Euler-Lagrange Equation example I have been working on solving Euler-Lagrange Equation problems in differential equations, specifically in Calculus of Variations, but this one example has me stuck.  I am probably making mistakes in my integration.
I am supposed to solve the Euler-Lagrange equation given that $f(t,x(t),x^{\prime}(t))=f(t,u,v)=\frac{\sqrt{1+v^2}}{u}$.
I know that $f_u=-\frac{\sqrt{1+v^2}}{u^2}$ and $f_v=\frac{v\sqrt{1+v^2}}{u(1+v^2)}$.  Plugging this into the Euler-Lagrange Equation, $f_u-\frac{d}{dt}f_v$ gives
$$\frac{\sqrt{1+(x^{\prime})^2}}{x^2}-\left(\frac{x^{\prime}\sqrt{1+(x^{\prime})^2}}{x(1+(x^{\prime})^2)}\right)^{\prime}=0.$$
I cannot figure out from here how to solve it.  Have I made a mistake up to now?  Could someone help me out in actually solving this for x?
 A: You are on the right track.  Write
$$
  f_v = \frac{v}{u\sqrt{1+v^2}} \,.
$$
Then
$$
  \frac{df_v}{dt} = \frac{v}{\sqrt{1+v^2}}\frac{d}{dt}\left(\frac{1}{u}\right) + \frac{1}{u}\frac{d}{dt}\left(\frac{v}{\sqrt{1+v^2}}\right)
$$
Expand:
$$
  \frac{df_v}{dt} = -\frac{v}{u^2\sqrt{1+v^2}}\frac{du}{dt} +\left[\frac{1}{u\sqrt{1+v^2}}-\frac{v^2}{u(1+v^2)^{3/2}}\right]\frac{dv}{dt}
$$
Equate to $f_u$ to get
$$
 -\frac{1+v^2}{u} =  -\frac{v}{u}\frac{du}{dt} +\frac{1}{1+v^2}\frac{dv}{dt}
$$
Use $ v = du/dt$:
$$
 -\frac{1}{u} = \frac{1}{1+v^2}\frac{dv}{dt}
$$
The ODE becomes
$$
 u\frac{d^2u}{dt^2} + \left[\frac{du}{dt}\right]^2 +1 = 0
$$
Use ODE solution techniques to solve.
A: Since your $f(t,u,v)$ does not contain $t$ explicitly, there is a first integral of the Euler-Lagrange Equations that says that:
$$v\frac{df}{dv} - f = C$$
is constant. This leads to 
$$v = \frac{dx}{dt} = \pm \frac{\sqrt{C^2 - x^2}}x,$$
from which the solution can be found from integration to be
$$x(t) = \sqrt{C^2 - (K\pm t)^2},$$ where $K$ is another constant of integration. This can be seen to be equivalent to the solution 1 above, except you start from a first order DE instead of a second order one. The integral of the ODE at the end of Solution 1 is
$$u\sqrt{v^2 + 1} = C,$$ which can be solved for $v$ giving the same solution given here. 
