= Solution
{c}
With fixed endpoint values, the <Euler-Lagrange equation> is $F_y-\frac d{dx}F_{y'}=0$. If $F$ has no explicit $x$ dependence, differentiation gives
$$
\frac d{dx}(F-y'F_{y'})
=F_x+y'(F_y-\tfrac d{dx}F_{y'})=0.
$$
Thus the <Beltrami identity> is $F-y'F_{y'}=\text{constant}$.
For this integrand, $F_y=2y$ and $F_{y'}=1+2y'$, so $y''=y$. The extremals are
$$
\boxed{y(x)=Ae^x+Be^{-x},}
$$
where the two fixed endpoint values determine $A,B$. The Beltrami constant is $y^2-y'^2$, consistent with this solution.
For a nonzero smooth perturbation $\eta$ vanishing at the endpoints, expand the quadratic functional. The first variation vanishes at an extremal, while
$$
J[y+\epsilon\eta]-J[y]
=\epsilon^2\int_{x_1}^{x_2}(\eta^2+\eta'^2)\,dx>0.
$$
The linear term $\int y'$ depends only on the fixed endpoints. Consequently the extremal is a strict minimum and \b[cannot be a maximum].
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