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Integrating factor for exact differential equation

  1. Jun 2, 2012 #1
    xdy-ydx=(x2+y2)2(xdx+ydy)
    (Hint: consider d(x2+y2)2)


    2. Relevant equations
    d(x2+y2)2
    d(arctan(y/x))=xy|-y/(x2+y2)


    3. The attempt at a solution

    The answer is arctan(y/x)-(.25(x2+y2)2)=C

    I correctly solved other problems of this type, but this is the hardest one in the problem set and I have no idea how to solve it. This isn't a homework I have to turn in, I'm just trying to fully understand this concept, so I would really appreciate it if someone would just help me work through this. I don't understand how d(x2+y2)2 is relevant to anything, and I'm assuming the denominator of d(arctan(y/x)) somehow plays a role in 'cancelling' the (x2+y2)2? I guess I'm mostly confused because it seems like you have to use two integrating factors?

    Thanks!
     
  2. jcsd
  3. Jun 3, 2012 #2

    tiny-tim

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    hi giacomh! :smile:

    what is d[(x2+y2)2]?

    and what is d[y/x] ? :wink:
     
  4. Jun 3, 2012 #3
    d[arctan(y/x)]=xdy-ydx/(x2-y2)
    d(x2+y2)2=2(x2+y2)(2x+2y)

    I think the fact that d(x2+y2)2= 2(xdx+xdy+ydx+ydy) might have to do with something as well, now?
     
  5. Jun 3, 2012 #4

    tiny-tim

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    no … how can there be no d's on the RHS? :redface:

    try a simpler one … what is d(x2+y2) ?

    (and what is d(y/x) ?)​
     
  6. Jun 3, 2012 #5
    I rewrote the differential equation as:

    [itex]-y-x(x^{2}+y^{2})^{2}+(x-y(x^{2}+y^{2})^{2})\frac{dy}{dx}=0[/itex]

    Then I sort of just guessed from the hint that the integrating factor was [itex]μ(x,y)=\frac{1}{x^{2}+y^{2}}[/itex]. This indeed makes the equation exact.

    Using this integrating factor, I eventually found that the answer was:

    [itex]-tan^{-1}(\frac{x}{y})-\frac{1}{4}x^{4}-\frac{1}{2}x^{2}y^{2}-\frac{1}{4}y^{4}=C_{1}[/itex]

    Or, if we multiply both sides by -1, letting C2=-C1:

    [itex]tan^{-1}(\frac{x}{y})+\frac{1}{4}x^{4}+\frac{1}{2}x^{2}y^{2}+\frac{1}{4}y^{4}=C_{2}[/itex]

    Which is equivalent to the answer given by wolfram alpha. Simplified (I'll just rewrite C2 as C), it becomes similar to your answer (I wonder if they're equivalent?):

    [itex]tan^{-1}(\frac{x}{y})+\frac{1}{4}(x^{2}+y^{2})^{2}=C[/itex]

    However, the guess and check method of finding the integrating factor was very inefficient and I wonder if there is a more straightforward way of finding it. I doubt that I solved the problem the way you were expected to solve it.
     
    Last edited: Jun 3, 2012
  7. Jun 4, 2012 #6
    ermm..

    d(x2+y2)2=2(x2+y2)(2xdx+2ydy)?

    d(y/x)=xdy-ydx/x2
     
  8. Jun 4, 2012 #7
    Also, I have no idea how you rewrote the equation like this. Sorry, this thread is making me feel really dumb since its just basic algebra lol, but i can't rearrange it like that.


     
  9. Jun 4, 2012 #8

    tiny-tim

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    yup! :biggrin:

    now rearrange the original equation, using those two results …

    what do you get? :smile:
     
  10. Jun 4, 2012 #9
    think of it as df=(∂f/∂x)dx+(∂f/∂y)dy, there f(x,y)=(x2+y2)2.

    Let us know if you do not know how to compute this.

    Then try the harder dg for g(x,y)=arctan(y/x).

    No offense cjc0117, but I think finding integrating factor for this setup is not typically expected.

    EDIT: I see now that you took care of f, now try the g I wrote.
     
  11. Jun 4, 2012 #10
    d[y/x]=2d[(x2+y2)2](x2+y2)/(x2)
     
  12. Jun 4, 2012 #11
    algebrat,

    I think I solved your g(x,y) in a previous post:
    d(arctan(y/x))=xy|-y/(x2+y2)

    Also, arctan wasn't really given as a hint in the problem, it just appeared in the solution in the solution manual.
     
  13. Jun 4, 2012 #12

    tiny-tim

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    ok! :smile:

    (except it's 4, not 2 :wink:)

    now rewrite that as

    x2/(x2+y2) d[y/x] = 4d[(x2+y2)2],

    then fiddle about with the LHS so as to make it a function of y/x :wink:
     
  14. Jun 4, 2012 #13
    Ahh using arctan(y/x) I got:

    d[arctan(y/x)]=4d[(x2+y2)2]

    Thats closer...but still not right...
     
  15. Jun 4, 2012 #14

    tiny-tim

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    oops!

    oops! not 4, 1/4 ! :rolleyes:
     
  16. Jun 4, 2012 #15
    Oh, wait. That is right lol. Thanks so much everyone!
     
  17. Jun 4, 2012 #16
    Sorry, that wasn't my intention. If it's any consolation, this thread is making me feel dumb too :smile:

    No offense taken. As I said in my post, I doubt that's how they wanted you to solve the problem. I find the above method of solving very interesting, albeit confusing. I need more time to look over the solution.

    What I'm wondering is why both my answer and the answer given by giacomh in the first post both seem correct. Is it because my solution is wrong or because the solution is not unique?
     
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