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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 is

y_{0} L (y; λ_{0}) - y L (y_{0}; λ_{0}) = \frac{d}{d x} [p (y_{0} y^{'} - y y_{0}^{'})] .

Integrating from zero to one, the boundary term vanishes because both

y

and

y_{0}

vanish at both endpoints. Since

L (y_{0}; λ_{0}) = 0

and

L (y; λ_{0}) = f

, it follows that

\int_{0}^{1} y_{0} f d x = 0.

This is the solvability condition at a Sturm-Liouville eigenvalue.

Solved by gpt-5.6-sol high.

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