1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A non-exact nonlinear first ODE to solve

  1. Jun 14, 2016 #1
    1. The problem statement, all variables and given/known data

    Solve the following equation.

    2. Relevant equations

    ( 3x2y4 + 2xy ) dx + ( 2x3y3 - x2 ) dy = 0

    3. The attempt at a solution

    M = ( 3xy4 + 2xy )
    N = ( 2x3y3 - x2 )

    ∂M/∂y = 12x2y3 + 2x
    ∂N/∂x = 6x2y3 - 2x

    Then this equation looks like that the integrating factor is (xM-yN).
    IF = x3y4 + 3x2y

    then the new equation would be:

    (3xy^3+2)/(x^2y^3+3x)dx + (2xy^3-1)/(xy^4+3y)dy = 0

    but then if we find that this new equation is exact or not, it proves that this is not exact.
    If so what should I do?
    Please help...

    Note:
    The given answer is:
    x3y2 + x2/y = c
     
  2. jcsd
  3. Jun 14, 2016 #2

    Mark44

    Staff: Mentor

    The above should be ##M = 3x^2y^4 + 2xy##
    You have a mistake above.
    If ##M = 3x^2y^4 + 2xy## then ##M_y = 12x^2y^3 + 2x##
    Since My ≠ Nx, that's enough to show that the equation is not exact.

    I taught diff. equations a number of times, but this isn't a trick that I remember. Where does xM - yN come from?
    I'm not following this at all. If you can find an integrating factor, you multiply both sides of the original diff. equation by it. If it really is an integrating factor, then the new equation will be exact, and you can get the solution.
     
  4. Jun 14, 2016 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  5. Jun 14, 2016 #4

    Mark44

    Staff: Mentor

    I understand that he/she was trying to come up with an integrating factor. The link below presents the usual technique to start with when you're faced with an inexact equation: seeing if a function of x (##\mu(x)##) in the PDF below, and proceeding to a function of y alone (##\nu(y)##) if the previous try didn't work. What threw me was the immediate jump to xM - yN, which as you point out, is not the integrating factor.
     
  6. Jun 16, 2016 #5
    There was a misinterpretation in the question by me... I solved it by myself. Thanks for the concern.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A non-exact nonlinear first ODE to solve
Loading...