Answer
See below
Work Step by Step
Given:
$\frac{d^2y}{dt^2}+4y=F_0\cos 3t$
At $t=5$
$Force=-4\delta(t-5)$
The equation becomes:
$\frac{d^2y}{dt^2}+4y=F_0\cos 3t -4\delta(t-5)$
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$[s^2(Y)-sY(0)-y'(0)]+4Y(s)=\frac{F_0s}{s^2+9}-4e^{-5s}\\
s(Y)(s^2+4)=\frac{F_0s}{s^2+9}-4e^{-5s}$
which implies that:
$Y(s)=\frac{F_0s}{s^2+9}-\frac{4e^{-5s}}{s^2+4}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{F_0s}{s^2+9}-\frac{4e^{-5s}}{s^2+4}]$
We finally obtain
$y(t)=\frac{-1}{5}F_0\cos 2t+\frac{1}{5}F_0\cos 3t-2\sin 3(t-5)u_5(t)$
