The <eigenvalues> of $L$ are $\pi/\sqrt2$, $-\pi/\sqrt2$ and zero, none equal to one. On $E=\operatorname{span}\{1,\cos2x\}$ the map $L-I$ is an invertible two-dimensional operator; on $E^\perp$ it is $-I$. Hence it is invertible on all of $L^2(0,\pi)$. Thus $(L-I)y=\sin x$ has \b[exactly one solution]. This is the <finite-rank Fredholm alternative>; no range obstruction survives the nonresonant identity term.

Solution

Past exam of the differential equations and applications course of the University of Oxford / 2009 / A / 2 / B / d / Solution

 Solution (source code)