= Solution
{c}
For $\sin l\ne0$, use <variation of parameters> with the <homogeneous solutions> $\cos x,\sin x$, whose <Wronskian> is one. Write $y=A(x)\cos x+B(x)\sin x$ and impose $A'\cos x+B'\sin x=0$. The <differential equation> gives $-A'\sin x+B'\cos x=f$, hence
$$
A'=-f\sin x,\qquad B'=f\cos x.
$$
After integrating from zero and using the left <boundary condition>,
$$
y(x)=C\sin x+\int_0^x\sin(x-t)f(t)\,dt.
$$
The right <boundary condition> determines
$$
C=-\frac1{\sin l}\int_0^l\sin(l-t)f(t)\,dt,
$$
so the unique solution is
$$
\boxed{y(x)=\int_0^x\sin(x-t)f(t)\,dt
-\frac{\sin x}{\sin l}\int_0^l\sin(l-t)f(t)\,dt.}
$$
Combining the two <integrals> gives $y(x)=\int_0^lG(x,t)f(t)\,dt$, with <Green's function>
$$
\boxed{G(x,t)=-\frac1{\sin l}
\begin{cases}
\sin x\sin(l-t),&0\leq x\leq t,\\
\sin t\sin(l-x),&t\leq x\leq l.
\end{cases}}
$$
It is symmetric in $x,t$, as expected from <self-adjointness>, and has zero endpoint values. At $x=t$ it is continuous, while its first <derivative> obeys the <jump condition>
$$
G_x(t^+,t)-G_x(t^-,t)
=\frac{\sin t\cos(l-t)+\cos t\sin(l-t)}{\sin l}=1.
$$
Thus $(\partial_x^2+1)G=\delta(x-t)$, with the positive normalization of the <Dirac delta function> required for $\mathcal Ly=f$.
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