*edit*
Solved, please delete topic.
Let's write the DE in general form as...
$\displaystyle \frac{dq}{dt} = a\cdot q^{2} + b = b\cdot (1+\frac{a}{b}\cdot q^{2}) $ (1)
... and it can be solved in standard fashion writing...
$\displaystyle \frac{dq}{1+ \frac{a}{b}\cdot q^{2}} = b\cdot dt $ (2)
... and then integrate both terms...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$