An ordinary differential equation While I was working on a variational problem, I met this equation as its Euler-Lagrange equation, but I cannot solve it:
$ x= \frac{af'(x)}{\sqrt{1+af'(x)^{2}}} + \frac{bf'(x)}{\sqrt{1+bf'(x)^{2}}} \ (a\neq b) $.
 A: You can solve it parametrically as follows:  Write 
$$
x(t) = \frac{at}{\sqrt{1+at^2}}+\frac{bt}{\sqrt{1+bt^2}}
$$
and
$$
y(t) = c - \frac{\sqrt{1+at^2}+\sqrt{1+bt^2}}{\sqrt{1+at^2}\sqrt{1+bt^2}}
$$
where $c$ is a constant.  Then this gives the general solution as $c$ varies.
If you want an explicit relation between $x$ and $y$, you can eliminate $t$, but it won't be pretty.
A: Set $p:=f'(x)$ and
$$
\Phi(p)=\Phi_{a,b}:=\frac{ap}{\sqrt{1+ap^2}}+\frac{bp}{\sqrt{1+bp^2}}.
$$
The differential equation you wrote  can be rewritten as
$$
x=\Phi(p).
$$
If we could invert $\Phi$, then we could write
$$
f'(x)= p=\Phi^{-1}(x).
$$
For $a, b>0$ the function $\Phi$ seems to be increasing. The animation below depicts the graphs of $\Phi_{1,t}$ for $t\in [0,6]$

It already shows that  the solution blows up in finite time. (Here I think of $x$ as time.)
The next animation depicts $\phi_{1,t}$  for $t=-1..0$ and you can see that the injectivity of $\Phi$ is lost  for some values of $t$.

Remark.  Here is an animation of the curve  described by Robert Bryant for $a=1$ and $b\in [-0.1,0.2]$, $t\in[-3,3]$

