= Solution
{c}
Applying the derivative rule twice and using both zero initial data gives
$$
(p^2+1)\bar x(p)=\frac1{p^2+1},
$$
so $\bar x(p)=1/(p^2+1)^2$. By the <convolution theorem for Laplace transforms>,
$$
x(t)=\int_0^t\sin s\,\sin(t-s)\,ds.
$$
Use the product-to-sum identity $\sin s\sin(t-s)=\frac12[\cos(2s-t)-\cos t]$ to integrate:
$$
\boxed{x(t)=\tfrac12(\sin t-t\cos t).}
$$
It satisfies $x(0)=x'(0)=0$, and differentiation gives $x''+x=\sin t$. The term proportional to $t\cos t$ is the resonant response of the forced <harmonic oscillator>.
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