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Differential Equations- homogeneous (I think)

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of the equation
    x*y*(dy/dx)=(x^2) + 3(y^2)


    2. Relevant equations



    3. The attempt at a solution
    So I start by realizing this is (likely) a homogeneous differential equation, and then rewrite it in the form required:
    dy/dx = (x/y) + 3(y/x)
    then, using the method of homgeneous equations substitute the variable v, where
    v = (y/x), y = vx, dy/dx= v + x(dv/dx)
    so I have:
    v + x(dv/dx)= (1/v) + 3v
    x(dv/dx) = (1/v) + 2v
    dx/x = (v + (1/(2v)))dv
    then after integrating both sides I get
    ln(x) = ((v^2)/2) + (1/2)*ln(v) + C
    then substituting back to the original variables I get
    ln(x) = ((y^2)/(x^2))*(1/2) + (1/2)*ln(y/x) + C
    Now my problem arrives when expressing in the form y(x) = (+ or -)sqrt(c_1 x^6-x^2/2)
    which is the solution wolfram alpha got, however wolfram alpha doesn't show how it arrived at this, so I was hoping someone here would, or spot a mistake I made during my solving, or whatever, thank you!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 7, 2010 #2

    tiny-tim

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    Hi Stuart! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    ooooh … :cry:

    (so good up till then! :redface:)
     
  4. Feb 7, 2010 #3
    thanks for your comment, but I'm not sure I understand
    I can see that I skipped a few steps during that process, but I will show how I arrived there
    x(dv/dx)= (1/v) + 2v
    dv/dx = ((1/v) + 2v)/x
    dv=((1/v)+2v)*dx*1/x
    dv/((1/v)+2v) = dx/x
    dv*(v+(1/2v))= dx/x
    Forgive my ignorance, but I fail to see where I made a mistake, if you can enlighten me I'd appreciate it
     
  5. Feb 7, 2010 #4

    tiny-tim

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    Hi Stuart! :smile:

    The steps you missed :redface:

    dv/((1/v)+2v) = vdv/(1+2v2)

    [≠ dv*(v+(1/2v)) = vdv*(1+(1/2v2))]

    1/(1 + a) ≠ 1 + 1/a :wink:
     
  6. Feb 7, 2010 #5
    How embarassing...I now see the error of my ways, thank you for revealing this to me.
    (someone should probably close this thread)
     
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