I'm trying to determine the extremals of the functional $F(x,y,y',y'') = (y')^2 + (y'')^2$.
I know that in this case the first integral satisfies
$$F - y' \left( F_{y'} - \frac{\mathrm{d}}{\mathrm{d}x}F_{y''}\right)-y''F_{y''}=k$$
where $k$ is a constant.
For the aforementioned functional, I can easily see that $F_{y'} = 2y'$ and that $F_{y''} = 2y''$, and furthermore $\frac{\mathrm{d}}{\mathrm{d}x}F_{y''}=2y'''$.
If I put this into the equation satisfied by the extremal, we get $$(y')^2 + (y'')^2 -y'(2y' - 2y''') - y''(2y'')=k$$
and it is then possible to re-write this as
$$(y')^2 + (y'')^2 = 2y'y''' - k$$
I am unsure what to do next. Is there a trick that I can use to solve this nonlinear third order ODE? I tried integrating both sides between 0 and 1 (as those are the two $x$-values that the boundary conditions specify) as I can see that the functional appears on the left hand side of the equation, but the exercise I am doing is asking for the extremals which implies that I need to find $y(x)$ rather than what the stationary value of the integral is.
Calling
$$ F(x,y,y',y'')= (y')^2+(y'')^2 $$
we have according to Euler-Lagrange conditions
$$ F_y-\left(F_{y'}\right)'+\left(F_{y''}\right)'' = -2(y^{(4)}-y'')=0 $$
hence
$$ y = C_1e^x+C_2e^{-x}+C_3 x+ C_4 $$