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
12.1 (Differentiation of f(t)
). If are
of exponential order, then
Example 12.13. Show that .
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 ,
Example 12.14. (a) Show that , (b) Show that .
Part (a). Use the fact that , and
apply Theorem 12.14, with and ,
Part (b). Now we can use this first result as a fact, . This time we apply Theorem 12.14, with and ,
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
with initial conditions and . We can use the linearity property of the Laplace transform to obtain
If we let
and apply Theorem 12.13 and Corollary 12.1 then we have
this in turn can be rearranged to obtain the form
The Laplace transform
of the solution
is easily found to be
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.
12.15. Solve the initial value
A graph of the solution.
The right side of the differential equation is ,
so we have . The
initial conditions yield and
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.
12.16. Solve the initial value
graph of the 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.
Exercises for Section 12.6. Laplace Transforms of Derivatives and Integrals
The Next Module is
Shifting Theorems and the Step Function
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This material is coordinated with our book Complex Analysis for Mathematics and Engineering.
(c) 2012 John H. Mathews, Russell W. Howell