= Solution
Use the weighted <inner product> $\langle u,v\rangle_r=\int_a^b r(x)u(x)v(x)\,dx$. For real <eigenfunctions>, integration by parts with the zero endpoint values gives
$$
\int_a^b y_m\mathcal Ly_n\,dx
=\int_a^b[p y_m'y_n'+q y_my_n]dx
=\int_a^b y_n\mathcal Ly_m\,dx.
$$
Thus $(\lambda_n-\lambda_m)\langle y_m,y_n\rangle_r=0$. The regular separated <Sturm-Liouville problem> has simple <eigenvalues>: for a fixed <eigenvalue>, prescribing $y(a)=0$ leaves only the initial flux $p(a)y'(a)$ free, so uniqueness of the associated first-order system makes any two <eigenfunctions> proportional. Distinct <eigenfunctions> are therefore orthogonal, and choosing their signs and scales so that $\int r y_n^2=1$ gives
$$
\boxed{\langle y_m,y_n\rangle_r=\delta_{mn}}.
$$
The positive <energy> form $\int(p y'^2+q y^2)$ realizes $r^{-1}\mathcal L$ as a positive self-adjoint operator in this weighted space, with compact inverse on the bounded interval. The <spectral theorem for compact self-adjoint operators> supplies a complete eigenbasis. This is the standard regular <Sturm-Liouville eigenfunction expansion>, also valid with continuous $p$ by working with the continuous flux $py'$.
For a source $F$, expand a zero-endpoint solution of $(\mathcal L-\mu r)y=F$ in that <basis>. Multiplication by $y_n$ and unweighted integration give
$$
(\lambda_n-\mu)\langle y_n,y\rangle_r=\int_a^b y_n(\xi)F(\xi)\,d\xi.
$$
Since $\mu$ is not an <eigenvalue>, every coefficient is determined. The <spectral Green function of a regular Sturm-Liouville problem> is consequently
$$
\boxed{G_\mu(x,\xi)=\sum_{n=1}^\infty\frac{y_n(x)y_n(\xi)}{\lambda_n-\mu}}.
$$
Weighted completeness means $r(x)\sum_n y_n(x)y_n(\xi)=\delta(x-\xi)$ as a distribution, so applying $\mathcal L-\mu r$ to this series produces the required delta source. The source integration is with $d\xi$, explaining the absence of an additional factor $r(\xi)$ in the stated kernel.
For the subsequent operator $L=d^2/dx^2$ the sign is explicitly reversed: $y_n=\sqrt{2/\pi}\sin(nx)$ and $\lambda_n=-n^2$. To cover every real non-<eigenvalue> $\mu$, define the homogeneous solution
$$
S_\mu(x)=\begin{cases}\sinh(\sqrt\mu\,x)/\sqrt\mu,&\mu>0,\\x,&\mu=0,\\\sin(\sqrt{-\mu}\,x)/\sqrt{-\mu},&\mu<0.\end{cases}
$$
It satisfies $S_\mu(0)=0$, $S_\mu'(0)=1$ and $S_\mu''=\mu S_\mu$. Matching the two endpoint solutions continuously at $x=\xi$ with derivative jump $G_x(\xi^+,\xi)-G_x(\xi^-,\xi)=1$ yields
$$
\boxed{G_\mu(x,\xi)=-\frac{S_\mu(x_<)S_\mu(\pi-x_>)}{S_\mu(\pi)},\qquad
x_<=\min(x,\xi),\quad x_>=\max(x,\xi)}.
$$
The denominator is nonzero precisely off the Dirichlet <eigenvalues> $\mu=-n^2$. In particular $G_0=-x_<(\pi-x_>)/\pi$. At $x=\xi=\pi/2$, its spectral expansion gives
$$
-\frac\pi4=-\frac2\pi\sum_{n=1}^\infty\frac{\sin^2(n\pi/2)}{n^2}
=-\frac2\pi\sum_{j=0}^\infty\frac1{(2j+1)^2}.
$$
This <odd reciprocal-square sum from a Dirichlet Green function> proves $\boxed{\sum_{j=0}^\infty(2j+1)^{-2}=\pi^2/8}$.
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