Separating differential equatons The initial equation:
$$
y''= g-((C*(y')^2)/m)
$$
and I am trying to separate it into two differential equations.
I also have that the aerodynamic force $F=C*(y')^2$.
The initial equation describes falling motion of a skydiver where I am eventually trying to use python integrate function to predict times for falls of differing heights but first I need to separate this into two differential equations.
I don't know what method to use I started by trying to solve implicitly but i struggled to separate the variables and then I tried to use Laplace transformations but I didn't get far with that either. 
 A: $$y''(x)=g-\frac{cy'(x)^2}{m}\Longleftrightarrow$$

Let $y'(x)=v(x)$, which gives $y''(x)=v'(x)$:

$$v'(x)=g-\frac{cv(x)^2}{m}\Longleftrightarrow$$
$$\frac{v'(x)}{g-\frac{cv(x)^2}{m}}=1\Longleftrightarrow$$
$$\int\frac{v'(x)}{g-\frac{cv(x)^2}{m}}\space\text{d}x=\int1\space\text{d}x\Longleftrightarrow$$
$$\frac{\sqrt{m}\tanh^{-1}\left(\frac{v(x)\sqrt{c}}{\sqrt{gm}}\right)}{\sqrt{gc}}=x+\text{k}\Longleftrightarrow$$
$$v(x)=\frac{\sqrt{gm}\tanh\left(\frac{\sqrt{c}(x+\text{k})}{\sqrt{m}}\right)}{\sqrt{c}}\Longleftrightarrow$$
$$y'(x)=\frac{\sqrt{gm}\tanh\left(\frac{\sqrt{c}(x+\text{k})}{\sqrt{m}}\right)}{\sqrt{c}}\Longleftrightarrow$$
$$y(x)=\int\frac{\sqrt{gm}\tanh\left(\frac{\sqrt{c}(x+\text{k})}{\sqrt{m}}\right)}{\sqrt{c}}\space\text{d}x\Longleftrightarrow$$
$$y(x)=\frac{m\ln\left(\cosh\left(\frac{\sqrt{c}(x+\text{k})}{\sqrt{m}}\right)\right)}{c}+\text{k}_2$$


$$\int\frac{v'(x)}{g-\frac{cv(x)^2}{m}}\space\text{d}x=$$

Substitute $u=v(x)$ and $\text{d}u=v'(x)\space\text{d}x$:

$$\int\frac{1}{g-\frac{cu^2}{m}}\space\text{d}u=$$
$$\frac{1}{g}\int\frac{1}{1-\frac{cu^2}{m}}\space\text{d}u=$$

Substitute $s=\frac{u\sqrt{c}}{\sqrt{gm}}$ and $\text{d}s=\frac{\sqrt{c}}{\sqrt{gm}}\space\text{d}x$:

$$\frac{\sqrt{m}}{\sqrt{cg}}\int\frac{1}{1-s^2}\space\text{d}s$$
A: so I have ended up doing this which I think is write 
set z=dy/dx
so z=y'
then 
z'=g-((c*z^2)/m)
