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If $F=y+y'$, then Euler-Lagrange equation $\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y'})=0$ is not satisfied. Hence, we may conclude that there is no extremal for the functional concerned.

However, since $F$ does not contain $x$ explicitly, we can use the Beltrami identity $F-y'\frac{\partial F}{\partial y'}=C$ and obtain $y=C$ as an extremal.

Can anyone please explain what is the error in this process?

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If the Lagrangian $F$ does not depend explicitly on the independent variable $x$, then the Beltrami identity (BI) is a necessary (but not a sufficient condition) for a solution to the Euler-Lagrange (EL) equation.

The BI can alternatively be viewed as a special case of Noether's theorem: The energy function is conserved along a solution.

In OP's example it turns out that there are no solutions to the EL equation.

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  • $\begingroup$ Look it's Qmechanic! big fan of your answers on physics stack $\endgroup$
    – Bagaringa
    Feb 22 at 18:50
  • $\begingroup$ Hi @Bagaringa. Thanks. $\endgroup$
    – Qmechanic
    Feb 22 at 19:00

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