# First integrals of the Euler-Lagrange equation

Consider $y=y(x)$ that extremises the functional

$I[y]=\int_{a}^{b} f(x, y, y', y'')dx$

so that $y$ satisfies the E-L equation:

$\dfrac{\partial f}{\partial y}-\dfrac{d}{dx}(\dfrac{\partial f}{\partial y'})+\dfrac{d^2}{dx^2}(\dfrac{\partial f}{\partial y''})=0$

And as an exercise we were meant to show:

$\dfrac{d}{dx}[f-y'(\dfrac{\partial f}{\partial y'}-\dfrac{d}{dx}(\dfrac{\partial f}{\partial y'}))-y'\dfrac{\partial f}{\partial y'}]=\dfrac{\partial f}{\partial x} (1)$

I've tried for a while to get the above but can't somehow...I might have copied incorrectly, can someone first tell me whether (1) is correct in the first place (dashes in the right places etc.)? Thanks!

EDIT: Okay, seems like I've copied incorrectly in lectures, (1) should be:

$\dfrac{d}{dx}[f-y'(\dfrac{\partial f}{\partial y'}-\dfrac{d}{dx}(\dfrac{\partial f}{\partial y''}))-y''\dfrac{\partial f}{\partial y''}]=\dfrac{\partial f}{\partial x}$

then it's just straightforward computation,as Avitus says. Solved.

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What did you try? It looks to me a quite long but straightforward computation with repeated chain rules...is this what you did? For example, can you explicitly write $\frac{df}{dx}$ using the chain rule? –  Avitus Jul 21 '13 at 18:19