When trying to solve the tipical problem of free fall of an object with air drag we arrive at the differential equation:
$$ mx'' = -mg - k_1 x' + k_2 (x')^2 $$
with initial conditions: $x(0)=0$ and $x'(0)=0$, where $x=x(t)$.
By trying $\beta = (x')^2$ and re writting $x''= \frac{1}{2} (\frac{d}{dt} x'^2)$ I arrived at:
$$ \frac{1}{2}m \beta' = -mg - k_1 \sqrt{\beta} + k_2 \beta $$
It looks like a Ricatti equation but with the square root there. How do I manage this?