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Homework Help: How to find an integrating factor for this problem?

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Find an integrating factor for:
    xdy - (y + x^2 + 9y^2)dx = 0


    2. Relevant equations
    P(x,y)dx + Q(x,y)dy = 0
    Δμ = μyP - μxQ.
    where μ is the integrating factor.
    3. The attempt at a solution
    well, I don't know what I should do. I can use the formula I wrote but that would lead to solving a PDE. any ideas?
     
  2. jcsd
  3. Oct 25, 2011 #2
    No ideas?
     
  4. Oct 25, 2011 #3
    It would take a pretty lucky guess to come up with an integrating factor directly. However, you can use a Riccati transform and then back-track to find an integrating factor if that's your pleasure.

    If you substitute y(x) = - x (dw/dx) / (9 w) where w is a function of x, you'll wind up with a linear 2nd order ODE in w.

    Wikipedia has a decent description: http://en.wikipedia.org/wiki/Riccati_equation
     
  5. Oct 25, 2011 #4
    I don't know how to solve a linear 2nd order ODE yet. I mean I know but our professor hasn't started the second chapter which is about 2nd order linear equations. so I can't transform it into a 2nd order linear ODE. Is there a way to solve it without transforming it into a 2nd order linear ODE by finding an integrating factor?
     
  6. Oct 25, 2011 #5
    I sure don't see an obvious way. Is it possible that the equation should be

    y dy - (y + x^2 + 9 y^2) dy = 0 ?

    An integrating factor would be simple to find in that case.
     
  7. Oct 25, 2011 #6
    No :(
     
  8. Oct 25, 2011 #7

    lurflurf

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    Homework Helper

    group in terms of 3y/x to make the integrating factor easy to see
    x dy - (y + x^2 + 9y^2)dx = 0
    (x2/3)d(3y/x)-x2((3y/x)2+1)dx=0
     
  9. Oct 25, 2011 #8
    Now It's clear what I should do next. but I don't understand how you obtained the second equation from the first that directly.
     
  10. Oct 25, 2011 #9

    lurflurf

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    Homework Helper

    It helps that I knew that
    x dy-y dx=x2d(y/x)
    from there it was clear x and y/x would be better variables that x and y.
    These take a bit of practice.
    There are a few others, but I keep in mind these common examples
    d(xy)=x dy+y dx
    y2 d(x/y)=y dx-x dy
    x2 d(y/x)=x dy-y dx
    (x2+y2) d(arctan(y/x))=x dy-y dx
    [tex] x^{1-p}y^{1-q}{d}(x^py^q)=pydx+qxdy[/tex]
     
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