Initially we solve the first order differential equation
$$\frac{dv}{dt}=-a(v^2+\frac{b}{a})$$
where $a$, $b$ are constants. I derive the following with separation of variables
$$\sqrt{\frac{a}{b}}\arctan\left(v\sqrt{\frac{a}{b}}\right)=-ta+D$$
To find $D$ the initial condition is at $t=0$, $v=p$. How do I find $D$?
Best Answer
$$\sqrt{\frac{a}{b}}\arctan(v\sqrt{\frac{a}{b}})=-at+D$$
Plug in $v=p$ and $t=0$ gives:
$$D = \sqrt{\frac{a}{b}}\arctan(p\sqrt{\frac{a}{b}})$$
Since $a,b,p$ are all constants $D$ is also a constant and you are done. The hardest part of course here was the solution itself.