= Solution
For $t<0$ both inputs vanish. Eliminating $y$ gives $x''+4x=0$, and the data at $t=-\pi$ give
$$
x(t)=\cos2t.
$$
For all $t$, differentiating the first equation and using the second gives
$$
x''+4x=\delta(t)-2H(t-\pi).
$$
Thus $x$ is continuous at zero while $x'$ jumps by one. For $t>0$ the resulting solution is
$$
x(t)=\cos2t+\frac12\sin2t
+\frac12H(t-\pi)(\cos2t-1).
$$
Since $\cos2t=1-2\sin^2t$, this is
$$
x(t)=\frac12\sin2t+1+q(t)\sin^2t,
$$
where
$$
q(t)=\begin{cases}-2,&0<t<\pi,\\-3,&t\geq\pi.
\end{cases}
$$
\b[Thus $a=1/2$, $b=1$, and the sketch of $q$ is a downward unit step at $t=\pi$.]
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