How to simplify the Euler-Lagrange equation of Brachistochrone in this way?

Question

I already know that in the Brachistochrone problem, we have Euler-Lagrange equation: $$\frac{1}{2y}\sqrt{\frac{1+y'^2}{y}}+\frac{d}{dx}[\frac{y'}{\sqrt{y(1+y'^2)}}]=0$$ To solve this equation, we simplify the above equation and get: $$\frac{d}{dx}[\frac{1}{\sqrt{y(1+y'^2)}}]=0$$ How to get the second equation from the first equation?

Answer 1

I figured it out, it is just the application of the following equation: $$F-y'\frac{\partial F}{\partial y'}=C$$ To see the proof, please go to another questions:Euler-Lagrange formula

Answer 2

The integral $t=\int\sqrt{\frac{1+\frac{dy}{dx}^2}{y}}dx$ must be expressed as $t=\int\sqrt{\frac{1+\frac{dx}{dy}^2}{y}}dy$.

You can get it by thinking, instead, that $y=y(x)$ and by the chain rule $\frac{dx}{dx}=\frac{dx}{dy}\frac{dy}{dx}=1$ so $y'=\left({\frac{dx}{dy}}\right)^{-1}$.

For $t=\int\sqrt{\frac{1+\frac{dx}{dy}^2}{y}}dy$ the integrand is of the form $L=L\left(y,\frac{dx}{dy}\right)$ and the E - L equation is

$$\frac{d}{dy}\frac{\partial L}{\partial x'}=0,$$ because $\dfrac{\partial L}{\partial x}=0$.