Solving this ODE: $mv'(t)=mg-kv^2(t)$

Question

I was working on a model for both drag and gravity acting on an object dropped from a certain height, as given by the ODE below: $$mv'(t)=mg-kv^2(t)$$ Where $m$ is the item's weight, $v(t)$ is its velocity (positive is downward), $g$ is gravity, and $k$ is the item's drag.

I got this fairly nasty integral: $$\frac1m \int \frac{dv}{mg-kv^2(t)}=\int dt=t+c$$

Any idea on how to proceed?

EDIT: I can factor out $\left (\sqrt{mg}+\sqrt kv(t) \right)\left (\sqrt{mg}-\sqrt kv(t) \right)$, so the integral becomes a question of partial fractions. My question has been resolved.

Answer 1

Another way to evaluate $\int \frac{dv}{mg-kv^2}$ is via substitution:

$$\begin{align} \int \frac{dv}{mg-kv^2} &= \frac{1}{mg} \int \frac{dv}{1-\frac{k}{mg}v^2} \\ &=\frac{1}{\sqrt{mgk}}\int \frac{du}{1-u^2} \\&=\frac{1}{\sqrt{mgk}} \tanh^{-1}(u)=\frac{1}{\sqrt{mgk}} \tanh^{-1} \left(v\sqrt{\frac{k}{mg}}\right)\end{align} $$

where we have done the substitution $u=v\sqrt{\frac{k}{mg}}$ in the second line.

Answer 2

By a suitable rescaling of time, you can rewrite as

$$v'=a^2-v^2,\\ \frac{v'}{a^2-v^2}=1.$$

You should recognize the derivative of the inverse of the hyperbolic tangent, and

$$\frac1a\text{artanh}\left(\frac va\right)=t$$

(as the object is dropped, the integration constant is $0$).

Then

$$v=a\tanh(at)$$ and by another integration

$$y-y_0=\log(\cosh(at)).$$