= Solution
For a variation $y+\varepsilon\eta$,
$$
\left.\frac d{d\varepsilon}I[y+\varepsilon\eta]\right|_{\varepsilon=0}
=\int_0^{x_0}\left(F_y\eta+F_{y'}\eta'\right)dx.
$$
Integration by parts gives
$$
\delta I=\left[F_{y'}\eta\right]_0^{x_0}
+\int_0^{x_0}\left(F_y-\frac d{dx}F_{y'}\right)\eta\,dx.
$$
For fixed endpoints, $\eta(0)=\eta(x_0)=0$. The <fundamental lemma of the calculus of variations> therefore gives the <Euler-Lagrange equation>
$$
\boxed{\frac d{dx}F_{y'}-F_y=0}.
$$
A solution makes the first variation vanish for every admissible variation, so it is a stationary candidate; whether it is a minimum or maximum is decided by higher variations. If endpoint values are free, $\eta$ is arbitrary there, and the boundary term instead vanishes under the natural conditions
$$
\boxed{F_{y'}(0)=F_{y'}(x_0)=0}.
$$
Thus the same Euler-Lagrange solution is stationary for all free-endpoint variations. These are the <natural boundary conditions for a free endpoint>.
For
$$
F=y'^2+z'^2+2yz,
$$
the two equations are
$$
\boxed{y''=z,\qquad z''=y}.
$$
Set $u=y+z$ and $v=y-z$. Then
$$
u''=u,
\qquad
v''=-v.
$$
The conditions $y(0)=z(0)=0$ give
$$
u=A\sinh x,
\qquad
v=B\sin x,
$$
and hence the most general solution is
$$
\boxed{
y=\frac12(A\sinh x+B\sin x),
\qquad
z=\frac12(A\sinh x-B\sin x)}.
$$
Free conditions at $x_0$ are $y'(x_0)=z'(x_0)=0$. Adding and subtracting them gives
$$
A\cosh x_0=0,
\qquad
B\cos x_0=0.
$$
Thus $A=0$. The zero solution exists for every $x_0$, while nonzero solutions exist precisely when
$$
\boxed{x_0=\left(k+\frac12\right)\pi,
\qquad k=0,1,2,\ldots}.
$$
For those values they form the one-parameter family
$$
\boxed{y=C\sin x,
\qquad z=-C\sin x},
$$
where $C$ is arbitrary. This is the <free-endpoint normal mode of a coupled variational functional>.
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