= Solution
{c}
Set $a=\pi/2$. To construct the <Green's function> from a <Dirac delta function>, require $\mathcal L_xg(x,\xi)=\delta(x-\xi)$ and zero endpoint values. Away from $x=\xi$, the <homogeneous solutions> satisfying the left and right <boundary conditions> are respectively
$$
u(x)=e^{-x}\sin x,\qquad v(x)=e^{-x}\cos x.
$$
Write $g=A(\xi)u(x)$ for $x<\xi$ and $g=B(\xi)v(x)$ for $x>\xi$. A discontinuity in $g$ would generate a derivative of the <Dirac delta function>, which is absent from the desired equation. Thus $g$ is continuous. Integrating that equation through $\xi$ then gives the <jump condition>
$$
g_x(\xi^+,\xi)-g_x(\xi^-,\xi)=1.
$$
The resulting two equations are $Au(\xi)=Bv(\xi)$ and $Bv'(\xi)-Au'(\xi)=1$. Since the <Wronskian> $uv'-u'v=-e^{-2\xi}$, they give
$$
\boxed{g(x,\xi)=-e^{\xi-x}
\begin{cases}
\sin x\cos\xi,&0\leq x\leq\xi,\\
\sin\xi\cos x,&\xi\leq x\leq\pi/2.
\end{cases}}
$$
Both endpoint values vanish. At $x=\xi$, subtracting the two first <derivatives> gives $\sin^2\xi+\cos^2\xi=1$, verifying the sign and normalization of the <Dirac delta function>.
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