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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 $D$?

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    $\begingroup$ Plug in $t=0$ and $v=p$, it seems to give you $D$ directly in this case (no algebra required). Of course $D$ will depend on $p$ but that's OK. $\endgroup$ – Ian Nov 8 '18 at 16:10
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$$\sqrt{\frac{a}{b}}\arctan(v\sqrt{\frac{a}{b}})=-at+D$$

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.

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