Answer
$x(t)=-4+4t^{2}+2\cos \sqrt 2t$
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$L[x(t)]=2.\frac{2}{s^3}+L[\sin 2t * x(t)]\\
=\frac{4}{s^3}+L[t].L[x(t)]\\
=\frac{4}{s^3}+\frac{2L[x(t)]}{s^2+4}$
then $L[x(t)]=\frac{4(s^2+4)}{s^3(s^2+2)}=-\frac{2}{s}+\frac{8}{s^3}+\frac{2s}{s^2+2}$
The general solution to the given equation is:
$x(t)=-4+4t^{2}+2\cos \sqrt 2t$
