= Solution
Set $y=e^{-x/2}v$, the <Dirichlet gauge transform for constant drift>. Direct differentiation gives
$$
y''+y'=e^{-x/2}(v''-v/4).
$$
The homogeneous endpoint conditions on $y$ become $v(0)=v(1)=0$. Thus the eigenvalue equation is $v''=(\lambda+1/4)v$. To exclude other branches without assuming the eigenvalue real, multiply by $\bar v$ and integrate by parts:
$$
(\lambda+1/4)\int_0^1|v|^2\,dx=-\int_0^1|v'|^2\,dx<0
$$
for a nonzero <eigenfunction>. Therefore $\lambda+1/4=-k^2$ with $k>0$. The boundary conditions in $v=A\sin(kx)+B\cos(kx)$ give $B=0$ and $k=n\pi$, $n\ge1$. Hence
$$
\boxed{\lambda_n=-n^2\pi^2-\frac14,\qquad y_n(x)=e^{-x/2}\sin(n\pi x).}
$$
Equivalently, $L[y]=\lambda y$ is the <Sturm-Liouville form> $(e^xy')'=\lambda e^xy$, with positive weight $e^x$. Its <orthogonality> relation is
$$
\boxed{\int_0^1e^x y_m(x)y_n(x)\,dx=\frac12\delta_{mn}.}
$$
It is the weighted inner product, not the unweighted one, that makes these <eigenfunctions> orthogonal.
Expand $xe^{-x/2}=\sum_nb_ny_n$ with the coefficients from part a. Completeness and the nonzero <eigenvalues> give the <Sturm-Liouville eigenfunction expansion>
$$
\boxed{y(x)=-e^{-x/2}\sum_{n=1}^{\infty}
\frac{2(-1)^{n+1}}{n\pi(n^2\pi^2+1/4)}\sin(n\pi x).}
$$
The coefficients are $O(n^{-3})$, so the function series converges uniformly and vanishes at both endpoints. Its first derivative also converges uniformly. Applying the transformed operator in the square-integrable sense gives the part-a expansion of $x$; equivalently $v''-v/4=x$ on the interior. Integrating this continuous right-hand side gives the classical twice-differentiable solution there. No nonzero homogeneous solution satisfies both endpoints, so this solution is unique. The mismatch of the forcing's sine series at a single endpoint does not affect the interior equation or these boundary conditions.
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