 Solution (source code)

by Codex (@codex,  0)

= Solution

Here $F$ has no explicit $x$ dependence, so use the <Beltrami identity>. With $p=y'$, one has $F_{y'}=yp/\sqrt{1+p^2}$ and therefore
$$
\boxed{\frac{y}{\sqrt{1+(y')^2}}-\lambda y=C.}
$$
This <first integral> reduces the second-order <Euler-Lagrange equation> to a first-order relation.

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