How do I solve $dv/dt=ov^2+a$?

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How do I solve $\cfrac{dv}{dt}=ov^2+a$?

Here, $a$ is the acceleration from gravity, $v$ is the speed, $t$ is the time and $o$ is a constant.

I am trying to make a formula for quadratic drag but it seems that integrating both sides doesn't work since $v$ appears on both sides.

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2 Answers

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HINT

It is a separable ODE

\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}t} = kv^{2} + a \Longleftrightarrow \int\frac{\mathrm{d}v}{kv^{2} + a} = \int\mathrm{d}t \end{align*}

Can you take it from here?

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Rewrite the ode as follows,

$$\frac{dv}{\rho v^2+a} = dt$$

which can be integrated as

$$ \frac{1}{\sqrt{a\rho}} \tan^{-1} \left( \sqrt{\frac{\rho}{a}}v\right) = t + C$$

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