform of , and write
To solve differential equations with the Laplace transform, we must be able to obtain
from its transform . There’s a formula for doing this, but we can’t use it because it
requires the theory of functions of a complex variable. Fortunately, we can
use the table of Laplace transforms to find inverse transforms that we’ll
need.

Use the table of Laplace transforms to find

(a)

(b)

.

item:8.2.1a Setting in the transform pair
shows that

item:8.2.1b Setting in the transform pair
shows that

The next theorem enables us to find inverse transforms of linear combinations of
transforms in the table. We omit the proof.

Linearity Property If are Laplace transforms and are constants then

Completing the square in the denominator yields
Because of the form of the denominator, we consider the transform pairs
and write

We’ll often write inverse Laplace transforms of specific functions without explicitly
stating how they are obtained. In such cases you should refer to the table of Laplace
transforms in Trench 8.8.

Inverse Laplace Transforms of Rational Functions

Using the Laplace transform to solve differential equations often requires finding the
inverse transform of a rational function
where and are polynomials in with no common factors. Since it can be shown that
if is a Laplace transform, we need only consider the case where . To obtain , we find
the partial fraction expansion of , obtain inverse transforms of the individual
terms in the expansion from the table of Laplace transforms, and use the
linearity property of the inverse transform. The next two examples illustrate
this.

Find the inverse Laplace transform of

(Method 1:) Factoring the denominator in (eq:8.2.1) yields
The form for the partial fraction expansion is
Multiplying this by yields
Setting yields and setting yields . Therefore
and

(Method 2:) We don’t really have to multiply (eq:8.2.3) by to compute and . We can
obtain by simply ignoring the factor in the denominator of (eq:8.2.2) and setting elsewhere;
thus,
Similarly, we can obtain by ignoring the factor in the denominator of (eq:8.2.2) and setting
elsewhere; thus,
To justify this, we observe that multiplying (eq:8.2.3) by yields
and setting leads to (eq:8.2.4). Similarly, multiplying (eq:8.2.3) by yields
and setting leads to (eq:8.2.5). (It isn’t necessary to write the last two equations.
We wrote them only to justify the shortcut procedure indicated in (eq:8.2.4) and
(eq:8.2.5).)

The shortcut employed in the second solution of Example example:8.2.4 is Heaviside’s method.
The next theorem states this method formally.

Suppose
where are distinct and is a polynomial of degree less than . Then
where can be computed from (eq:8.2.6) by ignoring the factor and setting elsewhere.

Find the inverse Laplace transform of

The partial fraction expansion of (eq:8.2.7) is of the form
To find , we ignore the factor in the denominator of (eq:8.2.7) and set elsewhere. This
yields
Similarly, the other coefficients are given by
and
Therefore
and

We didn’t “multiply out” the numerator in (eq:8.2.7) before computing the coefficients in (eq:8.2.8),
since it wouldn’t simplify the computations.

Find the inverse Laplace transform of

The form for the partial fraction expansion is
Because of the repeated factor in (eq:8.2.9), Heaviside’s method doesn’t work. Instead, we
find a common denominator in (eq:8.2.10). This yields
If (eq:8.2.9) and (eq:8.2.11) are to be equivalent, then
The two sides of this equation are polynomials of degree two. From a theorem of
algebra, they will be equal for all if they are equal for any three distinct values of .
We may determine , and by choosing convenient values of .

The left side of (eq:8.2.12) suggests that we take to obtain , and to obtain . We can now
choose any third value of to determine . Taking yields . Since and this implies that
. Therefore
and

Find the inverse Laplace transform of

The form for the partial fraction expansion is
The easiest way to obtain , , and is to expand the numerator in powers of . This
yields
Therefore

and

Find the inverse Laplace transform of

One form for the partial fraction expansion of is
However, we see from the table of Laplace transforms that the inverse transform of
the second fraction on the right of (eq:8.2.14) will be a linear combination of the inverse
transforms
of
respectively. Therefore, instead of (eq:8.2.14) we write
Finding a common denominator yields
If (eq:8.2.13) and (eq:8.2.16) are to be equivalent, then
This is true for all if it’s true for three distinct values of . Choosing , , and yields
the system
Solving this system yields
Hence, from (eq:8.2.15),
Therefore

Find the inverse Laplace transform of

The form for the partial fraction expansion is
The coefficients , , and can be obtained by finding a common denominator and
equating the resulting numerator to the numerator in (eq:8.2.17). However, since there’s no
first power of in the denominator of (eq:8.2.17), there’s an easier way: the expansion
of
can be obtained quickly by using Heaviside’s method to expand
and then setting to obtain
Multiplying this by yields
Therefore

Some software packages that do symbolic algebra can find partial fraction expansions
very easily. We recommend that you use such a package if one is available to you, but
only after you’ve done enough partial fraction expansions on your own to master the
technique.

Text Source

Trench, William F., ”Elementary Differential Equations” (2013). Faculty Authored
and Edited Books & CDs. 8. (CC-BY-NC-SA)