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$\begingroup$ thank you for your response . I agree with you, but i can transfer the equation to this one $\frac{\dot x^2}{(\dot x^2+\dot y^2)\,y}=k\quad\Rightarrow\quad \frac{1}{(1+y'^2)\,y(x)}=k$ and solve it for $~y(x)$ $\endgroup$
– Eli
Commented Jul 3, 2023 at 17:44
$\begingroup$ @Eli You can, but isn't that ultimately the same equation that comes from the Beltrami identity? I thought you were trying to avoid using it. $\endgroup$
– Michael Seifert
Commented Jul 3, 2023 at 17:48

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