Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 13A / a / Solution

The Lagrange identity for this Sturm-Liouville theory operator isIntegrating from zero to one, the boundary term vanishes because both $y$ and $y_0$ vanish at both endpoints. Since $L(y_0;{\lambda }_0)=0$ and $L(y;{\lambda }_0)=f$, it follows thatThis is the solvability condition at a Sturm-Liouville eigenvalue.