Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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 by a moderator: 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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook