I am in the process of learning Euler-Lagrange Equation and the Brachistochrone problem, and came across this ordinary differential equation, which was derived by applying the Euler-Lagrange equation to the Brachistochrone problem, where a and b are constants.
$$\tag{1} \frac{dy}{dx} = \left(\frac{b + y}{a - y}\right)^\frac{1}{2} $$
To solve this equation, the document I am looking at does a change of variable from $y = y(x)$ to $\theta = \theta(x)$ by the following equation.
$$\tag{2} y = \frac{1}{2}(a-b) - \frac{1}{2}(a+b)\cos\theta $$
$$\tag{3} \frac{dy}{dx} = \frac{1}{2}(a+b)\sin\theta \frac{d\theta}{dx} $$
By using these particular substitutions to (1), the author solves the problem by outputting parametric representation of a cycloid.
Is there a particular reason why this change of variable was used here? I am wondering how did the author came up with this idea of change of variable. Thanks!
One can first observe that the equation is only defined in the real domain for $y\in(-b,a)$. This might motivate to parametrize $y=(1-s)(-b)+sa=(a+b)s-b$, $s\in(0,1)$. The equation then transforms to $$ (a+b)·s'(x)=\sqrt{\frac{(a+b)s}{(a+b)(1-s)}}=\sqrt{\frac{s}{(1-s)}} $$ This now invites to have $s$ to be a sine square, as then $1-s$ is the corresponding cosine square. $s(x)=\sin^2\phi(x)$ then gives $$ 2(a+b)\sin\phi\cos\phi·\phi'=\frac{\sin\phi}{\cos\phi}\implies 2\cos^2\phi·\phi'= \frac1{a+b}. $$ You get to your parametrization by using the double-angle formula $\cos(2\phi)=1-2\sin^2\phi$, so that $\theta=2\phi$.