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Homework Help: Differential Equation

  1. Feb 22, 2004 #1
    I'm having trouble finding the general solution for the following problem:

    (1-xy)y' + y^2 + 3xy^3

    It's obviously not an exact equation.

    I multiply through by an integrating factor (x^m)(y^n) and get

    (x^m)(y^(n+2)) + [(3x^(m+1))(y^(n+3))] + [(x^m)(y^n) - (x^(m+1)y^(n+1)]y' (the forms aren't similar, I can't multiply one side to make them similar and have an equivalent equation)

    The partial derivative with respect to y:
    (n+2)[(x^m)(y^(n+1))] + 3(n+3)[x^(m+1)y^(n+2)]

    The partial derivative with respect to x:
    m(x^(m-1)y^n) - (m+1)[(x^m)y^(n+1)]

    They are not of a similar form so it doesn't do any good to solve

    n + 2 = m
    3n + 9 = m + 1

    What am I missing? Do I need to manipulate the equations somehow?

    The General Solution in the book is:

    y = [x+-(4x^2 + c)^(1/2)]^(-1) with an integrating factor of y^(-3)

    Thanks in advance for the help.
  2. jcsd
  3. Feb 23, 2004 #2


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    The first thing I would see is that the very nice
    (n+2)[(x^m)(y^(n+1))] and - (m+1)[(x^m)y^(n+1)] terms match up as long as n+2= -(m+1).

    The second thing I would see is that in order to have
    3(n+3)[x^(m+1)y^(n+2)] and m(x^(m-1)y^n) match up, we would need to have either m+1= m-1, which is impossible, or have those terms equal to 0: n+3= 0 and m= 0. That, of course gives m=0, n= -3 which, fortunately, also satisfy -3+2= -1= -(0+1).

    Taking m=0, n= -3, the integrating factor is x0y-3= y-3 just as your book said.
  4. Feb 23, 2004 #3
    Thanks for the help.

    The mechanical approach the instructor took didn't make this clear, but I understand your explanation.
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