Need Help with Differential Equation (integrating factor)

  1. minger

    minger 1,498
    Science Advisor

    Hi there, I am having a bit of difficulty finding the integration factor for the following problem. The problem lies in taking the integral of a function of two variables. Anyways, heres what I have:

    [itex] y^2+y-xy'=0[/itex]
    I then divided by x (i prefer it this way), so

    [itex] \frac{y^2+y}{x} + y' = 0 [/itex]

    then, letting u be a integration factor:

    [itex] \frac{duM}{dy} = \frac{duN}{dx}
    \\\\\Rightarrow
    \frac{du(y^2+y)x^{-1}}{dy} = -\frac{du}{dx}
    [/itex]

    [itex]
    (\frac{du}{dy})(\frac{y^2+y}{x}) + u\frac{2y+1}{x} = -\frac{du}{dx}
    [/itex]
    Assuming that u is a function of only x, we get rid of the u partial with y term, so:
    [itex]
    \frac{du}{dx} = -u\frac{2y+1}{x}
    \\\mbox{now doing some dividing on both sides and integrating}
    \\
    \int\frac{\delta\mu}{\mu} = -\int\frac{2y+1}{x}dx
    [/itex]

    Now this is where I'm stuck. I guess I can't remember how to integrate that right side. Can I just pull the (2y+1) term out of the integral assuming that its contant? Thanks for the help

    edit: wow, this latex is really messed up. I'm working on it. The last equation, the integral is correct, and that's the most important part though.
     
    Last edited: Sep 5, 2006
  2. jcsd
  3. Why not just make the case separable? Seems to me a lot easier this way
    [tex] y^2+y - x y^{\prime} = 0 \Rightarrow \frac{y^{\prime}}{y^2+y}=\frac{1}{x} \Rightarrow \frac{dy}{y^2+y} = \frac{dx}{x} [/tex]
    and this is an easy integral to do.
     
  4. HallsofIvy

    HallsofIvy 40,678
    Staff Emeritus
    Science Advisor

    You can't integrate that last integral precisely because in has y in it. Your assumption that the integrating factor u could be a function of x only is incorrect.

    As xman pointed out, this is obviously a separable equation. (Which is the same as saying that
    [tex]\frac{1}{x(y^2+ y)}[/tex]
    is an integrating factor.)
     
  5. minger

    minger 1,498
    Science Advisor

    wow, I cannot believe that I missed that. It was in the integrating factor section, so I just assumed it to be of that form.

    I feel like an idiot, thanks a lot.
     
  6. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    Note that usually, non-linear diff.eqs rarely can be solved by methods of finding integrating factors.
    That is, the method of finding an integrating factor is mainly of use for solving linear diff. eqs.

    On special occasions, though, you may succeed..
     
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