= Solution
{c}
The <homogeneous solution> is $A\cos x+B\sin x$. The left <boundary condition> removes $A$, and the right <boundary condition> gives $B\sin l=0$. Thus the <kernel of a linear map> associated with this <boundary value problem> is
$$
\boxed{\ker\mathcal L=
\begin{cases}
\{0\},&l\notin\pi\mathbb N_{>0},\\
\operatorname{span}\{\sin x\},&l=m\pi,\quad m\in\mathbb N_{>0}.
\end{cases}}
$$
The <Fredholm alternative> now gives the complete classification:
$$
\boxed{\begin{array}{c|c|c}
l&\text{condition on }f&\text{solutions}\\\hline
l\notin\pi\mathbb N_{>0}&\text{any prescribed }f&\text{exactly one}\\
l=m\pi&\displaystyle\int_0^{m\pi}f(x)\sin x\,dx=0&\text{infinitely many}\\
l=m\pi&\displaystyle\int_0^{m\pi}f(x)\sin x\,dx\ne0&\text{none}.
\end{array}}
$$
For necessity at $l=m\pi$, multiply by the <homogeneous solution> $\sin x$ and use the <Lagrange identity for a Sturm-Liouville operator> to obtain $\int_0^lf\sin x\,dx=0$. For sufficiency, the <particular solution>
$$
z(x)=\int_0^x\sin(x-t)f(t)\,dt
$$
satisfies $z(0)=0$ and $z''+z=f$. At the other endpoint,
$$
z(m\pi)=(-1)^{m+1}\int_0^{m\pi}\sin t\,f(t)\,dt,
$$
which is zero under the compatibility condition. All solutions are then $z+B\sin x$, with $B$ arbitrary. The assumed <Lipschitz continuity> of $f$ in particular makes $f$ continuous, so the displayed <integral> defines a classical solution. The length $l=0$ is excluded: it would not define a nonempty interval for this <boundary value problem>.
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