= Solution
Denote the unshifted differential expression by $L_0y=(py')'+qy$. The supplied <Green's function> satisfies $L_0G(x,\xi)=\delta(x-\xi)$, whereas the inhomogeneous equation is $L_0y=\lambda wy+wf$.
Multiply the latter equation by $G$ and the Green equation by $y$, subtract and integrate in $x$:
$$
\int_a^b\left[G L_0y-yL_0G\right]dx
=\lambda\int_a^bwGy\,dx+\int_a^bwGf\,dx-y(\xi).
$$
The left-hand side is the boundary term
$$
\left[p(Gy'-yG_x)\right]_a^b.
$$
It vanishes because $G$ and $y$ satisfy the same homogeneous <Robin boundary conditions>. For example, where $\alpha\ne0$, substituting $y'=-\beta y/\alpha$ and $G_x=-\beta G/\alpha$ cancels the endpoint expression; where $\alpha=0$, both values vanish. The other endpoint works identically. Hence
$$
\boxed{\int_a^b f(x)w(x)G(x,\xi)\,dx
+\lambda\int_a^b w(x)G(x,\xi)y(x)\,dx-y(\xi)=0.}
$$
The <Dirac delta function> contribution is valid for an interior evaluation point $\xi$; equivalently, $G$ is continuous there and has flux-derivative jump $[pG_x]_{\xi^-}^{\xi^+}=1$.
This uses the existence of the unshifted <Green's function> stated in the question. Such a function is not automatic for every choice of $p,q$ and endpoint conditions: a nontrivial <zero mode> of $L_0$ would obstruct an ordinary inverse and would require a compatibility condition and a modified <Green's function>. Under the stated inverse assumption, the <differential equation> and the displayed <integral equation> are equivalent for functions in the boundary-condition domain.
Back to article page
