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(Differential Equations) General and singular solutions

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the general solution and any singular solutions to [itex](2xy^3+4x)y'=x^2y^2+y^2[/itex].


    2. Relevant equations



    3. The attempt at a solution
    [itex]2x(y^3+2)y'=y^2(x^2+1)[/itex]
    [itex]\int\frac{y^3+2}{y^2}\,dy=\int\frac{x^2+1}{2x}\,dx[/itex]
    [itex]\frac{y^3-4}{2y}=\frac{x^2+2\ln x}{4}+C[/itex]

    Is this correct?
    To find the singular solution, do I set [itex]y'=0[/itex] and see if its solutions fit into the general solution? i.e.

    [itex]y'=\frac{x^2y^2+y^2}{2x^3+4x}=0 \Rightarrow y=0[/itex]​

    Which in this example does not fit into the general solution since the left side becomes [itex]\frac{(0)^3-4}{2(0)}[/itex]. Therefore [itex]y=0[/itex] is a singular solution?
     
    Last edited: Feb 26, 2013
  2. jcsd
  3. Feb 27, 2013 #2

    LCKurtz

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    That looks OK, at least for ##x>0##.

    In your solution, you divided by ##y^2##. That rules out finding a potential solution of ##y=0##. So you plug that into the original equation and see if it works. You don't set ##y'=0## to do it.
     
  4. Feb 27, 2013 #3
    Ah, I get it now. Based my singular solution attempt off the one given example we had, and admittedly what I tried didn't make intuitive sense to me. Thanks. :)
     
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