Solve the differential equation of brachistochrone I'm solving the brachistochrone problem. 
My solution got as far as $y'=\sqrt{k-y\over y}, k={1\over 2gC^2}$. 
From https://math.berkeley.edu/~strain/170.S13/cov.pdf  page 12, I found that the expression can be written as $\sqrt{y\over k-y}=tanφ$ , and it says " function $φ$ is a new function of $x$" but I'm not sure what this $φ$ is. Is it just a definition? Has this $φ$ got any geometrical meaning? 
The other way to solve this is by letting $ 2A={1\over 2gC^2}$ , and the equation can be written as $y=A-Acosθ$.  I assume this is a coordinate transformation. But how do I get from $y=rsinθ$ to this? And is there any relation between $φ$ and $θ$? 
Thank you in advance.
 A: Concerning the first part, it is just a change of variable. If you define $$\sqrt{y(x)\over k-y(x)}=\tan(\phi(x))$$, you then have $$y(x)=\frac{k \tan ^2(\phi (x))}{1+\tan ^2(\phi (x))}$$ $$y'(x)=k\, \phi '(x)\, \sin (2 \phi (x))$$ Replacing in the original equation, you then arrive to $$k\, \phi '(x)\, \sin (2 \phi (x))=\frac{1}{2} \sqrt{k^2\, \sin ^2(2 \phi (x))}$$ which beautifully simplifies to $\phi (x)=C\pm\frac x 2$.
A: $\phi$ depends on $y$, but $y$ depends on $x$, so I think it means that $phi$ depends on $x$ implicitly, and as far as I can tell, there is no relation between $\phi$ and $\theta$; it's just a integration technique. 
A: The equation $$\frac {dy}{dx}=\sqrt {\frac {k-y}y}$$ is equivalent to $$\frac {dx}{dy}=\sqrt {\frac y{k-y}}$$ so you can find $x$ by an integration with respect to $y\,$.
Use the substitution $$\sqrt {\frac y{k-y}}=\tan \phi$$ so $$y=k\, \frac {\tan^2 \phi}{1+\tan^2 \phi}=k\, \sin^2 \phi$$ Then $$x= \int \sqrt {\frac y{k-y}}\,dy= \int \tan \phi \cdot 2k \sin \phi \cos \phi\;d\phi=2k \int \sin^2 \phi\,d\phi$$ and so on.
No geometrical meaning for $\phi$ is required in this computation.
