254 Ordinary Differential Equations
SOLUTION
By the partial fraction expansion, there exist constants A, B , and C such
that
1 A B C
--,,-,--= - + - + ---.
s^2 ( s + 1) s^2 s ( s + 1)
Multiplying by s^2 (s + 1), we obtain
(4) 1=A(s+1) + Bs(s + 1) + Cs^2.
Settings= 0 in (4), we find 1 =A.
Settings= -1 in (4), we find 1 = C.
Then setting s = 1, A = 1, and C = 1 in ( 4), we see that B must satisfy1 = 2 + 2B + 1. Hence, B = -1 and
1 1 -1 1
.C[h(x)] = H(s) = s2(s + 1) = s2 +--;-+ (s + 1).
Taking the inverse Laplace transform, we obtain
h(x)=.C-1 []__] +.c-1 [-1] +.c-1 [-1-] =x-l+e-x.
s^2 s (s + 1)Examples 1 and 2 illustrate that using the convolution to find an inverse
Laplace transform is often simpler than using partial fraction expansion.
EXAMPLE 3 Using the Convolution Theorem
to Solve an Initial Value ProblemUse the Laplace transform method and the convolution theorem to so lve
the initial value problem y' + y = x; y(O) = 0.
SOLUTION
Taking the Laplace transform of the given differential equation, we obtain.C[y' + y] = .C[x]
.C[y'(x)] + .C[y(x)] = .C[x]
1-y(O) + s.C[y(x)] + .C[y(x)] = 2·
sImposing the initial condition y(O) = 0 and combining like terms, we get
1
(s + 1).C[y(x)] = 2.
s