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Differential equation help

  1. May 9, 2007 #1

    ssb

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    1. The problem statement, all variables and given/known data
    Consider the differential equation given by

    [tex]y' = Ay - By^2 [/tex]

    List the equilibrium solutions in increasing order and classify them as stable, semistable, or unstable.

    2. Relevant equations



    3. The attempt at a solution

    Im just looking for a point in the right direction on this one. I guess my attempt would be to possibly factor out one of the y's on the right hand side. Then maybe move all the y's to the left side of the equation. Tell me... is this what I should be doing? The overall goal is to find up to 3 equilibrium solutions.
     
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  3. May 9, 2007 #2

    Hootenanny

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    Well you know at equilibrium that y' = 0 so....
     
  4. May 9, 2007 #3

    ssb

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    Thankyou.

    [tex] 0 = Ay - By^2 [/tex]

    [tex] By^2 = Ay [/tex]

    [tex] By = A [/tex]

    [tex] y = A/B [/tex]

    the a/b is one equilibrium solution. Did I solve this the correct way or did I just luck out by doing this? Also how would I know that another existed (if it did)?
     
  5. May 9, 2007 #4
    it should be [tex] y = \frac{a}{b} [/tex]. And look at the end behavior to look at stability(i.e. going away from eq. pt, converging to eq pt, or one part going away and one converging)
     
    Last edited: May 9, 2007
  6. May 9, 2007 #5

    Hootenanny

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    I'm not sure, I'd have solve it like this;

    [tex]\begin{array}{l}
    y(A - By) = 0\\
    \Rightarrow y = \left\{ 0,A/B\right\}
    \end{array}
    [/tex]

    But you got one of the answers right :smile:
     
    Last edited: May 9, 2007
  7. May 9, 2007 #6

    ssb

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    omg its like basic algebra all over again. Im convinced that after so much math gets pushed inside your head, you start forgetting the earlier stuff. Thanks a ton.
     
    Last edited: May 9, 2007
  8. May 9, 2007 #7

    Office_Shredder

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    Ok, I'll be the one to throw it out there... where did the +/- come from? A-By=0 only if y is A/B, it's not if y=-A/B.
     
  9. May 9, 2007 #8

    Hootenanny

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    Wait, correction! The negative root isn't a solution. I really don't know what I was thinking! There are only two solutions (as the should be with a quadratic :mad:).

    Edit: Shredder beat me to it. Sorry guys, this has happened twice now; I guess I should stop posting so late.
     
    Last edited: May 9, 2007
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