Separating differential equatons

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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.

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5
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$$y''(x)=g-\frac{cy'(x)^2}{m}\Longleftrightarrow$$


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


$$\frac{\sqrt{m}}{\sqrt{cg}}\int\frac{1}{1-s^2}\space\text{d}s$$

0
On

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)