I need to find the extremal for:
$$ J[y] = \int_{0}^{1}(1+x)(y')^2 ~ dx $$ with $y(0) = 0$ and $y(1)=1$. I then need to find the extremal if the boundary condition at $x = 1$ changed to $y'(1) = 0$.
So far I've created an Euler-Lagrange equation and gotten it down to
$$ 2y''+x2y''+ 2y' = 0.$$ $$y''=(-2y')/(x+2)$$
this is where I think I'm going wrong, because I don't think I should be trying to integrate a side that has both $y'$ and $x$, I got this but I don't think it's right:
$$y'=(-2y'x)/(x+2)+c$$
If you could help me solve the Euler properly I think I can do the other bit fine by myself
EDIT:
$$v=y'$$
$$y''=v'=dv/dx=dv/dy* dy/dx = V*dv/dy $$
$$v'(1+x)=-1$$
$$v'=-1/(1+x) = y'' $$
$$ y' = -ln(x+1) + c $$
$$ y = x -(x+1)ln(x+1)+cx+d$$
From $y(0) =0$ and $y(1)=1$
=> $$y = x -(x+1)(ln(x+1))+2xln2$$
Lagrangian $$L~=~(1+x)\dot{y}^2.$$ Momentum $$p~=~\frac{\partial L}{\partial\dot{y}}~=~2(1+x)\dot{y}$$ is conserved since $y$ is a cyclic variable. So $$\dot{y}~=~\frac{p}{2(1+x)},$$ or $$y(x)~=~ \frac{p}{2}\ln|1+x|+C.$$ The lower BC $y(0)=0$ leads to $C=0$.
The upper BC $y(1)=1$ leads to $p=2/\ln 2$.
The upper BC $\dot{y}(1)=0$ leads to $p=0$.