Past exam of the differential equations and applications course of the University of Oxford / 2008 / A / 3 / d / Solution

The nonzero kernel of $L$ rules out a unique solution for every value of $c$. By the Fredholm alternative, solvability requires orthogonality of the forcing to $\ker L^{\ast }=\mathrm{span}\{3a-5b\}$. Its pairing with $6a+cb$ isThus only $c=-4$ is compatible. In that case $6a-4b=2y_{-4}$ and a particular solution is $-y_{-4}/2$. All solutions areTherefore infinitely many solutions exist for $c=-4$, no solution exists for $c\ne -4$, and no parameter value gives a unique solution.