# Integrating factor of (a+1)ydx + (b+1)xdy = 0

1. Oct 3, 2005

### asdf1

for the equation, (a+1)ydx+(b+1)xdy=0,
i am wondering how to get (x^a)(y^b) as an integrating factor~

the following is my work:

(1/F)(dF/dx)=(a-b)/[(b+1)x]
=> F=cx^[(a-b)/(b+1)]

why doesn't that method work?

2. Oct 3, 2005

### HallsofIvy

Staff Emeritus
It's not clear to me what you are doing. What happened to y?

The simplest way to get the integrating factor is just to try an integrating factor of the form xαyβ. The equation becomes
$$((a+1)x^{\alpha}y^{\beta+1})dx+ ((b+1)x^{\alpha+1}y^{\beta})dy$$

In order for that to be exact, we must have
$$((a+1)x^{\alpha}y^{\beta+1})_y= ((b+1)x^{\alpha+1}y^{\beta})_x$$
or
$$(a+1)(\beta+1)x^{\alpha}y^{\beta}= (b+1)(\alpha+1)xx^{\alpha}y^{\beta}$$
That will clearly be true if α= a and β= b. Therefore xayb is an integrating factor.

3. Oct 3, 2005

### asdf1

wow! how'd you think of trying the integrating factor of the form (x^α)(y^β)?

4. Oct 4, 2005

### saltydog

Well, shouldn't that be:

$$(a+1)(\beta+1)x^{\alpha}y^{\beta}= (b+1)(\alpha+1)x^{\alpha}y^{\beta}$$

Just want to be precise that's all.

5. Oct 5, 2005

### asdf1

hmmm... i looked up an edition of advanced engineering mathematics, and i saw a little description of that kind of substition, but it didn't explain why...
:P