Terminal velocity ODE solution A skydiver's velocity after jumping from an initial height can be modeled by the equation 
$$
m \frac{dv}{dt}=mg-k v^2.
$$
I am trying to solve this analytically given the initial condition $v(0)=0$, but don't know how. It's not doable by integrating factor or Bernoulli's substitution, nor is it exact. While it is separable, the standard trig sub $v=\sqrt{mg/k} \sec(\theta)$ doesn't make sense for the initial condition. If someone could give me the initial substitution, I'd be really grateful.
Thanks! 
 A: To start with, I would recommend to separate the variables:
\begin{align*}
m\frac{\mathrm{d}v}{\mathrm{d}t} = mg - kv^{2} \Longleftrightarrow \int\frac{\mathrm{d}v}{mg - kv^{2}} = \int\frac{\mathrm{d}t}{m}
\end{align*}
Then we can make the substitution $\displaystyle v = \sin(\theta)\sqrt{\frac{mg}{k}}$. Consequently, one has that
\begin{align*}
\int\frac{\mathrm{d}v}{mg-kv^{2}} & = \sqrt{\frac{1}{mgk}}\int\frac{\cos(\theta)}{1-\sin^{2}(\theta)}\mathrm{d}\theta\\\\
& = \sqrt{\frac{1}{mgk}}\int\frac{\mathrm{d}\theta}{\cos(\theta)}\\\\
& = \sqrt{\frac{1}{mgk}}\ln|\sec(\theta) + \tan(\theta)| + C
\end{align*}
If we denote $(mg/k)^{1/2}$ by $a$, we have that
\begin{align*}
\sec(\theta) + \tan(\theta) & = \frac{1+\sin(\theta)}{\cos(\theta)} = \frac{1+a^{-1}v}{\sqrt{1-a^{-2}v^{2}}} = \frac{a + v}{\sqrt{a^{2} - v^{2}}} = \sqrt{\frac{a+v}{a-v}} \Longrightarrow\\\\
\int\frac{\mathrm{d}v}{mg - kv^{2}} & = \frac{1}{2}\sqrt{\frac{1}{mgk}}\ln\left|\frac{a+v}{a-v}\right| = \frac{t}{m} + C
\end{align*}
Since $v(0) = 0$, we conclude that $C = 0$, and we are done.
