Solution
= Solution
For this forcing the <integral> in part (a) simplifies to
$$
y_p(x)=e^{-2x}\int_0^x\sin(x-t)\,dt=e^{-2x}(1-\cos x),\qquad C=-\int_0^{\pi/2}\cos t\,dt=-1.
$$
Consequently the unique <boundary value problem> solution is
$$
\boxed{y(x)=e^{-2x}(1-\sin x-\cos x).}
$$
It vanishes at zero and at $\pi/2$. To check the <differential equation> directly, set $y=e^{-2x}w$. Then $Ly=e^{-2x}(w''+w)$. For $w=1-\sin x-\cos x$, one has $w''+w=1$, as required.