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The example I'm doing gives an equation $L(y, y') = \frac{y'}{y}$

then $\frac{\partial L}{\partial y} = -\frac{y'}{y^2}$ and $\frac{\partial L}{\partial y'} = \frac{1}{y}$ ... $\frac{d}{dx}\frac{\partial L}{\partial y'} = \frac{d}{dx}\frac{1}{y} = -\frac{y'}{y^2}$

substituting into the Euler Lagrange $\frac{\partial L}{\partial y} = \frac{d}{dx} \frac{\partial L}{\partial y'}$ then yields $-\frac{y'}{y^2} = -\frac{y'}{y^2}$

Does this mean that any differentiable function $y$ is a solution to the Euler-Lagrange equation for that Lagrangian??

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The reason is that $ \mathcal{L}=\int_a^b L\big(y(t),y'(t)\big)\,dt =\int_a^b\frac{y'(t)}{y(t)}\,dt=\big[\ln y(t)\big]_a^b=\ln y(b)-\ln y(a) $ is independent of the path, and hence, all paths are stationary.

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    @Sother: That's right.2018-01-21