= Solution
Set $y=e^{-x}u$. Then
$$
y''+2y'=e^{-x}(u''-u),
$$
so the eigenvalue equation becomes
$$
u''+(\lambda-1)u=0,
\qquad u(0)=u(1)=0.
$$
The <Dirichlet eigenvalue>[Dirichlet eigenvalues] and corresponding eigenfunctions are therefore
$$
\boxed{\lambda_n=1+n^2\pi^2,\qquad
y_n(x)=e^{-x}\sin(n\pi x),\quad n\geq1}.
$$
They form an infinite discrete increasing set. Their weighted inner products satisfy
$$
\int_0^1w\,y_ny_m\,dx
=\int_0^1\sin(n\pi x)\sin(m\pi x)\,dx
=\frac12\delta_{nm}.
$$
Thus the <Sturm-Liouville eigenfunction expansion> coefficient is
$$
\boxed{
A_n=
\frac{\int_0^1e^{2x}(x-x^2)y_n(x)\,dx}
{\int_0^1e^{2x}y_n(x)^2\,dx}
=2\int_0^1e^x(x-x^2)\sin(n\pi x)\,dx }.
$$
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