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Solution to the exact differential equation

  1. Sep 22, 2004 #1
    (x^2)(y^3) + x(1 + y^2)y' = 0

    the integrating factor to make the above equation exact is (1)/(xy^3)

    i have worked this equation out and have c = .5x^2 as the solution; however, the textbook says the solution is c = x^2 - y^(-2) + 2lnlyl

    apparently they got this solution because h'(y) = y^(-3) + y(-1)
    i found h'(y) to be equal to zero.

    some sort of feedback would be greatly appreciated.
  2. jcsd
  3. Sep 23, 2004 #2


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    I get a positive sign with a factor of 2 for the middle term compared with the textbook solution. Are you sure you typed it correctly?

    I just did a direct integration of the ODE since it's separable.
  4. Sep 23, 2004 #3


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    Do you understand that "c= .5x^2" says that x is a constant? What happened to y in your solution?

    By the way, this equation is separable. It's easier to do it that way.
    Last edited: Sep 23, 2004
  5. Sep 23, 2004 #4
    This is a seperable equation from the get go. I got the book's answer.

    It seperates to [(1 + y^2) / y^3 ] dy = -x dx . Once you integrate both sides you get :

    ln(y) - 1/2y^2 = -(x^2 / 2) + c. Solve for c gives you :

    c = 2ln(y) - y^-2 + x^2 Don't worry about the fact that c would really be 2c, because it is just a constant.

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