Solution
= Solution
For
$$
F(x,y,y')=y'^2+yy'+y'+y^2+yx^2,
$$
the <Euler-Lagrange equation> is
$$
\frac d{dx}(2y'+y+1)-(y'+2y+x^2)=0,
$$
or
$$
y''-y=\frac{x^2}{2}.
$$
The general solution of this <inhomogeneous linear differential equation> is
$$
y(x)=Ce^x+De^{-x}-\frac{x^2}{2}-1.
$$
The condition $y(0)=-1$ gives $C+D=0$, and the condition at $x=1$ gives $C=1$. Therefore
$$
\boxed{y(x)=e^x-e^{-x}-\frac{x^2}{2}-1}.
$$