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Laplace Transforms of Derivatives and Integrals

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 .

Proof.

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

.

Proof.

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

.

Explore Solution 12.13.

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

.

Proof.

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

.

Explore Solution 12.14 (a).

Explore Solution 12.14 (b).

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

Explore Solution 12.15.

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

Explore Solution 12.16.

Laplace Transform

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