1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]


    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook