# Can't understand derivation multivariate integrating factor µ(x,y) for non-exact DE's

I was looking at how to derive an integrating factor for a non-exact DE that has multiple variable dependency, i.e. µ is xy-dependent, and I found the explanation at the link in the middle of the page at equation (22) (link: http://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html" [Broken]).

However three things in the explanation don't make sense to me:

1. Why do they do µ(x,y) = g(xy) instead of µ(x,y) = xy? What is the extra g for? Is it suppose to be a function?

2. When they say they combine equations (22) and (23) into (24) how do they get rid of the partial derivative g? Do they just divide equation (22) by (23) to get (24)? I'm not sure how they get (24).

3. In equation (29) it seems like they're employing the chain rule to get ∂µ / ∂t. However, they only do it half way. I thought if you did the chain rule it was suppose to be this:

∂µ / ∂t = ∂µ / ∂x * ∂x / ∂t + ∂µ / ∂y * ∂y / ∂t

Is (29) not the chain rule? I guess I'm confused at what they did at (29).

If anyone could help, I would be much appreciated. I've been racking my brain for hours on this.

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lurflurf
Homework Helper

This is a very confusingly written derivation.

1)Yes it is a function, the form of which on deduces from the problem. g(x y) is more general than x y and works in more cases.
2)They make use of the special form of u
ux=y g'(x y)
uy=x g'(x y)
(1/y) ux=g'(x y)=(1/x) uy
3)You form of the chain rule is correct
Here the functional dependencies are unclear
g'(t)=g'(x y)=(1/x) uy
then use line (27)

So, when you say g'(x,y) does that mean it's the derivative dg/dx and not the partial derivative?

And why when you take the partial derivative of the integrating factor µ in both cases:

∂u / ∂x = y * ∂g / ∂x

and

∂u / ∂y = x * ∂g / ∂y

Do you get the partial derivatives of g with respect to x and y multiplied by just one x and one y respectively? Since the integrating factor µ is the function g, wouldn't it be possible that you get more than one y and x?

You made everything else more clear except this part. Thanks a lot for helping me. I've looked for other sources that address integrating factors to make non-exact DE's exact, but they only discuss the µ(x) and µ(y) dependencies which are very easy to understand in terms of the derivation. However, this is the only site I could find that addresses an µ(x,y) dependent integrating factor.

lurflurf
Homework Helper

I think all you confusion involves partial derivatives of related function.
Strickly speaking it is an abuse of notation to talk about u(t),u(x,y) and so on. If we were being more careful we would use different leters for these functions, of at least use the thermodynamic notation to indicate the dependency.
By g' I mean gt because in a sense that is the real g, g only depends on
t=x y, then t depends on x and y so even though there is no point other than confusion we can write
tx=y
ty=x
ux=ut tx=g'(x y) y
uy=ut ty=g'(x y) x
It is a way to switch the variables around.

Your right, that is confusing. But I understand better now. Thank you.