242 Ordinary Differential Equations5.2 Using the Laplace Transform and Its Inverse
to Solve Initial Value Problems
The Laplace transform method for solving n-th order linear initial value
problems in which the differential equation is homogeneous or nonhomoge-
neous is a three-step process. First, the differential equation is transformedby the Laplace transform into an algebraic equation in s and .C[y(x )]-the
Laplace transform of the solution of the initial value problem. Next, the
unknown in the algebraic equation .C[y(x)] is solved for by algebraic manip-
ulation. And finally, the inverse Laplace transformation is appli ed to obtain
the solution of the initial value problem. The Laplace transform method
immediately yields the solution of n-th order linear homogeneous and nonho-
mogeneous differential equations and n-th order linear initial value problems
in which the differential equation is homogeneous or nonhomogeneous. That
is, one does not have to (1) find the general solution of the associated homo-
geneous differential equation, (2) find a particular solution to the nonhomo-
geneous differential equation, and (3) add these solutions to get the general
solution. In addition, if the problem is a linear initial value problem, the
initial conditions are incorporated in the transforming equations. Hence, one
does not have to find the general solution and then choose the constants to
satisfy the specified initial equations.
Let us formally calculate the Laplace transform of the derivative of the
function y(x). By definition,
(1) .C[y'(x)J = 1= y'(x)e-sxdx.
Letting u = e-sx and dv = y' ( x) dx, and differentiating and integrating, we
find du= -se-sxdx and v = y(x). Then applying the integration by parts
formula to equation (1), yields
If y(x) is of exponential order a as x---> +oo, then for s >a, y(x)e-sx---> 0
as x ---> +oo and, therefore,
.C[y'(x)] = -y(O) + s.C[y(x)] for s >a.
Next, we formally calculate the Laplace transform of the second derivative
of the function y( x). Again we use integration by parts. This time we set
u = e-sx and dv = y"(x) dx, and differentiating and integrating, we find