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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?

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  • $\begingroup$ For the record consider to include the Lagrangian. $\endgroup$ – Qmechanic Nov 17 '18 at 20:56
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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

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