What is purpose of an integrating factor and how do you get one?

[jenzao]

Homework Statement

Am I doing this correctly?
Is 2y^4dx +4xy^3dy exact or inexact?

Homework Equations

The Attempt at a Solution

So if its exact, then M(x,y)dx = N(x,y)dy right? (Euler)

But how can I take partial derivative of 2y^4 with respect to x, if there is no x to derive??

thats where I am stuck?

 

[rootX]

Am I doing this correctly?
Is 2y^4dx +4xy^3dy exact or inexact?

This is unclear.

So if its exact, then M(x,y)dx = N(x,y)dy right? (Euler)

Unclear

But how can I take partial derivative of 2y^4 with respect to x, if there is no x to derive??

2y^4 is constant does not change with respect to x

If the function f = 2y^4 does not change as x changes; it remains constant
d [f] / dx should be?

 

[dynamicsolo]

Homework Statement

Am I doing this correctly?
Is 2y^4dx +4xy^3dy exact or inexact?

Homework Equations

The Attempt at a Solution

So if its exact, then M(x,y)dx = N(x,y)dy right? (Euler)

You have omitted something here: this is supposed to be

M(x,y) dx + N(x,y) dy = 0

is an exact differential equation (in R) iff

M_y (x,y) = N_x (x,y) everywhere on R.

So you need to take partial derivatives to check this...

But how can I take partial derivative of 2y^4 with respect to x, if there is no x to derive??

... which you have reversed the order of in your work. You should be taking the partial derivative of 2y^4 with respect to y.

 

[HallsofIvy]

Homework Statement

Am I doing this correctly?
Is 2y^4dx +4xy^3dy exact or inexact?

Homework Equations

The Attempt at a Solution

So if its exact, then M(x,y)dx = N(x,y)dy right? (Euler)

No, you have it reversed. It is M(x,y)_y= N(x,y)_x- assuming you mean M(x,y)dx+ N(x,y)dy.

But how can I take partial derivative of 2y^4 with respect to x, if there is no x to derive??

Then it is a constant with respect to x and the derivative would be 0. However what you want is (2y⁴)_y and (4xy³)_x.

thats where I am stuck?

Remember WHY you are doing that. IF M(x,y)dx+ N(x,y)dy is an exact differential, then there exist some function f(x,y) such that df= M(x,y)dx+ N(x,y)dy. By the chain rule,
[tex]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y} dy[/itex]
so if M(x,y)dx + N(x,y)dy is exact, M(x,y)= f_x and N(x,y)= f_y. Taking the derivatives of M and N give the mixed derivatives: M(x,y)_y= f_xy and N(x,y)_x= f_yx. As long as the derivatives are all continuous, those mixed derivatives must be equal.

That does NOT answer your title questin "What is the purpose of an integrating factor and how do we find one".

Given a function f(x,y), it "differential" is
$$df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy$$
Writing arbitrary functions for M(x,y) and N(x,y) in M(x,y)dx+ N(x,y)dy may LOOK like a differential but as I pointed out above, such an f exists if and only if the above criterion is satisfied- if and only if the differential is exact. It can be shown that, given anything of the form M(x,y)dx+ N(x,y)dy you can always multiply it by some function of x and y to make it exact- an "integrating factor". There is, however, NO general method of finding an integrating factor.

In this particular case, since both M(x,y) and N(x,y) involve only powers of x and y, I might try xⁿy^m as an integrating factor. Multiplying by that, I have 2xⁿy^m+4dx+ 4xⁿ⁺¹y^m+3dy. Now, (2xⁿy^m+4)_y= 2(m+4)xⁿy^m+3 and (4xⁿ⁺¹y^m+3)_x= 4(n+1)xⁿy^m+3. The powers of x and y are correct! In order that those be equal we must have 2(m+4)= 4(n+1) or 2m+ 8= 4n+ 4 so m= 2n- 2. It is easest to take n= 1 so m= 0. That means that x itself is an integrating factor: (2xy⁴)<sub>y[/sup]= 8xy³ and (4x²y³)x= 8xy³ also.

That means there must exist some function f(x,y) so that df= fxdx+ fyydy= (2xy⁴)dx+ (4x²y³)dy. In particular, fx= 2xy⁴ so, integrating with respect to x (while treating y as a constant since this is a partial derivative) we have f(x,y)= x²y⁴+ C(y). Notice that, because y is treated as a constant in partial differentiation, the "constant of integration" might be a function of y. However, differentiating that with respect to y we have (x²y⁴+ C(y))y= 4x²y³+ C'(y)= 4x²y³ so C'(y)= 0- C really is a constant. f(x,y)= x² y⁴+ C.

Now, whether multiplying by that "integrating factor" disrupts the rest of the problem depends upon what the problem is! dynamicsolo is incorrect to suggest that there must be an "= 0" on that. A "differential" is not a "differential equation" and an "exact differential is not an "exact differential equation". However, if the problem really were, as he suggested, to solve the (non-exact) differential equation 2y⁴dx+ 4xy³= 0, then multiply both sides by x gives 2xy⁴dx+ 4x²y³= d(x²y⁴)= 0 which has the general solution x²y⁴= C.</sub>