Deriving an Integrating Factor for a Non-Exact DE w/ Multiple Dependencies

In summary, the author explains how to derive an integrating factor for a non-exact DE with multiple variable dependencies, but three parts of the explanation don't make sense to them. First, why do they use µ(x,y) = g(xy) instead of µ(x,y) = xy? Second, when they say they combine equations (22) and (23) into (24), how do they get rid of the partial derivative g? Third, in equation (29), it seems like they're employing the chain rule to get ∂µ / ∂t, but they only do it half way. If anyone could help, they would be much appreciated.
  • #1
jshowa
3
0
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" ).

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.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2


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)
 
  • #3


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.
 
  • #4


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.
 
  • #5


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

FAQ: Deriving an Integrating Factor for a Non-Exact DE w/ Multiple Dependencies

How do you determine if a differential equation is non-exact?

A differential equation is considered non-exact if its integrability condition, given by ∂M/∂y = ∂N/∂x, is not satisfied. If this condition is not met, then the equation cannot be solved using the standard method of separation of variables.

What are multiple dependencies in a differential equation?

Multiple dependencies in a differential equation refer to cases where the equation depends on multiple variables, such as x, y, and z. This can make the equation more complex and difficult to solve using traditional methods.

How do you derive an integrating factor for a non-exact differential equation?

To derive an integrating factor for a non-exact differential equation, you first need to identify the integrating factor μ(x,y) by equating the integrability condition to a function of x and y. Then, you can use this integrating factor to multiply both sides of the equation, transforming it into an exact equation that can be solved using separation of variables.

What is the purpose of finding an integrating factor for a non-exact differential equation?

The purpose of finding an integrating factor for a non-exact differential equation is to transform the equation into an exact one, which can be solved using separation of variables. This allows for a simpler and more straightforward solution to the equation.

Are there any limitations to using an integrating factor to solve a non-exact differential equation with multiple dependencies?

Yes, there are limitations to using an integrating factor to solve a non-exact differential equation with multiple dependencies. In some cases, it may not be possible to find an integrating factor, or the resulting exact equation may be too complex to solve. Additionally, this method may not be applicable to all types of non-exact differential equations.

Similar threads

Back
Top