Diffrential equations, integration factor with two vars

[barak]

1. I need to find a condition that the equation will have a integration factor from the shape K(x*y).
(K-integration factor sign)

2.the eq from the shape M(x,y)dx+N(x,y)dy=0 ,not have to be exact!3. i tried to open from the basics. d(k(x*y)M(x,y))/dy=d((k(x*y)N(x,y))/dx.
and i used the fact that d(k(x,y))/dy is x (exc. for dx)/

im hoping i was clear enough , thanks and sorry for my bad english.

 

[BiGyElLoWhAt]

well, normally an integration factor will be of the form $e^{\int P(x)}$
I'm confused by your notation, it seems as though you're setting up an exact function, or maybe that's what you're saying in 2: it's not an exact function?

 

[Mark44]

1. I need to find a condition that the equation will have a integration factor from the shape K(x*y).
(K-integration factor sign)

"the equation" -- What equation?

2.the eq from the shape M(x,y)dx+N(x,y)dy=0 ,not have to be exact!3. i tried to open from the basics. d(k(x*y)M(x,y))/dy=d((k(x*y)N(x,y))/dx.
and i used the fact that d(k(x,y))/dy is x (exc. for dx)/

im hoping i was clear enough , thanks and sorry for my bad english.

It's not clear to me at all. What is the equation you're trying to solve?

 

[LCKurtz]

@barak: Perhaps this link will help you:
http://www.cliffsnotes.com/math/differential-equations/first-order-equations/integrating-factors

 

[HallsofIvy]

The initial equation is M(x,y)dx+ N(x,y)dy= 0 and you multiply by k(xy) where k is some function of a single variable: k(xy)M(x, y)dx+ k(xy)N(x,y)dy= 0.
In order that this be "exact" we must have $(k(xy)M(x, y))_y= (k(xy)N(x, y))_x$.

xk'(xy)M(x,y)+ k(xy)M_y(x, y)= yk'(x,y)N(x,y)+ k(xy)N_x(x,y)

(xM(x,y)- yN(x,y))k'(xy)= k(xy)(N_x(x,y)- M_y(x,y)

 

[barak]

thanks a lot guys u all helped me