Simplifying the Euler-Lagrange Equation for Explicitly Independent Functions

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SUMMARY

The discussion focuses on simplifying the Euler-Lagrange equation when the integrand f(y, y', x) is independent of x, specifically when f = f(y, y'). The key equation derived is \(\frac{df}{dx} = \frac{d}{dx} (y' \frac{\partial f}{\partial y'})\), which is established by substituting \(\frac{\partial f}{\partial y}\) using the Euler-Lagrange equation. Participants encountered challenges with additional terms arising during the substitution process, indicating the need for careful manipulation of derivatives.

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Homework Statement



If the integrand f(y, y', x) does not depend explicitly on x, that is, f = f(y, y') then
[tex]\frac{df}{dx} = \frac{\partial f}{\partial y}y' + \frac{ \partial f } {\partial y' } y''[/tex]Use the Euler-Lagrange equation to replace [tex]\partial f / \partial y[/tex] on the right and hence show that [tex]\frac{df}{dx} = \frac{d}{dx} ( y' \frac{\partial f}{\partial y'} )[/tex]

Homework Equations



[tex]\frac{\partial f }{\partial y} = \frac{d}{dx} \frac{\partial f}{\partial y'}[/tex]

The Attempt at a Solution



By substituting in for df/df, I get an extra term that I can't seem to make go away.

[tex]\frac{df}{dx} = \frac{d}{dx} y' \frac{ \partial f }{\partial y'} + \frac{\partial f}{\partial y'} y''[/tex]

I can't seem to get rid of that extra term, it seems like it should be straight forward but...
 
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Maybe it helps if you work backwards... what is
[tex] \frac{df}{dx} = \frac{d}{dx} ( y' \frac{\partial f}{\partial y'} )[/tex]
?
 

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