I'm starting to read about Calculus of Variations and one of the first problem the text tackled using this tool was the Brachistochrone problem, which was given by the functional $$F[y(x)]=\int_a^b \frac{ds}{v}$$ with $ds=\sqrt{1+[y'(x)]^2}dx$ and $v=y'(x)$.
The author then solves the problem by invoking the equality of potential and kinetic energy, but do we need to do this? Just looking at the integrand $f$, we see that $$f=\sqrt{u^{-2}+1}$$ with $u=y'(x)$. Using the Beltrami identity, we obtain $$C=\sqrt{u^{-2}+1}+\frac{1}{u^2\sqrt{u^{-2}+1}}=\frac{u^2+2}{u^2\sqrt{u^{-2}+1}}$$ Can't we just solve this as an ordinary equation and then integrate to obtain $y(x)$? This is clearly wrong, since the solutions to the equation above are constants, so $y(x)$ becomes linear in $x$, which is not the solution to the Brachistochrone problem, but wherein lies the mistake?
Since A.G. didn't want to post an answer, I'll do it, just do get the question off the unanswered list - but all credit should go to A.G..
First of all, one can't write $$\frac{\sqrt{u^{2}+1}}{u}=\sqrt{u^{-2}+1}$$ since $u$ could be negative.
And second of all, there was a mistake in the problem formulation: $$v(x)=y'(t) \not=y'(x)$$ so the substitution using conservation of energy was necessary to get a $v(x)$ expressed as a function of $y(x)$.