= Solution
Let $K(p)=\mathcal L[k](p)$ and $F(p)=\mathcal L[f](p)$. Apply the <Laplace transform of a derivative> twice and use the two zero <initial conditions>. The <differential equation> becomes
$$
(p^2+1)F(p)=K(p),\qquad
\boxed{F(p)=\frac{K(p)}{p^2+1}.}
$$
The factor $1/(p^2+1)$ is the <Laplace transform> of $\sin x$. The <convolution theorem for Laplace transforms> therefore gives
$$
\boxed{f(x)=(\sin*k)(x)=\int_0^x\sin(x-t)k(t)\,dt.}
$$
This is a causal <convolution>: the value at $x$ uses only the forcing up to $x$. For ordinary continuous $k$, direct <differentiation> gives $f'(x)=\int_0^x\cos(x-t)k(t)\,dt$ and $f''(x)=k(x)-f(x)$, verifying both the <differential equation> and the <initial conditions>. For an impulsive $k$, the same formula is interpreted using <distributions>.
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