Do Equations of the Form P(x,y)dx+Q(x,y)dy=0 Always Have an Integrating Factor?

[kof9595995]

During one lecture it was mentioned that equations of the form P(x,y)dx+Q(x,y)dy=0 always have at least one integrating factor. But the lecturer didn't know the proof, I've tried using Google but no luck. Anybody can show me the proof? Thanks a lot.

 

[g_edgar]

If you know a solution of the differential equation dy/dx -P/Q, you can use that to find an integrating factor. Or, if you know an integrating factor you can solve the DE. Sometimes it is easy to find an integrating factor. But usually it is just as hard as solving the DE.
In any case, what you are looking for is just the existence theorem for solutions to a first-order DE.

 

[matematikawan]

During one lecture it was mentioned that equations of the form P(x,y)dx+Q(x,y)dy=0 always have at least one integrating factor. But the lecturer didn't know the proof, I've tried using Google but no luck. Anybody can show me the proof? Thanks a lot.

If it is true that there always exist the integrating $$\mu(x)$$, finding one is not easy. We need to solve
$$\frac{\partial}{\partial y}\mu(x,y)P(x,y) = \frac{\partial}{\partial x}\mu(x,y)Q(x,y)$$

I agree with g_edgar sometime it just easier to obtain the solution compare to finding the integrating factor.

 

[kof9595995]

If you know a solution of the differential equation dy/dx -P/Q, you can use that to find an integrating factor. Or, if you know an integrating factor you can solve the DE. Sometimes it is easy to find an integrating factor. But usually it is just as hard as solving the DE.
In any case, what you are looking for is just the existence theorem for solutions to a first-order DE.

Well, thanks man.It seems to be a easy transformation of the question, why didn't I think this way? Kind of embarrassing.