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Differential equation

  1. Sep 14, 2009 #1
    Give the general solution of:

    dy/dx = 6x y^2 -6x + 3 y^2 -3.

    No clue how to start. Guessing you would use Bernoulli's equation but i don't see how it is possible to put in the y'(x) + p(x)y(x) = q(x) y(x)^n form.
     
  2. jcsd
  3. Sep 14, 2009 #2
    Bernoulli Equation

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

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

    where

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

    [tex] y_1 = Ce^\phi [/tex]

    [tex] y_2 = (1-k)e^\phi \int e^{-\phi} g(x) dx [/tex]
     
  4. Sep 14, 2009 #3
    Actually don't bother, you can rearrange to get this:

    [tex]\frac{dy}{dx} = (6x+3)(y^2-1)[/tex]

    you can take it from there, no?
     
  5. Sep 14, 2009 #4
    Yes i can thank you
     
  6. Sep 14, 2009 #5
    actually im still having trouble do i still use the bernoulli method?

    so it is a simple separation problem?
     
  7. Sep 14, 2009 #6
    Sorry I didn't realise you replied.

    [tex] \frac{dy}{dx} = (6x+3)(y^2-1) [/tex]


    [tex] \int \frac{dy}{y^2-1} = \int (6x+3) dx [/tex]


    [tex] \frac{1}{2}log\left(\frac{1-y}{1+y}\right) = 3x^2+3x + C [/tex]


    [tex] \sqrt{\frac{1-y}{1+y}} = Ae^{3x^2+3x} [/tex]


    [tex] y=\frac{1-Ae^{6x^2+6x}}{1+Ae^{6x^2+6x}}[/tex]


    Something like this I think?
     
    Last edited: Sep 14, 2009
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