1550078481-Ordinary_Differential_Equations__Roberts_

The Laplace Transform Method 225

EXAMPLE 2 Laplace Transform of x^0 , .C[x^0 ]

Calculate the Laplace transform of f(x) = xn, where n is a positive integer.

SOLUTION

By definition

.C[xn] = 1= xne-sx dx.

R ecall from calculus the following formula for integration by parts

l udv~u{-l vdu

Letting u = xn and dv = e-sxdx and differentiating and integrating, we find

du= nxn-^1 dx and v = --e-sx.^1 Integrat10n. by parts, yields
s
(1)

.C[xn] = r= xne-sx dx = - xn e-sx 1= + ".:1'. r= xn-le-sx dx = ".:1'..c[xn-1]

lo s 0 s lo s

provided s > 0.

Letting n = 1, we find

1 1

.C[x ] = -.C[l] = 2 for s > 0.

s s

If n > 1, we repeatedly use the result of equation (1) to obtain

.C[xn] = ".:1'..c[xn-1] = n(n - 1) .C[xn-2] = ... = n! .C[l] = ~ for s > 0.

s 32 sn sn+l

EXAMPLE 3 Laplace Transform of sin bx, .C [sin bx]

Find .C[sinbx ] = J 000 (sinbx)e-sx dx.

SOLUTION

We use integration by parts. This time, we let u = sin bx and dv =
1
csxdx. Differentiation and integration yields du= bcosbx and v = --e-sx.
s
Consequently,
(2)

.C[sinbx] = -(-1 sinbx)e-sx 1= +-b 1= (cosbx)e-sx dx = -b 1= (cosbx)e-sx dx
s 0 s 0 s 0