**for**

**Chapter 12 Fourier Series and
the Laplace Transform**

**12.6 Laplace Transforms of
Derivatives and Integrals**

This section is a continuation of our
development of the Laplace Transform in __Section
12.5__.

**Theorem 12.13 (Differentiation
of f(t) ).** Let
be continuous for ,
and of exponential order. Then

,

where .

**Corollary
12.1 (Differentiation of f(t)
).** If are
of exponential order, then

.

**Example 12.13.** Show
that .

Solution.

If we let ,
then and , and
is to be determined. Because , we
have , and
Theorem 12.13 implies that

Thus, we have , and from
which it follows that

.

**Theorem 12.14 (Integration of
f(t) ).** Let
be continuous for ,
and of exponential order and ,
then

.

**Example
12.14.** **(a)** Show
that , **(b)** Show
that .

Solution.

Part (a). Use the fact that , and
apply Theorem 12.14, with and ,

then obtain

.

Part (b). Now we can use this first result as a
fact, . This
time we apply Theorem 12.14, with and ,

and obtain

.

One of the main uses of the Laplace transform is its role in the solution of differential equations. The utility of the Laplace transform lies in the fact that the transform of the derivative corresponds to multiplication of the transform by s and then the subtraction of . This permits us to replace the calculus operation of differentiation with simple algebraic operations on transforms.

This idea is used to develop a method for
solving linear differential equations with constant
coefficients. Let's consider the initial value
problem

,

with initial conditions
and . We
can use the linearity property of the Laplace transform to obtain

.

If we let
and
and apply Theorem 12.13 and Corollary 12.1 then we have

.

this in turn can be rearranged to obtain the form

(12.30) .

The Laplace transform
of the solution
is easily found to be

(12.31) .

For many physical problems involving mechanical systems and electrical circuits, the transform is known, and the inverse of can easily be computed. This process is referred to as operational calculus and has the advantage of changing problems in differential equations into problems in algebra. Then the solution obtained will satisfy the specific initial conditions.

**Example
12.15.** Solve the initial value
problem

A graph of the solution.

Solution.

The right side of the differential equation is ,
so we have . The
initial conditions yield and
Equation (12.30)

becomes which
simplifies, and we get

.

Solving we get .

We then solve
with the help of Table 12.2 to compute

**Example
12.16.** Solve the initial value
problem

A
graph of the solution.

Solution.

As in Example 12.15, the right side of the differential equation is , so we have . The initial conditions yield and , and Equation (12.30) becomes , which can be rewritten as which simplifies, and we get .

This time we use Equation (12.31) and obtain ,
which simplifies, and we get . Now
use the partial fraction expansion to
get the solution

**Exercises
for Section 12.6. Laplace Transforms of Derivatives and
Integrals**** **

**The Next Module
is**

**Shifting
Theorems and the Step Function**

**Return to the Complex
Analysis Modules**

__Return
to the Complex Analysis Project__

This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

(c) 2012 John H. Mathews, Russell W. Howell