Finding the Euler Lagrange equation - differentiation I'm teachin myself the basics of Calculus of variations. So far I know how to calculate the Euler Lagrange equation for simple functionals. 
I'm now stuck on how to compute the total differentiation of the following problem:
$$I[y]=\int_0^1 (y\frac{dy}{dx})^2 -\lambda y^2 \ dx$$
To calculate the Euler Lagrange equation I have the following:
$$\frac{d}{dx}(2y^2y')-2y^2\frac{dy}{dx}+2\lambda y=0$$.
Is this correct? If so I'm unsure of how to evaluate the total differentiation part. That is taking the total derivative in this case. 
 A: 
For a functional
  $$
I[y]=\int_a^bf(x,y,y')
$$
  the Euler-Lagrange equation is defined as
  $$
\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)=0
$$

For the $I$ you gave above, $f(x,y,y')=(yy')^2-\lambda y^2$, hence
the E-L equation for $I$ is
$$
2y\left(\frac{dy}{dx}\right)^2-2\lambda y-\frac{d}{dx}\left(2y^2\frac{dy}{dx}\right)=0
$$
Your only mistake is that $\frac{\partial f}{\partial y}=2\color{red}{y(y')^2}-2\lambda y$, not $2\color{red}{y^2y'}-2\lambda y$ as you have written. Does this help you go from here?

Edit. Since $f$ does not depend upon $x$, we can simplify the problem.


Lemma. For a functional
  $$
I[y]=\int_a^b f(y,y')
$$
  which is independent of $x$, we have that
  $$
\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=0
$$

Proof. Just calculate the derivative
$$
\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=y''\frac{\partial f}{\partial y'}+y'\frac{d}{x}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial x}-y'\frac{\partial f}{\partial y}-y''\frac{\partial y}{\partial y'}
$$
Since $f$ is independent of $x$, $\frac{\partial f}{\partial x}=0$, and we can take out a factor of $y'$
$$
\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=y'\left(\frac{d}{dx}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y}\right)
$$
which is clearly equal to $0$ by E-L. QED

This means that
$$
y'\frac{\partial f}{\partial y'}-f=\text{constant}
$$
Applying this to your function $I$ gives
$$
\begin{align*}
y'(2y^2y')-(yy')^2+\lambda y^2&=C \\
y^2\left(\left(\frac{dy}{dx}\right)^2+\lambda\right)&=C \\
\frac{dy}{dx}&=\sqrt{\frac{C}{y^2}-\lambda}
\end{align*}
$$
