Brachistochrone - Solution of a Cycloid - Parametric Equations I am trying to understand the math behind the Brachistochrone.
I could understand all the technical intricacies of the mathematical treatment of the topic found at Wolfram-Mathworld|Brachistochrone Problem.
At the last part, they say, 
$$
\boxed{\left[1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\right]y = \frac{1}{2gC^2} = k^2} \tag*{(1)}
$$
and then they just say, This equation is solved by the parametric equations and write two equations, 
$$
x = \frac{1}{2}k^2(\theta-\sin \theta)\\ \tag*{(2)}
y=\frac{1}{2}k^2(1-\cos \theta)
$$
How did this come? 
I plotted them on Desmos and I could clearly see a cycloid (Can be see here).
Note:


*

*I have not shown my steps here because, I am just studying the problem straight from Wolfram. If you wish to have a look at the steps, I have mentioned the link above.

*Learning Calculus of Variations on my own. I will be grateful if you could direct me towards some good resources. 

*I know how to derive the parametric equation of a cycloid, I learnt it from Math.Stackexchange|How to find the parametric equation of a cycloid?. I just don't know how to solve $(1)$ using the two equations in $(2)$.
Thank you.
 A: Note that we can use the trigonometric identities, $\sin(\theta)=2\sin(\theta/2)\cos(\theta/2)$ and $\frac12(1-\cos(\theta))=\sin^2(\theta/2)$, to write $x(\theta)$ and $y(\theta)$ as
$$\begin{align}
&x(\theta)=k^2(\theta/2 -\sin(\theta/2)\cos(\theta/2))\tag1\\\\
&y(\theta)=k^2 \sin^2(\theta/2)\tag2
\end{align}$$
Then, it is easy to see that we have 
$$\frac{dy}{dx}=\cot(\theta/2) \tag3$$
Finally, using $(2)$ and $(3)$ it is straightforward to show that
$$\left(1+\left(\frac{dy}{dx}\right)^2\right)y=k^2$$
as was to be shown!
A: I suppose that this is probably the simplest parametrization.
If we consider the differential equation $$1+\left(\frac{dy}{dx}\right)^2=\frac{k^2} y$$ I suppose that we cannot find $y$ as a function of $x$. However, using the positive solution, the equation can write $$\frac{dx}{dy}=\sqrt{\frac y {k^2-y}}$$ which leads to $$x=-\sqrt{y(k^2-y)}+k^2 \tan^{-1}\left(\sqrt{\frac y {k^2-y}} \right)$$ In a first step, use $y=k^2 z$ to get $$x=-k^2\sqrt{z(1-z)}+k^2 \tan^{-1}\left(\sqrt{\frac z {1-z}} \right)$$ and now, just play with trigonometric identities. To get rid of the $\tan^{-1}(.)$, let $$\sqrt{\frac z {1-z}}=\tan(\theta)\implies z=\sin^2(\theta)$$
A: This is an old post, but there will still be people wanting to know how the parametric equations of the cycloid were obtained from the solution to the DE. Historically, the people who solved this problem already knew that the answer was the cycloid, so they knew the parametric equations, so they just had to show that they satisfied the DE.
You can obtain the parametric equations by using the method demonstrated above by Claude (Though I think there is an error in what he has written above). Once you have ArcTan in the solution, the natural thing to do is to let its argument equal Tan(theta), then go on to obtain Sin and Cos, which requires a substitution, again demonstrated by Claude.  From there, with the double angle formulae, the parametric equations fall out fairly easily.
I can show full working if anyone is interested.
