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Finding integrading factor for differential equation

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data
    The differential equation:

    (1-y2) dx + (1-x2) dy = 0

    the integrating factor is of the form μ(x,y) = f(xy). Find f

    2. The attempt at a solution

    I have tried many times to get a function of (xy) such that the LHS of the differential equation to be exact.

    I have tried the integrating factor 1/[(1-y^2)(1-x^2)] and it works..... but it is definitely not a function of the term (xy)...

    Any hints on how should I find it? Or is there any formal way to find it?
     
  2. jcsd
  3. Mar 17, 2012 #2

    vela

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    You know that ##f(xy)(1-y^2)\,dx + f(xy)(1-x^2)\,dy## is supposed to be an exact differential, so what does that tell you about how ##M(x,y) = f(xy)(1-y^2)## and ##N(x,y) = f(xy)(1-x^2)## are related?
     
  4. Mar 17, 2012 #3
    This means that the partial differentiation My(x,y)=Nx(x,y).... right?
     
  5. Mar 17, 2012 #4

    vela

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    Yes. If you do that, you'll eventually end up with a differential equation for f(xy) you can solve.
     
  6. Mar 17, 2012 #5
    That is actually the part where I am stuck. The differential equation will be a partial differential equation right? But I hv not learnt that before... Or it is the same as the ordinary differential equation?
     
  7. Mar 17, 2012 #6

    vela

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    No, you should end up with an ODE. Show us your work.
     
  8. Mar 17, 2012 #7
    My(x,y)=Nx(x,y)

    fy(x,y)(1-y2)-2yf(xy) = fx(x,y)(1-x2)-2xf(x,y)

    Is the above correct? As I don't know how to continue from here onwards.
     
  9. Mar 17, 2012 #8

    vela

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    You need to evaluate the partial derivatives. Think of it as f(u) where u=xy. What's ##\partial f/\partial x## equal to?
     
  10. Mar 17, 2012 #9
    will ∂f/∂x be (df/du)y?
     
  11. Mar 17, 2012 #10

    vela

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    Right, which is f'(xy)y.
     
  12. Mar 17, 2012 #11
    fy(x,y)(1-y2)-2yf(xy) = fx(x,y)(1-x2)-2xf(x,y)

    f'(xy)(x-xy2)-2yf(xy) =f'(xy)(y-x2y)-2xf(x,y)

    f'(xy)+[itex]\frac{2x-2y}{x-xy^{2}-y+x^{2}y}[/itex]f(xy)=0

    Is this the ODE that I need to solve to find f(xy)? It feels like I'm getting something but I don't know how to start :bugeye:
     
  13. Mar 17, 2012 #12

    vela

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    Simplify.
     
  14. Mar 17, 2012 #13
    Oh, so i get

    f'(xy) + [itex]\frac{2}{1+xy}[/itex]f(xy)=0


    So I'll get f(xy)=[itex]\frac{1}{(1+xy)^{2}}[/itex]

    YAY!!! Then I found the integrating factor...

    Thanks A LOT!!!
     
  15. Mar 17, 2012 #14
    For my own practice, I tried to solve the differential equation

    (1-y[itex]^{2}[/itex]) dx + (1-x[itex]^{2}[/itex]) dy = 0

    And the solution I got is

    [itex]\frac{y^{2}-1}{y(1+xy)}[/itex]+[itex]\frac{1}{y}[/itex]=D
    where D is some arbitrary constant

    Is this correct?
     
  16. Mar 17, 2012 #15

    tiny-tim

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    Hi Lily@pie! :smile:
    (I can see how you got it, but it only works for the dx, not the dy :redface:)

    nooo …

    the solution obviously is going to be symmetric in x and y …

    so what's the simplest symmetric numerator you can think of? :wink:
     
  17. Mar 17, 2012 #16
    What do u mean by it only works for the dx and not dy?

    What do u mean by symmetric numerator? y=x?
     
  18. Mar 17, 2012 #17

    HallsofIvy

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    No, that's not what "symmetric in x and y" means. It means that if you were to swap "x" and "y" you would get exactly the same formula. For example, both [itex]x^3+ y^3[/itex] and [itex]x^2+ 3xy+ y^2[/itex] is "symmetric in x and y". [itex]x^3+ y^2[/itex] and [itex]x^2+ 3x+ y^2[/itex] are not.
     
  19. Mar 17, 2012 #18
    Ohhh.... so the simplest one will be something like x+y?

    But i still don't understand what does it mean by the solution only works for dx and not dy...
     
  20. Mar 17, 2012 #19

    vela

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    I think your solution is correct, but you can simplify it a bit by putting everything over a common denominator.
     
  21. Mar 18, 2012 #20
    Okay.. Thanks a lot =)
     
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