= Solution
{c}
Let $\eta$ be a smooth <variation> with $\eta=\eta'=0$ at both endpoints, and consider $y_\epsilon=y+\epsilon\eta$. <stationary point> gives the <first variation>
$$
0=\left.\frac d{d\epsilon}J[y_\epsilon]\right|_{\epsilon=0}
=\int_{x_1}^{x_2}
\left(F_y\eta+F_{y'}\eta'+F_{y''}\eta''\right)dx.
$$
Integrate the middle term once by parts and the last term twice. All boundary terms vanish because both $\eta$ and $\eta'$ vanish there. Assuming the derivatives needed for the classical formula, the given <fundamental lemma of the calculus of variations> yields
$$
\boxed{F_y-\frac d{dx}F_{y'}+
\frac{d^2}{dx^2}F_{y''}=0.}
$$
Without assuming those classical derivatives initially, the same calculation with compactly supported variations states this equation weakly; sufficient regularity then gives the classical equation.
For the specific integrand, $F_y=-2y$, $F_{y'}=0$, $F_{y''}=2y''$, so the equation is $y^{(4)}-y=0$. Its <weak solutions> here are smooth, and its general classical solution is
$$
y=A\cos x+B\sin x+C\cosh x+D\sinh x.
$$
Put $\ell=\pi/2$, $c=\cosh\ell$, $s=\sinh\ell$. The conditions at zero give $C=-A$ and $D=1-B$. Those at $\ell$ give
$$
B-Ac+(1-B)s=0,\qquad
-A-As+(1-B)c=0.
$$
Using $c^2-s^2=1$, solve to obtain $A=c/2$, $B=(1-s)/2$, $C=-c/2$, $D=(1+s)/2$. Thus an <extremal> is
$$
\boxed{y(x)=\frac12\left[
c(\cos x-\cosh x)+(1-s)\sin x+(1+s)\sinh x
\right].}
$$
Substitution checks all four boundary values. The additive term $x^2$ in the integrand contributes no <variation> in $y$, so it does not alter this equation.
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