Classical differential equation $mg-kv^3=m\frac{dv}{dt}$? How to solve the differential equation $mg-kv^3=m\frac{dv}{dt}$?
The equation is easily solvable when it is not $-kv^3$ but $-kv$ (linear) or $-kv^2$ (use trig identity). 
Strangely, I didn't find any information on the web for this equation.
$m$,$g$,$k$ are constants
 A: We are interested in integrating
$$\int \frac{dv}{\alpha^3-\beta^3 v^3}$$ Where $\beta =k^{1/3}$  and $\alpha=(mg)^{1/3}$
Notice that the denominator can be factored as: $$\alpha^3-\beta^3 v^3=\left(v-\frac{\alpha}{\beta}\right)\left(v-\frac{\alpha}{\beta}\omega\right)\left(v-\frac{\alpha}{\beta}\omega^2\right)$$
Where $\omega=-\frac{1}{2}+i\frac{\sqrt3}{2},\omega^2=-\frac{1}{2}-i\frac{\sqrt3}{2}$ and $i=\sqrt{-1}$
So our problem becomes:
$$\int\frac{dv}{\left(v-\frac{\alpha}{\beta}\right)\left(v-\frac{\alpha}{\beta}\omega\right)\left(v-\frac{\alpha}{\beta}\omega^2\right)}$$
Which can be done by partial fractions (if you're willing to)
A: You can clean the problem up a little by scaling: let $v=au$, $t=bs$, then
$$mg-ka^3u^3=m\frac ab\frac{du}{ds}$$
And then if $mg=ka^3=m\frac ab$ so that $a=\left(\frac{mg}k\right)^{1/3}$ and $b=\left(\frac m{kg^2}\right)^{1/3}$ then the equation reads $1-u^3=\frac{du}{ds}$. This can be expanded by partial fractions
$$\begin{align}\left(\frac1{u^3-1}\right)du&=\left(\frac A{u-1}+\frac{Bu+C}{u^2+u+1}\right)du\\
&=\left(\frac{\frac13}{u-1}+\frac{-\frac13u-\frac23}{u^2+u+1}\right)du\\
&=\left(\frac{\frac13}{u-1}+\frac{-\frac16(2u+1)-\frac12}{u^2+u+1}\right)du=-ds\end{align}$$
We can integrate this, noting that $u^2+u+1=\left(u+\frac12\right)^2+\left(\frac{\sqrt3}2\right)^2$, to get
$$\frac13\ln|u-1|-\frac16\ln(u^2+u+1)-\frac1{\sqrt3}\tan^{-1}\left(\frac{u+\frac12}{\frac{\sqrt3}2}\right)=-s+C$$
$$\frac16\ln\left(\frac{(v-a)^2}{v^2+av+a^2}\right)-\frac1{\sqrt3}\tan^{-1}\left(\frac{2v+a}{a\sqrt3}\right)=C-\frac tb$$
And if you want a rather messy expression you may now substitute in the values of $a$ and $b$. But now if you want to further integrate to find displacement as a function of time I am afraid that you are going to have to go numerical.
A: To the best of my knowledge, this equation has no (simple) closed-form solution; this is the equation for the velocity of a falling object with air resistance (drag force), correct?
I remember that we had to solve it numerically in my computational physics class for this reason. See Chapter 2 of Giordano and Nakanishi, Computational Physics.
