# Euler's Equation/Independent of x

## Theorem

Let $y$ be a mapping.

Let $J$ be a functional such that:

$\ds J \sqbrk y = \int_a^b \map F {y, y'} \rd x$

Then the corresponding Euler's Equation can be reduced to:

$F - y' F_{y'} = C$

where $C$ is an arbitrary constant.

## Proof

Assume that:

$\ds J \sqbrk y = \int_a^b \map F {y, y'} \rd x$

Then:

 $\ds \delta J$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds F_y - \dfrac \d {\d x} F_{y'}$ $=$ $\ds F_y - \paren {\dfrac {\d y} {\d x} \dfrac {\partial F_{y'} } {\partial y} + \dfrac {\d y'} {\d x} \dfrac {\partial F_{y'} } {\partial y'} }$ $\ds$ $=$ $\ds F_y - y' F_{y'y} - y'' F_{y'y'}$

Multiply this differential equation by $y'$.

This gives:

 $\ds F_y y' - {y'}^2 F_{y'y} - y'y'' F_{y'y'}$ $=$ $\ds \paren {F_{y} y' + F_{y'} y''} - F_{y'} y'' - y' \paren {y' F_{y'y} + y'' F_{y'y'} }$ $\ds$ $=$ $\ds \dfrac {\d F} {\d x} - y' \paren {\dfrac {\d y} {\d x} \dfrac {\partial F_{y'} } {\partial y} + \dfrac {\d y'} {\d x} \dfrac {\partial F_{y'} } {\partial y'} } - F_{y'} y''$ $\ds$ $=$ $\ds \dfrac {\d F} {\d x} - y' \dfrac {\d F} {\d x} - \dfrac {\d y'} {\d x} F_{y'}$ $\ds$ $=$ $\ds \dfrac \d {\d x} \paren {F - y' F_{y'} }$ $\ds$ $=$ $\ds 0$

Integration yields the desired result.

$\blacksquare$