The Laplace Transform MethodEXAMPLE 2 Solving an Initial Value Problem with an
Impulse Forcing FunctionSolve the initial value problemy" + 2y' + 5y = M(x - 2); y(O) = 4, y' (0) = 0.
SOLUTION
271The solution process is the same as for example l. Taking the Laplace
transform of the given differential equation and imposing the initial conditions,
we find successively
L'.[y" + 2y' + 5y] = L'.[M(x - 2)]
L'.[y"(x)] + 2£[y'(x)] + 5£[y(x)] = 5e-^28
-y'(O) - sy(O) + s^2 L'.[y(x)] - 2y(O) + 2sl'.[y(x)] + 5£[y(x)] = 5e-^28
-4s - 8 + (s^2 + 2s + 5)£[y(x)] = 5e-^28.
Solving for L'. [y( x) J and using partial fraction expansion and information found
in Table 5.1, we see
£[ (x)] = 5C
28
+ 4s + 8
y s^2 + 2s + 5= ~e-^28 L'.[e-x sin 2x] + 4L'.[e-x cos 2x] + 2£[e-x sin 2x].
Applying the second translation property yields
L'.[y(x)] = L'.[-u(^5 x - 2)e-(x-^2 ) sin2(x - 2) + 4e-x cos2x + 2e-x sin2x].
2Hence, the solution of the initial value problem is