1550078481-Ordinary_Differential_Equations__Roberts_

250 Ordinary Differential Equations

The following three MAPLE statements calculate and print the general

solution of the differential equation, DE2, using the Laplace transform method

without displaying the results of any intermediate steps.

with(inttrans):

DE2:=diff(y(x), x$2)+diff(y(x), x) - 2 * y(x) = x /\ 2 - l;

dsolve( {DE2, y(O) =A, D(y)(O) = B}, y(x), method=laplace);

5.3 Convolution and the Laplace Transform

One can often write a function H(s) as the product of two functions F(s)

and G(s) in such a way that both F(s) and G(s) are the Laplace transforms

of known functions, say f(x) and g(x), respectively. That is, one sometimes

encounters the situation in which H(s) = F(s)G(s), where F(s) = .C[f(x)]

and G(s) = .C[g(x)]. Hence, H(s) = .C[f(x)].C[g(x)]. Momentarily, we might

expect that H(s) = .C[f(x)].C[g(x)] = .C[f(x)g(x)]. Stated verbally, we might

anticipate that the Laplace transform distributes over the multiplication of

two functions. However, we know from calculus that, in general, the integral

of the product of two functions is not equal to the product of the integrals of

the two functions. And since the Laplace transform is defined in terms of an

integral, we should not expect the Laplace transform to distribute over the

multiplication of functions. The following example illustrates that, in general,

.C[f(x)].C[g(x)]-/= .C[f(x)g(x)].

Let f(x) = x and let g(x) =ex. Then

and

Clearly,

1 1

.C[f(x)].C[g(x)] = .C[x].C[ex] = 2 -

s s - 1

1

.C[f(x)g(x)] = .C[xex] = ( ) 2.

s- 1

.C[f(x)].C[g(x)]-/= .C[f(x)g(x)] for any reals.

Two questions should immediately come to mind: "Is there any function
h(x) such that .C[h(x)] = H(s) = .C[f(x)].C[g(x)]?" and "If so, how is h(x)
related to f(x) and g(x)?" We shall now answer these two questions.