First Order Non-linear Differential Equation Question I am looking for a (hopefully closed form) solution to the non-linear DE of the form 
$$\frac{dv}{dt} = A v^2 + B.$$ 
Other than being real there are no constraints on $A$ and $B$. I would assume that this is a well-explored equation as it is the equation of motion ($v$ = velocity) of an object experiencing aerodynamic drag, frictional drag, and a constant propelling force. 
Be kind - I have not tried to work in this world since the early 1970's :-)
Thanks.
dave
 A: I assume $AB \neq 0$.
Changing the variables according to $$t=\frac 1{\sqrt {|AB|}}\,s$$and $$v=\sqrt {\left|\frac BA \right|}\,w$$ one has to solve $$\frac {dw}{ds}=\pm1\pm w^2$$ so $$s=\pm \int \frac 1{1+w^2}\,dw=\pm \arctan w +c$$ or $$s=\pm \int \frac 1{1-w^2}\,dw=\pm \text {arctanh}\, w +c$$ i.e., renaming the constant $$w=\pm\tan\,(s+c)$$ or $$w=\pm\tanh\, (s+c)$$ Finally, if $AB>0$ $$v=\frac {\sqrt {AB}}A \tan\,(\sqrt {AB}\,t+c)$$ otherwise $$v=-\frac {\sqrt {-AB}}A \tanh\,(\sqrt {-AB}\,t+c)$$ Maple confirms, using the assume command.
A: HINT: rewrite $$\frac{dv}{Av^2+B}=dt$$ integrating gives $$v(t)=\frac{\sqrt{B} \tan \left(\sqrt{A} \sqrt{B} c_1+\sqrt{A} \sqrt{B}
   t\right)}{\sqrt{A}}$$
A: Beside the fact that the equation is separable, as 
@Dr.SonnhardGraubner answered, the key idea is to arrive to something close to a well known formula (this is something I suggest you to always do if you lost contact with this area for too long - there are very good tables for many integrals). 
The closest I can see is $$\frac {1}{1+z^2}=\Big(\tan^{-1}(z)\Big)'$$ So, assuming for simplicity that $A>0$, $B>0$, which ensures the square roots cause no problem, try changing variable $$v=\sqrt{\frac BA}z\qquad dv=\sqrt{\frac BA}dz$$ After some minor simplifications, one gets $$\frac{dv}{Av^2+B}=dt=\frac 1{\sqrt{AB}}\frac {dz}{1+z^2}$$ that is to say $$\frac {dz}{1+z^2}=\sqrt{AB}dt$$ Now integrate both sides $$\tan ^{-1}(z)=\sqrt{AB}t+C$$ So,$$z=\tan(\sqrt{AB}t+C)$$ Now, going back from $z$ to $v$, $$v=\sqrt{\frac BA}\tan(\sqrt{AB}t+C)$$ the integration constant $C$ being probably fixed by some condition.
