= Solution
Write $Y=\sum_{n\geq0}b_ny_n$ in the complete <Sturm-Liouville eigenfunction expansion>. To determine $b_m$, multiply the differential equation by $y_m$ and integrate. The <Sturm-Liouville operator> has the symmetry identity
$$
\int_0^1\big(y_m\mathcal LY-Y\mathcal Ly_m\big)\,dx=\big[p(Yy_m'-y_mY')\big]_0^1=0,
$$
because both functions obey the homogeneous <Dirichlet boundary conditions>. Consequently
$$
\lambda_m\int_0^1wYy_m\,dx=\alpha\int_0^1wYy_m\,dx+\int_0^1wfy_m\,dx,
$$
so that $(\lambda_m-\alpha)b_m=a_m$. The stipulated exclusion of every <eigenvalue> makes all denominators nonzero, and gives
$$
\boxed{Y(x)=\sum_{n=0}^\infty\frac{a_n}{\lambda_n-\alpha}y_n(x).}
$$
This is the <resolvent operator> applied to $f$ in the weighted space. The same projection shows <uniqueness>: a homogeneous solution would have $(\lambda_m-\alpha)b_m=0$ for every $m$, hence would vanish by completeness. For the regular problem the formula gives the solution in the operator domain when $f$ is square-integrable.
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