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Ugh Bernoulli Equations

  1. Sep 13, 2009 #1
    Ugh Bernoulli Equations!!!

    1. The problem statement, all variables and given/known data

    dy/dx = y(xy^3-1)

    I tried to set it up and use the bernoulli equation method as a substitution but it didn't work. Any tips?





    3. The attempt at a solution

    I set u=y^(-3) and had it set up like this

    dy/dx + y = xy^4


    Ok HELP!

    Thanks...
     
  2. jcsd
  3. Sep 13, 2009 #2
    Re: Ugh Bernoulli Equations!!!

    I've never done this before, but:

    [tex] \frac{dy}{dx} = f(x)y + g(x)y^k [/tex]

    The solution is

    [tex]y^{1-k} = y_1 + y_2 [/tex]

    Where

    [tex]y_1= Ce^{\phi(x)}[/tex]

    [tex]y_2=(1-k)e^{\phi(x)}\int e^{-\phi(x)}g(x)dx[/tex]

    [tex] \phi(x) = (1-k)\int f(x)dx [/tex]



    Here,

    [tex]k=4[/tex]

    [tex]g(x)=x[/tex]

    [tex]f(x)=-1[/tex]



    So,


    [tex] \phi(x) = 3x [/tex]

    Then,

    [tex]y_1= Ce^{3x}[/tex]

    [tex]y_2=-3e^{3x}\int xe^{-3x}dx=\frac{1}{3}+x[/tex]


    Finally,

    [tex]y = (Ce^{3x} + \frac{1}{3} + x)^{-\frac{1}{3}}[/tex]


    I think...
     
  4. Sep 13, 2009 #3

    LCKurtz

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    Re: Ugh Bernoulli Equations!!!

    OK, writing it as y' + y = xy^4 is a good start. What you want to do next is divide by y^4:

    y^(-4) y' + y^(-3) = x. Now let v = y^(-3) so v' = -3y^(-4) y'

    This gives you expressions for y^(-4)y' and y^(-3) in terms of v and it gives you a first order linear equation in v which you can solve. Then back substitute to get y.
     
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