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Bernoulli Differential Equation

  1. Aug 26, 2010 #1
    [tex]\frac{dy}{dx} - \frac{1}{x}*y + \frac{1}{y^2}*x = 0[/tex]

    This is a bernoulli differential equation which follows the following procedure
    http://en.wikipedia.org/wiki/Bernoulli_differential_equation
    Can someone provide the full steps because somewhere I am mistaken?
     
  2. jcsd
  3. Aug 26, 2010 #2
    See attachment :
     

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  4. Aug 26, 2010 #3
    Thank you a lot. One more question:
    In ODE I reach in a place where i have that
    [tex]\frac{dz}{dx}*x^3- 3x^2*z + 3x^4 = 0[/tex]

    I cannot find the derivatve on that
     
    Last edited: Aug 26, 2010
  5. Aug 26, 2010 #4
    I cannot understand what you mean : The derivative of what function ?
    Well, the equation is
    x(dz/dx)-3xz = -3x²
    So , solving x(dZ/dx)-3x = 0 leads to Z = C*(x^3)
    Remplace C by f(x)
    then z = f(x)*(x^3)
    dz/dx = (x^3)*(df/dx) +3*x² f
    x( (x^3)*(df/dx) +3*x² f ) -3x (x^3)*f = -3x²
    (x^4)*(df/dx) = -3x²
    df/dx = -3/x²
    f = (3/x) + C
    z = 3 x² + C (x^3)
     
  6. Aug 26, 2010 #5
    x(dz/dx)-3xz = -3x²
    how did you find that?
     
  7. Aug 27, 2010 #6
    Sorry, there was a typo. The correct equation is
    x(dz/dx) - 3z = -3x²
    and there was other typo in my message #4.
    As a consequence, my message #4 is wrong.
    Nevertheless, it shows the method to solve this kind of equations.
     
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