How to nondimensionalize a second order differential equation 
A body of mass $m$ is thrown upwards in a vertical direction from the
  earth's surface with a velocity $v$. The air resistance is supposed to
  be taken into account by Stoke's law $F_R= - cv$ for the flow
  resistance in viscous fluids, which is reasonable for small
  velocities. Here $c$ is a coefficient depending on the shape and size
  of the body. The motion is supposed to depend on the mass $m$, the
  velocity $v$, the gravitaional acceleration $g$ and the friction
  coefficient $c$ with dimension $[c]= \frac{M}{T}$.
The initial value problem for the height is assumed to take the form
  $$ mx''+cx'=-mg$$
$$x(0)=0$$
$$x'(0)=v$$
  Non dimensionalize the differential equation.

First I rewrote it to
$$ x'' = -g-\frac{c}{m} x'$$
$$x(0)=0$$
$$x'(0)=v$$
Than I wrote down the fundamental dimensions, mass $M$, time $T$ and height $H$.
After that I determined the dimensions of the involved quantities
$[x] = H$
$[g]=\frac{H}{T^2}$
$[v]=\frac{H}{T}$
$[c]=\frac{M}{T}$
$[m]=M$
$[t]= T$
I thought to nondimensionalize  I to define $\tau = \frac{t}{\bar{t}}$, $y=\frac{x}{\bar{x}}$ and  $z= \frac{m}{\bar{m}}$ where $\bar{t},\bar{x} and \bar{m}$ are characteristic quantities.
Now I wanted to do a change of variables, and write $x''$ as $y''(\tau)$,but that is where it went wrong.
$$\frac{d^2x}{dt^2}=\frac{d}{dt}(\bar{x}\frac{dy}{dt}) = \bar{x}\frac{d^2y}{dt^2}=0$$
I'm pretty new to this and I tried to look up similar problems and try the methods they used, but I just don't know what to do. Any nod in the right direction is much appreciated :)

Calling 
$$
\cases{
\tau = \frac{t}{t_o}\\
\eta = \frac{x}{x_o}
}
$$
we have
$$
\frac{d^n}{dt^n} = \frac{1}{t_o^n}\frac{d^n}{d\tau^n}
$$
and after substitution
$$
m\frac{x_o}{t_o^2}\eta''+ c\frac{x_o}{t_o}\eta' + m g = 0
$$
or
$$
\eta'' + \frac{c t_o}{m}\eta'+\frac{t_o^2 g}{x_o} = 0
$$
now determining $x_o,t_o$ such that
$$
\cases{
\frac{c t_o}{m} = 1\\
\frac{t_o^2 g}{x_o} = 1
}
$$
we have
$$
t_o = \frac mc,\ \ \ x_o = \frac{m^2}{c^2}g
$$
and consequently
$$
\eta''+\eta'+1=0
$$
etc.
