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Bernoulli Equation - Help?

  1. Feb 5, 2007 #1
    1. The problem statement, all variables and given/known data
    x^2y' + 2xy = 5y^3
    2. Relevant equations
    Bernoulli Equation
    3. The attempt at a solution
    so... v=y^-2, y=v^(-1/2) y' = -1/2v'v^(-3/2)

    moving everything over...
    y'/y^3 + 2/(xy^2) = 5/x^2

    and plugging everything in...
    -1/2v' + 2v/x = 5/x^2
    v' - 4v/x = -10/x^2

    v' - 4v/x = 0
    dv/dx = 4v/x
    dv/v = 4dx/x
    ln v = 4 ln x + C
    v = C(x)x^4
    v' = C'(x)x^4 + C(x)4x^3

    C'(x)x^4 + C(x)4x^3 - 4C(x)x^4/x = 5/x^2
    C'(x)x^4 = 5/x^2
    C'(x) = 5/x^6
    C(x) = -x^-5 + C

    v = (-x^-5 + C)x^4

    y = 1/(v)^(1/2)

    y^2 = 1/((-x^-5 + C)x^4)
    ... which is wrong. Could someone point me in the right direction?
     
  2. jcsd
  3. Feb 5, 2007 #2

    Dick

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    I think you are doing great up to the last line here. Why the x dependence in C?
    I would just say v=C*x^4. This the general solution to the homogeneous equation v'-4v/x=0. Now you just want to add this to a particular solution (any one) of the inhomogeneous equation v' - 4v/x = -10/x^2. Can you find one?
     
  4. Feb 9, 2007 #3
    I think 2/x is a particular solution?

    cause... you got, let's see....

    -2/x^2 - 4*2/x^2 = -10/x^2

    Okay so now what...??

    Cx^4 + 2/x?

    v=Cx^4 + 2/x

    y = v^(-1/2)

    y^2 = 1/(Cx^4+2/x)

    or...

    y^2 = x/(Cx^5 + 2) which is... CORRECT!!! thanks a bunch!!!

    So all you have to do is find the general solution, then a particular solution? When is this applicable (only in cases where there is one derivative??)?
     
    Last edited: Feb 9, 2007
  5. Feb 9, 2007 #4

    HallsofIvy

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    Finding a "general" solution (to the corresponding homogeneous equation), then adding a specific solution (to the entire equation), works only for linear equations. The Bernoulli equation is not linear (because of the y3) but the equation you get for v,v' - 4v/x = -10/x2 is linear.
     
  6. Feb 9, 2007 #5

    Dick

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    Nope, any number of derivatives. You only need that the ODE is linear. So you CAN add solutions and still get a solution.
     
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