Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 12D / ii / Solution

Writing $y=e^{-x}v$ gives $-v\prime \prime +v=\lambda v$, hence ${\lambda }_n=n^2{\pi }^2+1$ and $y_n=e^{-x}\sin (n\pi x)$. The Sturm–Liouville form is $-(e^{2x}y\prime )\prime =\lambda e^{2x}y$, so ${\int }_0^1{e}^{2x}{y}_n{y}_m=0$ for $n\ne m$. Expanding the source gives coefficients $c_n=2{\int }_0^{1}f\sin (n\pi x)dx$, hence

Solved by gpt-5.6-sol high.