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ODE Integrating factors found by inspection

  1. Jul 20, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm reading a chapter out of Elem. DE 6th Edition by Rainville and Bedient (Ch 4 pg 61) titled integrating factors found by inspection. To explain it, the authors start with an equation, which is grouped to become:

    [itex]y dx + x dy + x^3y^2 dy = 0[/itex]

    which then becomes:

    [itex]\frac{d(xy)}{(xy)^3} + \frac{dy}{y} = 0[/itex]

    I am unsure of how they get from there to the next step which is this....

    [itex]-\frac{1}{2x^2y^2} + ln|y| = -ln|c|[/itex]

    3. The attempt at a solution

    Trying to figure out the intermediate steps I start by doing this...

    [itex]\frac{d(xy)}{x^3y^2} = - \frac{dy}{y}[/itex]

    And then I'm guessing they integrate both sides. However, if that's the case, what would I be integrating the left side with respect to? I expected the equation to become this...

    [itex]\frac{1}{x^2y^2} + ln|y| = c[/itex]

    Thank you.
     
    Last edited: Jul 20, 2012
  2. jcsd
  3. Jul 20, 2012 #2

    LCKurtz

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    No, it doesn't. There is either a typo in the text or in your post. If your original equation was$$
    y dx + x dy + x^3y^2 dy = 0$$ and you divided by ##x^3y^3## it would become$$
    \frac {d(xy)}{(xy)^3} + \frac {dy}{y}=0$$which gives your desired answer.
     
  4. Jul 20, 2012 #3
    The typo was in my post. I corrected it. Sorry about that. With that now corrected, I'm still not sure how they get from the one step to the next.
     
  5. Jul 20, 2012 #4

    LCKurtz

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    It is of the form ##\frac {du}{u^3} = u^{-3}du## which integrates to ##\frac{u^{-2}}{-2}##. And calling the constant of integration ##-\ln c## instead of ##c## is just a convenience. They can both be anything.
     
  6. Jul 20, 2012 #5
    ok, and in this case u = xy. I was thinking that they had to be dealt with separately, but now I see it.

    Thank you very much.
     
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