Finding solution of non linear DE $x\ddot{x}-\dot{x}^2=1$ I am looking for help on how to find the solution of the following differential equation,
$$x\ddot{x}-\dot{x}^2=1,$$
which comes from solving the Euler-Lagrange equations for the lagrangian $L=x\sqrt{\dot{x}^2+1}$. It is from an exercise in my field theory course. My professor provided me with the solution $x(t)=c_1\cosh{\left(\frac{t-c_2}{c_1}\right)}$, which is a solution when plugged into the equation, but does not give any derivation. 
Is there a way to solve this DE other than an educated guess? If so how? 
 A: Yes, there is a way. Dividing both sides of the differential equation by $x^2$ gives
$$\frac{x\ddot{x} - \dot{x}^2}{x^2} = \frac{1}{x^2}$$
or
$$\frac{d}{dt}\left(\frac{\dot{x}}{x}\right) = \frac{1}{x^2}.$$
Since $\displaystyle \frac{\dot{x}}{x} =  \frac{\mathrm{d}}{\mathrm{d}t}(\log x)$, we have 
$$\frac{d^2}{dt^2}(\log x) = \frac{1}{x^2}.$$
Now set $x = e^u$, so that 
$$\ddot{u} = e^{-2u}.$$
Multiplying both sides by $\dot{u}$,
$$\ddot{u}\dot{u} = e^{-2u}\dot{u}$$
or
$$\frac{d}{dt}\left(\frac{\dot{u}^2}{2}\right) = -\frac{d}{dt}\frac{e^{-2u}}{2}.$$
By integration,
$$ \dot{u}^2 = C - e^{-2u}$$
where $C$ is a constant. Hence $\dot{u} = \pm \sqrt{C - e^{-2u}}$.
By separation of variables,
$$\int \frac{du}{\sqrt{C - e^{-2u}}} = \pm\int dt.$$
Now
$$\int \frac{du}{\sqrt{C - e^{-2u}}} = \int \frac{e^u\, du}{\sqrt{Ce^{2u} - 1}}.$$
Setting $\sqrt{C}e^u = \cosh \theta$, we get
$$\int \frac{e^u\, du}{\sqrt{Ce^{2u} -1}} = \int \frac{\sinh\theta \,d\theta}{\sqrt{C}\sinh \theta} = \frac{1}{\sqrt{C}}(\theta + D) = \frac{1}{\sqrt{C}}\cosh^{-1}(\sqrt{C}e^u) + \frac{D}{\sqrt{C}}.$$
It follows that 
$$\frac{1}{\sqrt{C}}\cosh^{-1}(\sqrt{C}e^u) = \pm t + A,$$
where $A$ is a constant. Since $e^u = x$,
$$\frac{1}{\sqrt{C}}\cosh^{-1}(\sqrt{C}x) = \pm t + A$$
or
$$x(t) = \frac1{\sqrt{C}}\cosh(\sqrt{C}(t + A))$$
which we can write in the form
$$x(t) = c_1 \cosh\left(\frac{t - c_2}{c_1}\right) .$$
A: Let $\dot x=p$. Then
$$
\ddot x=\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=p\,\frac{dp}{dx}.
$$
The equation becomes
$$
x\,p\,\frac{dp}{dx}-p^2=1\implies\frac{p\,dp}{p^2+1}=\frac{dx}{x}.
$$
Integration results in
$$
p^2+1=C\,x^2\implies {\dot x\,}^2=c_1\,x^2-1\implies \frac{dx}{\sqrt{c_1\,x^2-1}}=\frac{dt}{t}.
$$
The solution is
$$
\int\frac{dx}{\sqrt{c_1\,x^2-1}}=\log t+c_2.
$$
A: The ODE is of the autonomous kind :
http://mathworld.wolfram.com/Autonomous.html
The usual method to solve it is shown below :

