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Homework Help: Bernoulli Equation

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

    [tex]t^2y'+2ty-y^3=0[/tex]

    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].

    [tex]-\frac{1}{2}t^2v^{-\frac{3}{2}}+2tv^{-\frac{1}{2}}-v^{-\frac{3}{2}}=0[/tex]

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

    [tex]-\frac{1}{2}t^2v^{3}+2t-v^{3}=0=2t-\left(\frac{1}{2}t^2+1\right)v^3[/tex]

    For which I would say:

    [tex]v=\left(\frac{2t}{\frac{1}{2}t^2+1}\right)^{\frac{1}{3}}[/tex]

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

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

    HallsofIvy

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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

    saltydog

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    And what happen to the constant of integration?

    So you got:

    [tex]t^2y^{'}+2ty-y^3=0[/tex]

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

    [tex]y^{-3}dy+\frac{2}{t}y^{-2}t=\frac{1}{t^2}dt[/tex]

    Noting that:

    [tex]d(y^{-2})=-2y^{-3}dy[/tex]

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

    [tex]v=y^{-2}[/tex]

    and:

    [tex]dv=-2y^{-3}dy[/tex]

    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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