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Homework Statement:

For ##y"+{\omega}^2y=f(t)## and if ##f(t)=1## for ##0<t<a##; ##f(t)=0## for ##t<0## and ##t>a##; show that:
1. For t < a: ##y=1/\omega^2(1\cos {\omega t)}##
2. For t > a: ##y=1/\omega^2[\cos{\omega(ta)  \cos{\omega t}]##
If intial condition for y=y'=0
Relevant Equations:

Let G(P) be the Laplace transformation of g(t) and H(p) be the Laplace transformation of h(t), then; invers transformation of the product of two Laplace transformation is the convolution of two correspondence function. We define convolution of g(t) and h(t) as follow:
$$y=g*h=\int_{0}^{t} [g(t\tau)h(\tau)] d\tau$$
So, ##L(y)=Y=G(P) H(p)## and therefore ##L^{1} (G(p) H(p))=g*h##
I can use the convolution integrals and get the idea of this concept for t<a. But, I can't get the answer for t>a.
MY idea is substitute ##f(t) = 0## to the ODE, then I have second order linear differential equations with right hand is zero. So, the solution is
$$y=Ae^{i\omega t} + Be^{i\omega t}$$. From the initial condition, I get A = B = 0.
The answer should be:
$$y=\frac{1}{\omega^2} [\cos {\omega (ta)}  \cos {\omega t}]$$
What should I do to get the answer? Thanks
MY idea is substitute ##f(t) = 0## to the ODE, then I have second order linear differential equations with right hand is zero. So, the solution is
$$y=Ae^{i\omega t} + Be^{i\omega t}$$. From the initial condition, I get A = B = 0.
The answer should be:
$$y=\frac{1}{\omega^2} [\cos {\omega (ta)}  \cos {\omega t}]$$
What should I do to get the answer? Thanks