= Solution
{c}
The <homogeneous solution> $h(x)=e^{-x}\cos x$ already has the prescribed endpoint values. Thus $z=y-h$ has zero <boundary conditions> and satisfies $\mathcal Lz=f$. The <Green's function> gives
$$
\boxed{y(x)=e^{-x}\cos x+\int_0^{\pi/2}g(x,\xi)f(\xi)\,d\xi}.
$$
Equivalently, splitting the <integral> at $\xi=x$ gives the explicit formula
$$
y(x)=e^{-x}\left[
\cos x\left(1-\int_0^xe^\xi\sin\xi\,f(\xi)\,d\xi\right)
-\sin x\int_x^{\pi/2}e^\xi\cos\xi\,f(\xi)\,d\xi
\right].
$$
Any difference between two solutions would be $e^{-x}(C\cos x+D\sin x)$ with zero endpoint values, forcing $C=D=0$. The displayed solution is therefore unique.
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