How to recover free fall equation from equations of motion with linear drag For a particle of mass $m$, initial velocity $v = 0$, and initial height $h$, integrate the equation of motion to find $y(t)$, assuming that the particle  experiences  a  frictional  force  that  is  linearly  proportional  to  its  velocity,  namely $F=−αv$. Demonstrate explicitly that  as $\alpha\rightarrow0$ you recover the free fall equation.
I started with $m\frac{dv}{dt} = -mg - \alpha v$, and I arrived at the following equation:
$$y(t) = h - \frac{m^2g}{\alpha^2}e^{-\frac{\alpha}{m}t} - \frac{mg}{\alpha}t$$
It is easy to find that the velocity approaches terminal velocity, and the unbounded position approaches $-\infty$. However, I can't see a readily apparent way to recover the free fall equation from the position equation. Simply setting $\alpha$ to $0$ doesn't help because I don't have correct units, and I arrive at $y(t) = h - 0 - \infty$. I don't see any simple algebra tricks, and a Taylor expansion doesn't seem to help either. Could someone help me find the error in my approach please?
 A: The trouble comes from a mistake in solving the equation $$m\frac{d^2y}{dt^2} = -mg - \alpha \frac{dy}{dt}\quad\text{with conditions}\quad y(0)=h\text{ and }y'(0)=0.$$
You wrote :
$$y(t) = h - \frac{m^2g}{\alpha^2}e^{-\frac{\alpha}{m}t} - \frac{mg}{\alpha}t$$
In fact, a term is missing. The correct solution is :
$$y(t) = h - \frac{m^2g}{\alpha^2}e^{-\frac{\alpha}{m}t} - \frac{mg}{\alpha}t +\frac{m^2g}{a^2}$$
Expending to series with respect to $\alpha$ leads to :
$$y=h-\frac12 gt^2+\left(\frac{g}{6m}\right)t^3\alpha+O\left(t^4\alpha^2\right)$$
In case of $\alpha=0$, one recovers $y=h-\frac12 gt^2$.
A: $$
v'(t)=-\frac{\alpha}{m}v(t)-g
$$
Using as integrating factor $Q=e^{\alpha t/m}$ we have
$$
\frac{(Qv)'}{Q}=-g
$$
and integrating
$$
Qv=-\frac{mgQ}{\alpha}+\frac{c}{Q}
$$
and then 
$$
v(t)=-\frac{mg}{\alpha}+\left(v(0)+\frac{mg}{\alpha}\right)e^{-\alpha t/m}
$$
from which we have
$$
y(t)=y(0)-\frac{mg}{\alpha}t+\frac{m}{\alpha}\left(v(0)+\frac{mg}{\alpha}\right)\left(1-e^{-\alpha t/m}\right)
$$
With $y(0)=h,\,v(0)=0$ we have
$$
y(t)=h-\frac{mg}{\alpha}t+\frac{m^2g}{\alpha^2}\left(1-e^{-\alpha t/m}\right)
$$
Using $e^{-\alpha t/m}\sim 1-\frac{\alpha t}{m}+\frac{\alpha^2 t^2}{2m^2}+o(\alpha^3)$ as $\alpha\to 0$
we have
\begin{align}
y(t)&\sim h-\frac{mg}{\alpha}t+\frac{m^2g}{\alpha^2}\left(1-1+\frac{\alpha t}{m}-\frac{\alpha^2 t^2}{2m^2}+o(\alpha^3)\right)\\
&\sim h-\frac{g}{2}t^2
\end{align}
