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.