Homogeneous equations always do have an integrating factor

Printable View

October 1st 2010, 08:31 PM
Krizalid

Homogeneous equations always do have an integrating factor

I'm looking for some proofs about the following statement:

let $M(x,y)\,dx+N(x,y)\,dy=0,$ be a homogeneous ODE.

find an integrating factor for the above equation.
October 2nd 2010, 05:29 AM
Danny

If the ODE is homogeneous then it admits the symmetry

$ \bar{x} = e^{\varepsilon \alpha}x,\;\;\;\bar{y} = e^{\varepsilon \alpha}y, $

or infinitesimal transformation

$ \bar{x} = x + \varspsilon \alpha x + O(\varepsilon^2),\;\;\; \bar{y} = y + \varspsilon \alpha y + O(\varepsilon^2) $

and the original ODE has the integrating factor

$ \mu = \dfrac{1}{\alpha x M + \alpha y N} $ (we can set $\alpha = 1$ wlog)

This was show by Lie in the late 1800's.
October 3rd 2010, 04:38 AM
HallsofIvy

Let me point out that "find an integrating factor" is NOT a statement and you are not trying to prove it.

In fact, any first order differential equation has an integrating factor- its just simpler to find if the equation is homogeneous.
October 3rd 2010, 07:11 AM
Krizalid

Quote:

Originally Posted by Danny

If the ODE is homogeneous then it admits the symmetry

$ \bar{x} = e^{\varepsilon \alpha}x,\;\;\;\bar{y} = e^{\varepsilon \alpha}y, $

or infinitesimal transformation

$ \bar{x} = x + \varspsilon \alpha x + O(\varepsilon^2),\;\;\; \bar{y} = y + \varspsilon \alpha y + O(\varepsilon^2) $

and the original ODE has the integrating factor

$ \mu = \dfrac{1}{\alpha x M + \alpha y N} $ (we can set $\alpha = 1$ wlog)

This was show by Lie in the late 1800's.

two advanced for me! but thanks!

Quote:

Originally Posted by HallsofIvy

Let me point out that "find an integrating factor" is NOT a statement and you are not trying to prove it.

In fact, any first order differential equation has an integrating factor- its just simpler to find if the equation is homogeneous.

i never said the above ODE was linear or something, it could have quadratic or cubic terms and can still be homogeneous.