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A hard differential equation - where is the error in my logic?

  1. Jun 1, 2012 #1
    [SOLVED] A hard differential equation - where is the error in my logic?

    1. The problem statement, all variables and given/known data
    2xyy' + y2, where y is a function of x

    3. The attempt at a solution

    2xyy' + y2 = 12x2 |*(1/x)
    => (y2)' + y2/x =12x |*ln(x), 2yy' = (y2)'
    => (y2ln(x))' = 12x*ln(x)
    => y2ln(x) = ∫12x*ln(x)
    => y2ln(x) = 6x2ln(x) -3x2 + C
    => y = ±sqrt[3x2(2-1/ln(x)) + c/ln(x)]

    Well, my result is wrong. Can somebody tell me the error in my logic? HELP, my math exam is on the 4th of June...
     
    Last edited: Jun 1, 2012
  2. jcsd
  3. Jun 1, 2012 #2

    sharks

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    Gold Member

    What is that on the R.H.S. with the vertical bar, etc? Your problem isn't clear.


    Assuming that your original ODE is: ##2xyy'+y^2=12x^2##
    Re-arranging gives: [itex]y'+\frac{1}{2x}y=6xy^{-1}[/itex]. You get a Bernoulli equation.
     
    Last edited: Jun 1, 2012
  4. Jun 1, 2012 #3

    => (y2)' + y2/x =12x |*ln(x), 2yy' = (y2)'

    => (y2ln(x))' = 12x*ln(x)
    (y^2)'*ln(X)+(y^2/x)*ln(x)=/=(y^2*ln(x))'
    I believe
     
  5. Jun 1, 2012 #4
    Oh crap, I'm such an idiot! thanks, TT. In addition, I found out how to solve this, so forget about this thread guys!

    sharks: I have no idea what a bernoulli equation is, sorry. I haven't even started uni yet.
     
  6. Jun 1, 2012 #5

    Ray Vickson

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    Science Advisor
    Homework Helper

    Re: [SOLVED] A hard differential equation - where is the error in my logic?

    [tex] \frac{d}{dx}(x y^2) = 2 x y y' + y^2.[/tex]

    RGV
     
  7. Jun 2, 2012 #6

    Then just find out!
     
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