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

  1. Jun 29, 2008 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    3. 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 im stuck?
  2. jcsd
  3. Jun 29, 2008 #2
    This is unclear.


    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?
  4. Jun 29, 2008 #3


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    Homework Helper

    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...

    ... which you have reversed the order of in your work. You should be taking the partial derivative of 2y^4 with respect to y.
  5. Jun 30, 2008 #4


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    Science Advisor

    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.

    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.

    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)= fx and N(x,y)= fy. Taking the derivatives of M and N give the mixed derivatives: M(x,y)y= fxy and N(x,y)x= fyx. 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
    [tex]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy[/tex]
    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 xnym as an integrating factor. Multiplying by that, I have 2xnym+4dx+ 4xn+1ym+3dy. Now, (2xnym+4)y= 2(m+4)xnym+3 and (4xn+1ym+3)x= 4(n+1)xnym+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)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.
    Last edited by a moderator: Jun 30, 2008
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