Past exam of the differential equations and applications course of the University of Oxford / 2011 / A / 2 / b / ii

Since ${\lambda }_1=-4/5$, the kernel of a linear map $L+\frac{4}{5}I$ is the span of $\psi (x)=\cos (3x/2)$. The self-adjoint Fredholm alternative requires $⟨f,\psi ⟩=0$ for solvability. If it holds, an arbitrary multiple of $\psi$ can be added, so a solution is not unique.

For $f=\cos x$,so case (A) has no solution. For $f=\sin x\sin (3x/2)$,by sine-function orthogonality. Thus case (B) has solutions, and they are nonunique.