 Solution

Write $Y={\sum }_{n\ge 0}{b}_n{y}_n$ in the complete Sturm-Liouville eigenfunction expansion. To determine $b_m$, multiply the differential equation by $y_m$ and integrate. The Sturm-Liouville operator has the symmetry identitybecause both functions obey the homogeneous Dirichlet boundary conditions. Consequentlyso that $({\lambda }_m-\alpha )b_m=a_m$. The stipulated exclusion of every eigenvalue makes all denominators nonzero, and givesThis is the resolvent operator applied to $f$ in the weighted space. The same projection shows uniqueness: a homogeneous solution would have $({\lambda }_m-\alpha )b_m=0$ for every $m$, hence would vanish by completeness. For the regular problem the formula gives the solution in the operator domain when $f$ is square-integrable.