# Finding the Integrating Factor

1. Jul 16, 2008

### Happydog

Hi ok so te question is asking for me to fin the integrating factor of (y2x+y)dy + (x2y+2x)dy = 0 the only thing i know is that the integrating factor should be a function of xy

Can some one pleas explain how to do this? Thank you.

2. Jul 16, 2008

### Defennder

You appear to have mistyped your problem. There 2 "dy"s here but no "dx".

3. Jul 17, 2008

### HallsofIvy

Staff Emeritus
Assuming that you mean $(y^2x+ y)dx+ (x^2y+ 2x)dy= 0$, if you know that the integrating factor is a function of xy, the obvious thing to do is to put multiply that equation by u(xy) and see what happens. In order that $u(xy)(y^2x+ y)dx+ u(xy)(x^2y+ 2x)dy= 0$ be exact, we must have $(u(y^2x+ y))_y$ and $(u(x^2y+ 2x)_x)$ equal. That is, $u_x(y^2x+ y)+ u(2xy+1)= u_x(x^2y+ 2x)+ u(2xy+ 2)$. Since u is a function of xy specifically, $u_x= yu'$ and $u_y= xu'$. That is, u must satisfy $xu(y^2x+ y)+ u(2xy+1)= yu'(x^2y+ 2x)+ u(2xy+ 2)$. That gives you a differential equation for u which may or may not be solvable, depending upon whether there really is an integrating factor that is a function of xy only.

4. Jul 17, 2008

### Happydog

Yes the original equation is (y2x+y)dx + (x2y+2x)dy = 0, i have taken your sugestion but it did not work, i tryed a more general form of xnym and multiplied it through to see if i could solve for the n and m and equate the two of them but the two are not equal i end up getting (m+2)=(n+2) and (m+1)=2(n+1).

so can any one also give me some advice on how to solve this problem i am having. Thank you.

5. Jul 18, 2008

### Defennder

I think HallsofIvy's advice is spot on. Equate the two derivatives after differentiating the product of u and the expression preceding the dx and dy. Use the chain rule when differentiating u(xy) with respect to x and y only by denoting product xy as v. Simplify the resulting equation and you'll notice it reduces to a simple separable differential equation. Solve that and express u in terms of x,y only and you're done. I tried it out and it works like a charm.

6. Jul 18, 2008

### HallsofIvy

Staff Emeritus
You told us, originally, that the integrating factor was a function of xy. Why are you now trying xmyn? How do you know the integrating factor is either a function of xy or of the form xmyn?

7. Jul 18, 2008

### Happydog

Yeah it did work sorry about that, and as for havening it as xmyn other times you can use it to find out what power the x and y are to making it easer. mind you you have to know xy are the integrating factor.