226 Ordinary Differential Equations
provided s > 0. Now, we use integration by parts a second time by letting u =
cos bx and dv = e-sxdx. Then du= -bsinbx and v = -~e-sx. Integrating
s
the integral appearing on the right-hand side of (2) by parts, we find
(3)
b{-1 loo b1
00
.C[sinbx] = - -(cosbx)e-sx - - (sinbx)e-sxdx } = b b2.
SS O So^2 SS -^2 .C[smbx]
provided s > 0. Solving equation (3) algebraically for .C[sinbx], we obtain
. b/ s^2 b
.C [ sm bx] = 1 + b2 / s2 s2 + b2 for s > 0.
In the previous three examples, we used the definition of the Laplace trans-
form to calculate the transform of three different functions- eax, xn, and
sin bx. In elementary calculus, we learned the definition of the derivative of a
function, calculated the derivative of a few functions from the definition, and
then learned rules for the differentiation of the sum, difference, product, and
quotient of two functions. These rules allowed us to differentiate a variety of
functions without explicitly using the definition of the derivative. The opera-
tion of differentiating a function transforms the function f ( x) into the function
f^1 ( x). If the operation of differentiation is denoted by D, then the transforma-
t ion can be written as D [f ( x)] = f' ( x). The function f' ( x) is the transform of
f(x) under the transformation D. Thus, for example, D[x^3 ] = 3x^2. Another
transformation we encountered in calculus was integration. The operation of
integration transforms the function j(x) into the function F(x) =fox f(t) dt.
If we let I denote integration, then this transformation can be written as
I[f(x)] =fox f(t)dt = F(x). For instance, J[x^3 ] = x^4 /4.
DEFINITION Linear OperatorAn operator Tis linear if for every pair of functions f(x) and g(x) and
for every pair of constants c 1 and c2,T[cif(x) + c2g(x)] = c1T[f(x) ] + c2T[g(x)].
Differentiation and integration are both linear operators. That is, for any two
differentiable functions f(x) and g(x) and any two constants c 1 and c 2
D[cif(x) + c2g(x)] = c1D[f(x)] + c2D[g(x)].
Likewise, for any two integrable functions f(x) and g(x) and any two constants
c1 and c2
I[cif(x) + c2g(x)] = c1J[f(x)] + c2J[g(x)].