TITLE: What if the Euler Lagrange equation yields a 'trivial' answer
QUESTION [5 upvotes]: 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??

REPLY [9 votes]: 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.