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Bernoulli's Equation

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]
    2xy'+y^3e^(-2x)=2xy
    [/itex]



    2. Relevant equations
    [itex]
    dy/dx + P(x)y=Q(x)y^n
    [/itex]
    [itex]
    v=y^(1-n)
    [/itex]



    3. The attempt at a solution
    [itex]
    dy/dx-y=-y^3e^(-2x)/2x
    [/itex]

    [itex]
    P(x)=-1
    Q(x)=-e^(-2x)/2x
    [/itex]

    [itex]
    n=3
    [/itex]
    [itex]
    v=1/y^2
    [/itex]

    [itex]
    dy/dx=dy/dv*dv/dx
    [/itex]

    [itex]
    dy/dx=-1/2v^-(3/2)*dv/dx
    [/itex]

    [itex]
    -1/2v^(-3/2)*dv/dx-v^(-1/2)=-v^(-3/2)e^(-2x)/2x
    [/itex]

    [itex]
    dv/dx + 2v=e^(-2x)/x
    [/itex]

    [itex]
    e^(\int P(x)\,dx)=e^(\int -1\,dx)=e^(-x)
    [/itex]

    [itex]
    dv/dx*e^(-x)+2ve^(-x)=e^(-3x)/x
    [/itex]


    This is supposed to look live reverse chain rule so I get [itex] (e^(-x)*v)'=e^(-3x)/x [/itex]
    but it doesn't look right and I have no idea how to do the integral required to then solve the rest of this problem.
     
  2. jcsd
  3. Sep 15, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That should be ##\frac{dy}{dx}-y = -y^3e^{-2x}##
     
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