= Solution
Discard the constant factor $2\pi$ when forming the <Euler-Lagrange equation>. For $F=y\sqrt{1+y'^2}$ it becomes
$$
\sqrt{1+y'^2}-\frac d{dx}\frac{yy'}{\sqrt{1+y'^2}}=0,
\qquad yy''=1+y'^2.
$$
The <Beltrami identity>, since $F$ has no explicit $x$ dependence, gives $F-y'F_{y'}=y/\sqrt{1+y'^2}=C>0$. Combining with the differential equation yields $y''=y/C^2$. Its exponential solution has coefficients whose product is $C^2/4$, by $y'^2=y^2/C^2-1$. Positivity therefore puts it in the form
$$
\boxed{y(x)=C\cosh((x-K)/C),\qquad C>0}.
$$
Its constants must satisfy $y_1=C\cosh((x_1-K)/C)$ and $y_2=C\cosh((x_2-K)/C)$. Depending on the endpoint data, these equations may have no solution or more than one. A constant function can satisfy a formal first-integral equation but fails the full <Euler-Lagrange equation>; it is not an extra extremal.
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