270 Ordinary Differential Equations
by an impulse force. Taking the Laplace transform of the given differential
equation and imposing the initial conditions, we find successively
.C[y" + 2y' + 5y] = £[38(x - 1)]
.C[y"(x)J + 2.C[y'(x)] + 5.C[y(x)J = 3e-s
-y' (0) - sy(O) + s^2 .C[y(x )] - 2y(O) + 2s.C[y(x )] + 5.C[y(x )] = 3e-s
(s^2 + 2s + 5).C[y(x)] = 3e-s.
Solving for .C [y( x) J and using information found in Table 5 .1, we see
3e-s
.C[y(x)] = s2 + 2s + 53e-s 2 3 _ 8 £[ -x. 2 J
2 (s+l)2+22 = 2e e sm x.
Applying Theorem 5.3, yields
.C[y(x)] = ~e-s .C[e- x sin2x] = .C[~u(x - l)e-(x-1) sin2(x - 1)].
2 2Hence, the solution of t he initial value problem is
y(x) = ~u(x - l)e-(x-l) sin 2(x - 1) =
{^30,2
2e-(x-l) sin2(x -1), l<x
A graph of t his solution is displayed in Figure 5.8. For 0 ::; x < 1, the initial
1 y0.5
x0
4 5
-0.5
- 1
3
Figure 5.8 Graph of t he Solution y(x) = -u(x - l )e-Cx-l) sin2(x - 1)
2value problem is y" + 2y' + 5y = O; y(O) = 0, y' (0) = 0 and it obviously
has the unique solution y(x) = 0, which can easily be seen in Figure 5.8. The
portion of the graph to the right of x = 1 is due to the impulse force which was
applied to the system at the instant x = 1. The "solution" y(x) is continuous
for x 2". 0 but the first derivative has a jump discontinuity at x = 1 and the
second derivative has an infinite discontinuity there.