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Differential Equations: Bernoulli Equation

  1. Nov 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the general solution:

    y'-3y=(y^2)

    2. Relevant equations


    3. The attempt at a solution

    divide both sides by y^2

    y'(y^-2) -3(y^-1) = 1

    we know v=y^(n-1)
    v=y^-1
    v'=d/dx(y^-1)
    v'=-(y^-2) y'

    plug it back into

    y'(y^-2) -3(y^-1) = 1

    -v'-3v=1

    this is where I think I am making a mistake

    im putting it into a 1st ODE by making v' positive

    v'+3v=-1

    then u(x)=e^(3x)

    =-{[e^(3x)]/3+C}/e^(3x)

    =-1/3+C/e^(3x)

    the solution given to me is

    1/((Ce^(-3x))-1/3)

    Any help is appreciated. thanks!
     
  2. jcsd
  3. Nov 2, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    which is equal to ##-\frac 1 3 + Ce^{-3x}##

    Isn't that the reciprocal of what you have? Remember ##v = \frac 1 y##.
     
  4. Nov 2, 2014 #3
    i always feel so stupid when i come on the site. yes it is the reciprocal. thank you for your help
     
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