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Solving a homogeneous diff eq

  1. Jan 25, 2005 #1
    The book only has one example of this and it's really confusing me.


    I can see that it's homogeneous of degree 2

    They then let [tex]y = ux[/tex]
    From there they state that [tex]dy = udx+xdu[/tex] (I'm not sure where this is coming from, but can just accept it on faith if I have to)

    I'm fine with making the subs.

    [tex](x^2+u^2x^2)dx+(x^2-ux^2)(u dx+x du)=0[/tex]

    This is the part that really screws me up.


    Where did the [tex]x^3[/tex] come from? All I see is [tex]x^2[/tex]. Or I guess I should ask, how did [tex]udx+xdu[/tex] become just [tex]xdu[/tex]? That would explain the [tex]x^3[/tex] they have.
  2. jcsd
  3. Jan 25, 2005 #2
    Just noticed something else odd.

    How did they turn [tex](x^2+x^2u^2)[/tex] into [tex]x^2(1+u)[/tex] what happend to the other u? Shouldn't it be [tex]x^2(1+u^2)[/tex]?
  4. Jan 25, 2005 #3
    your book is correct

    [tex](x^2+u^2x^2)dx+(x^2-ux^2)(u dx+x du)=0[/tex]

    [tex] (x^2+u^2x^2)dx + ux^2dx -u^2x^2dx + x^3du -ux^3du=0[/tex]
    just calculate the above formula...you must have mage an error there

    how do you do this ???

    Last edited: Jan 25, 2005
  5. Jan 25, 2005 #4
    Thanks, I got it now, I just wasn't carrying things far enough.
  6. Jan 25, 2005 #5
    It's just the product rule of differentiation:

    [tex]y = ux[/tex]

    [tex]\frac{d}{du}\left[ y\right] =\frac{d}{du}\left[ ux\right] [/tex]

    [tex]\frac{dy}{du} = u\frac{dx}{du}+x\frac{du}{du}[/tex]

    [tex]dy = udx+xdu[/tex]
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