= Solution
Write $\omega=\sqrt2$ and $Y(p)=\overline y(p)$. The zero initial data and the <Laplace transform of a derivative> give $\mathcal L[y'']=p^2Y$. Transforming the forcing with part (b),
$$
\begin{aligned}
(p^2+2)Y
&=\sqrt2\,\frac{p^2-2}{(p^2+2)^2}
+\frac12\frac{\sqrt2}{p^2+2}\\
&=\frac{\sqrt2(3p^2-2)}{2(p^2+2)^2}.
\end{aligned}
$$
Therefore
$$
\boxed{Y(p)=\frac{\sqrt2(6p^2-4)}{4(p^2+2)^3}.}
$$
Part (b) with $\omega=\sqrt2$ gives $\mathcal L[t^2\sin(\sqrt2t)]=2\sqrt2(3p^2-2)/(p^2+2)^3$. Comparing coefficients and using uniqueness of the <Laplace transform>,
$$
\boxed{y(t)=\frac{t^2}{4}\sin(\sqrt2\,t).}
$$
As a direct check, $y'=\tfrac t2\sin(\sqrt2t)+\tfrac{\sqrt2t^2}4\cos(\sqrt2t)$, so both initial values vanish. Differentiating once more cancels the $t^2\sin(\sqrt2t)$ terms in $y''+2y$, leaving $\sqrt2t\cos(\sqrt2t)+\tfrac12\sin(\sqrt2t)$ as required. This also checks the factor $1/4$ in the inverse transform.
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