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Bernoulli Equation

  1. Sep 22, 2005 #1
    I cannot get the correct answer to this for some reason:


    I use the substitution [itex]v=y^{1-n}=y^{-2}\implies y=v^{-\frac{1}{2}}[/itex] and come up with [itex]y'=-\frac{1}{2}v^{-\frac{3}{2}}[/itex] and [itex]y^3=v^{-\frac{3}{2}}[/itex].


    Then multiplying through by [itex]v^{-2}[/itex] gives:


    For which I would say:


    But according to the back of my book, that answer is incorrect. What did I do wrong?

    Thanks for your help.
  2. jcsd
  3. Sep 23, 2005 #2


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    If [itex] y= v^{-\frac{1}{2}}[/itex] then [itex]y'= -\frac{1}{2}v^{-\frac{3}{2}}v'[/itex].

    You seem to have dropped v' throughout. Your differential equation became an algebraic equation!
  4. Sep 23, 2005 #3


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    And what happen to the constant of integration?

    So you got:


    Divide by [itex]t^2[/itex] and then divide by [itex]y^3[/itex], leads to:


    Noting that:


    and that's what you have up there on the left, we make the substitution:




    Plug that in up there, rearrange to a first-order ODE in terms of v and t, find the integrating factor, integrate and not forgetting about the constant of integration, solve for v(t), then converting back to y, you should get:

    [tex]y(t)=\pm \sqrt{\frac{5t}{2+5ct^5}}[/tex]

    Now what?
  5. Sep 23, 2005 #4
    Ahhh, I can't believe I forgot the v'! Thanks for your help.
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