= Solution
At a regular constrained extremal, the first variations satisfy $\delta I=\lambda\delta J$ for a constant <Lagrange multiplier> $\lambda$. One way to justify this is to choose a variation for which $\delta J\neq0$, use it to correct any other variation to preserve the constraint by the <implicit function theorem>, and differentiate $I$ along that corrected family. This gives the same ratio $\lambda$ for all variations. The exceptional case $\delta J=0$ for every variation requires the usual separate abnormal-extremal analysis.
Set $L=F-\lambda G$. Integration by parts in its <first variation> gives
$$
\delta(I-\lambda J)
=\left[L_{y'}\eta\right]_a^b+
\int_a^b\left(L_y-\frac d{dx}L_{y'}\right)\eta\,dx.
$$
Taking compactly supported variations and applying the <fundamental lemma of the calculus of variations> yields the <Euler-Lagrange equation>. For the quadratic integrands, it is
$$
\boxed{(py')'+(q+\lambda r)y=0,}
$$
or $\mathcal Ly=\lambda ry$ with $\mathcal L y=-(py')'-qy$. If endpoint values are free, independent endpoint variations give the natural conditions
$$
\boxed{p(a)y'(a)=p(b)y'(b)=0.}
$$
At a prescribed endpoint the variation instead vanishes and no natural condition is added there.
Under these natural conditions, integration by parts gives
$$
2I=\int_a^b(py'^2-qy^2)\,dx
=[py'y]_a^b+\int_a^by\mathcal Ly\,dx
=\lambda\int_a^bry^2\,dx=2\lambda J.
$$
Hence $\boxed{I=\lambda J}$. For two solutions satisfying the same natural conditions, subtract their cross-multiplied equations and integrate:
$$
(\lambda_m-\lambda_n)\int_a^bry_my_n\,dx
=\int_a^b(y_n\mathcal Ly_m-y_m\mathcal Ly_n)\,dx=0.
$$
The final equality is the vanishing endpoint boundary form. Thus distinct <eigenvalue>[eigenvalues] give
$$
\boxed{\int_a^b r y_m y_n\,dx=0.}
$$
The same argument works for other common homogeneous boundary conditions making that boundary form zero.
In the Legendre case, $((1-x^2)y_n')'+\lambda_ny_n=0$. Comparing coefficients of $x^n$ in a degree-$n$ solution gives $\lambda_n=n(n+1)$, so different degrees have different <eigenvalue>[eigenvalues]. A degree-zero solution is $y_0=C_0$. A linear solution $Ax+B$ must be orthogonal to the nonzero constant solution, so $B=0$ and $y_1=C_1x$. Write $y_2=Ax^2+Bx+C$. <Orthogonality> to $x$ gives $B=0$, and <orthogonality> to $1$ gives $2A/3+2C=0$. Therefore
$$
\boxed{y_2=C_2\frac{3x^2-1}{2}.}
$$
The constants are free <normalization> factors; for solutions of the stated exact degrees they are nonzero. The natural endpoint conditions hold automatically for these <polynomial>[polynomials] because $p(\pm1)=0$.
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