Solution
= 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 $\int_0^1e^{2x}y_ny_m=0$ for $n\ne m$. Expanding the source gives coefficients $c_n=2\int_0^1f\sin(n\pi x)dx$, hence
$$
\boxed{a_n=\frac{2[1+(-1)^n]}{n\pi(n^2\pi^2+1)}.}
$$