= Solution
Take a <variation> $y+\varepsilon\eta$, where the <differentiable function> $\eta$ obeys
$$
\eta(a)=\eta(b)=\eta'(a)=\eta'(b)=0
$$
because both $y$ and its first <derivative> have fixed endpoint values. The <first variation> of the <functional> is
$$
\delta L
=\int_a^b
\left(F_y\eta+F_{y'}\eta'+F_{y''}\eta''\right)\,dx.
$$
Applying <integration by parts> once to the second term and twice to the third gives
$$
\begin{aligned}
\delta L
={}&\int_a^b
\left[
F_y-\frac d{dx}F_{y'}
+\frac{d^2}{dx^2}F_{y''}
\right]\eta\,dx\\
&+\left[
F_{y'}\eta+F_{y''}\eta'
-\frac d{dx}(F_{y''})\eta
\right]_a^b.
\end{aligned}
$$
The endpoint conditions on $\eta$ and $\eta'$ make every boundary term zero. Since the remaining <integral> vanishes for every admissible variation, the <fundamental lemma of the calculus of variations> yields the <higher-order Euler-Lagrange equation>
$$
\boxed{
F_y-\frac d{dx}F_{y'}
+\frac{d^2}{dx^2}F_{y''}=0
}.
$$
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