Buoyancy Nonlinear second order differential equation Working on equations of motion of an expandable gaz-filled body immersed uderwater, I have reached the following equation : 
$$ \beta^{2} z'' - \alpha(z')^{2} + \beta z z'' - z = 0 $$
Could you give me some advices on how to find and analytical solution, if it exists ? If not, do we have one if we forget the drag, that is if :
$$ \beta^{2} z'' + \beta z z'' - z = 0 $$
Ultimately, what is really of interest for me is not really the entire solutions, but rather stability around the solution $z=0$
 A: Since there are no $x$-dependence, you can reduce the order by one.
Recall $z''=\frac12\frac{\mathrm{d}(z')^2}{\mathrm{d}z}$ and we get
$$ (\beta^2 + \beta z) \frac{\mathrm{d}(z'^2)}{\mathrm{d}z} - 2\alpha(z')^{2} = 2z $$
Integrating once gives
$$
z'^2 =
\begin{cases}
\frac{\beta^2}{\alpha(\beta-2\alpha)} + \frac{2z}{\beta-2\alpha}+ c (z+\beta)^{2\alpha/\beta} & \alpha\neq 0\\
c+\frac{2 z}{\beta} - 2 \log(z+\beta) & \alpha=0
\end{cases}
$$
but I don't think the final integral can be done in closed form.
A: First thing we notice is that the equation doesn't contain the independent variable, so we can immediately lower the order by one.
Introducing:
$$z'=u(z) \\ z''=u' z'=u u'$$
We get:
$$\beta^2 u u' - \alpha u^2 + \beta z u u' - z = 0$$
Collecting the derivative:
$$\beta (\beta+z) u u'-\alpha u^2-z=0$$
Let's introduce yet another function:
$$u^2=v(z) \\ v'=2u u'$$
We get:
$$\frac{1}{2} \beta(\beta+z) v'-\alpha v-z=0$$
This is now a linear inhomogeneous ODE, which is easy to solve.

I hope I haven't made any mistakes, but with ODEs it's always easy to check the final solution by direct substitution, so once you find $v(z), u(z)$ and then $z(x)$ from $z'=u(z)$ you can check.
