= Solution
At a regular constrained extremum, choose an admissible <variation> $k$ with $\delta J[k]\ne0$. The <implicit function theorem> applied to $J[y+\varepsilon h+s k]$ permits a compensating $s(\varepsilon)$ keeping $J$ fixed. Differentiation at zero gives $\delta I[h]=\lambda\delta J[h]$ for every $h$, where $\lambda=\delta I[k]/\delta J[k]$. Integrating the <first variation> of $I-\lambda J$ by parts and using arbitrary compactly supported $h$ gives
$$
\boxed{\frac d{dx}(F_{y'}-\lambda G_{y'})-(F_y-\lambda G_y)=0.}
$$
Smoothness alone does not exclude an abnormal constraint: if $\delta J$ vanishes identically, one needs the general multiplier form with two multipliers rather than automatically setting the coefficient of $I$ to one. The displayed normal form uses the usual regular-constraint hypothesis.
For the given quadratic functionals, the equation is $\boxed{(py')'+qy+\lambda ry=0}$. If the endpoint values of $y$ are free, the boundary <variation> $[py'h]_a^b$ gives the natural conditions $\boxed{p(a)y'(a)=p(b)y'(b)=0}$, interpreted by limits at singular endpoints. Multiply the equation by $y$ and integrate. These boundary terms vanish, leaving
$$
\int_a^b(p(y')^2-qy^2)dx=\lambda\int_a^b ry^2dx,\qquad\boxed{I=\lambda J}.
$$
For two eigenfunctions, multiply their equations by the other function and subtract. Integration gives
$$
[p(y_ny_m'-y_my_n')]_a^b+(\lambda_m-\lambda_n)\int_a^b ry_my_n\,dx=0.
$$
The natural conditions remove the bracket, so distinct eigenvalues imply $\boxed{\int_a^b ry_my_n\,dx=0}$.
In the Legendre case, comparison of leading <polynomial> coefficients gives $\lambda_n=n(n+1)$, so the degrees correspond to distinct eigenvalues. A <vertex degree>[degree]-zero <polynomial> is $y_0=C_0$. Write $y_1=Ax+B$; <orthogonality> to $1$ on $[-1,1]$ gives $B=0$. Write $y_2=Ax^2+Bx+C$. <Orthogonality> to $x$ gives $B=0$, and <orthogonality> to $1$ gives $2A/3+2C=0$. Thus
$$
\boxed{y_0=C_0,\quad y_1=C_1x,\quad y_2=C_2(3x^2-1)/2.}
$$
These <polynomials> satisfy the natural endpoint limits because $1-x^2$ vanishes at both endpoints.
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