My question concerns the functional $$I[x]=\int_{-4}^{4}\frac{1}{x} + \frac{(\dot{x})^2}{x} \ dt$$ and what $x$ makes the functional stationary, where $(t, x)$ ranges from $(-4, 3)$ to $(4, 3)$. I have obtained the following from Euler-Lagrange: $2x\ddot{x}=(\dot{x})^2-1$. I am told that the correct answer has the form $x(t) = \frac{1}{c}-\frac{ct^2}{4}$ for some constant $c$.
Where am I going wrong? The DE I obtained seems way too complicated and I cannot think of a way of solving it. I am fairly new to Calc. of Variations, so a full solution would be much appreciated.
The DGL, you have derive is correct. You question is how to solve it. For this it is import to know that given the fact that the DGL derives from a Lagrangian gives insights towards the solution.
In particular, in this case, the Lagrangian does not depend on $t$. With that you immediately can integrate the DGL once and reduce it to a first order DGL! In fact, what is called the `energy' $$E = \dot x \frac{\partial L}{\partial \dot x} - L = \frac{\dot x^2 -1}{x} \tag{1}$$ is conserved (that is it does not change over time).
This can be easily checked $$ \frac{d}{dt} E = \frac{d}{dt}\frac{\dot x^2 -1}{x} =\frac{2 x \ddot x}{x} - \frac{\dot x (\dot x^2 -1)}{x^2} =\frac{\dot x (2 x \ddot x -\dot x^2 +1)}{x^2} = 0$$ due to the Euler-Lagrange equation.
So the remaining task is to integrate (1) once. The equation is separable and we obtain $$\int_3^x \frac{dy}{\sqrt{E y -1}}= \frac{2}{E} \Bigl(\sqrt{1+ Ex}-\sqrt{1+3E}\Bigr) = t+4\;. \tag{2}$$ The boundary condition $(4,3)$ immediately determines the constant $$E=-\frac{1}{4}\;.$$
Solving (2) for $x$ yields the result $$ x= 4-\frac{1}{16} t^2\;. $$