Past exam of the differential equations and applications course of the University of Oxford / 2010 / A / 2 / ii / b

The forcing is $\frac{1}{2}\sin 2x$, which lies in the image of $L$. Since the eigenvalue on $\sin 2x$ is $\pi /4$, a particular solution is $y_p=(2/\pi )\sin 2x$. The full family isThus solutions exist and are not unique: adding any element of the infinite-dimensional kernel leaves $Ly$ unchanged. For example every constant can be added.

Every image of $L$ is a linear combination $a\cos 2x+b\sin 2x$. The function $\cos x$ is not such a combination. Indeed these functions are continuous, so equality almost everywhere would imply equality everywhere; evaluation at the endpoints would require simultaneously $a=1$ and $a=-1$. Therefore the forcing is outside the image and there is no solution. Orthogonality to the kernel is the equivalent image test for this self-adjoint finite-rank operator.