Okay, I have worked this problem a few times and I end up with a solution that I don't think is correct. Any feedback would be awsome.

If possible, find the stationary point of the functional

$\displaystyle \mathcal{J}(y)=\int_{0}^{1}(y^2+x^2y')dx, y(0)=0, y(1)=\alpha$

So I use the Euler-Lagrange Equation

$\displaystyle \mathcal{L}_y-\frac{d}{dx}\mathcal{L}_{y'}=0$ with

$\displaystyle \mathcal{L}(y,y',x)=y^2+x^2y'$

so $\displaystyle \frac{\partial}{\partial y}(y^2+x^2y')=2y$

and $\displaystyle \frac{\partial}{\partial y'}(y^2+x^2y')=x^2$

and $\displaystyle \frac{d}{dx}\bigg[\frac{\partial}{\partial y'}(y^2+x^2y') \bigg]=\frac{d}{dx}x^2=2x$

So now I end up with

$\displaystyle 2y-2x=0 \implies y=x$

I don't think this is correct and it would only be consistant with the boundry values if $\displaystyle \alpha=1$

Thanks