I have the following second order ODE of the form $$u''=-(u')^2 \left(\frac{1}{u-a}+\frac{1}{u-b}\right)$$ for boundary conditions $u(0)=n$, $u(x_{n})=s$
Is it possible to solve this equation analytically? If yes, can someone give me an idea how to do it?
There is a useful general method for solving equations that do not feature the IV explicitly.
Let $v = u'$ be the new dependent variable, and $u$ the new independent variable. Then $\frac{dv}{du} = \frac{dv/dx}{du/dx} = \frac{u''}{u'}$ so we can write a first-order ODE for $v(u)$ and solve it. The solution will be some family of functions $v(u)$ depending on an unknown constant $C_1$.
Now we see that $du/dx = v(u)$ for the function $v$ we got in the last step, keeping the constant of course. This is of course yet another first-order ODE which can be solved for the solution $u(x)$, now depending on two constants.
Plug in your initial or boundary conditions and you can find the particular solution you want!
Hint.
$$ \frac{u''}{u'}= -\left(\frac{1}{u-a}+\frac{1}{u-b}\right)u' $$
Hint
If you switch variables, the equation becomes $$-\frac {x''}{[x']^3}=-\frac {1}{[x']^2} \left(\frac{1}{u-a}+\frac{1}{u-b}\right)$$ that is to say $$\frac {x''}{x'}=\frac{1}{u-a}+\frac{1}{u-b}$$ which becomes simple (reduction of order first).
The problem is that for inversion you would need to solve a cubic equation in $u$. The result could be nasty.
