 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 $\delta J\ne 0$, 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 givesTaking compactly supported variations and applying the fundamental lemma of the calculus of variations yields the Euler-Lagrange equation. For the quadratic integrands, it isor $\mathcal{L}y=\lambda ry$ with $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }-qy$. If endpoint values are free, independent endpoint variations give the natural conditionsAt a prescribed endpoint the variation instead vanishes and no natural condition is added there.

Under these natural conditions, integration by parts givesHence $\boxed{I=\lambda J}$. For two solutions satisfying the same natural conditions, subtract their cross-multiplied equations and integrate:The final equality is the vanishing endpoint boundary form. Thus distinct eigenvalues giveThe same argument works for other common homogeneous boundary conditions making that boundary form zero.

In the Legendre case, $((1-x^2)y_n^{\prime }{)}^{\prime }+{\lambda }_n{y}_n=0$. Comparing coefficients of $x^n$ in a degree-$n$ solution gives ${\lambda }_n=n(n+1)$, so different degrees have different 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 $y_1=C_{1}x$. Write $y_2=A{x}^2+Bx+C$. Orthogonality to $x$ gives $B=0$, and orthogonality to $1$ gives $2A/3+2C=0$. ThereforeThe constants are free normalization factors; for solutions of the stated exact degrees they are nonzero. The natural endpoint conditions hold automatically for these polynomials because $p(\pm 1)=0$.