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

[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?

 

[HallsofIvy]

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 x^αy^β. 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 x^ay^b is an integrating factor.

 

[asdf1]

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

 

[saltydog]

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 x^αy^β. 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 x^ay^b is an integrating factor.

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.

 

[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