= Solution
An impulse at $a$ shifts the causal <sine> response. Substituting the <Dirac delta> in the <convolution> from part (i), or using the <Laplace transform time-shift rule>, gives
$$
\boxed{f(x)=H(x-a)\sin(x-a)=
\begin{cases}0,&0\le x<a,\\\sin(x-a),&x\ge a,\end{cases}}
$$
where $H$ is the <Heaviside step function>. The answer is continuous at $a$, while its first <derivative> jumps from $0$ to $1$. Consequently its second <distributional derivative> contains exactly one unit <Dirac delta> at $a$, giving $f''+f=\delta(x-a)$. Since $a>0$, both <initial conditions> vanish.
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