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Differential equation

  1. Aug 18, 2013 #1
    Sorry I couldn't think of any more relevant title. Here's the equation:

    [itex]{x^2} - 3{y^2} + 6xy\frac{{dy}}{{dx}} = 0[/itex]

    I'm thinking of rewriting the above to [itex]\frac{{dy}}{{dx}} = \frac{{3{y^2} - {x^2}}}{{6xy}}[/itex] followed by a change of variable u=y/x. But should I rule out the case when either x=0 or y=0 first? I'd also love to see if there's any alternative way to solve this ODE, thanks!
     
  2. jcsd
  3. Aug 18, 2013 #2

    D H

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    That term containing y*dy/dx suggests you should consider the substitution u=y2.
     
  4. Aug 18, 2013 #3
    Yes! And also let v = x2. Then see how these substitutions simplify things.

    Chet
     
  5. Aug 19, 2013 #4

    HallsofIvy

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    Notice that
    [tex]\frac{3y^2- x^2}{6xy}= \frac{\frac{3y^2- x^2}{xy}}{6}= \frac{1}{6}(3\frac{y}{x}- \frac{x}{y})[/tex]
    Which suggests that the substitution v= y/x would be useful (actually, the fact that both numerator and denominator are of second degree first suggested that).

    If v= y/x then y= xv so that y'= xv'+ v. The differential equation becomes
    [tex]x\frac{dv}{dx}+ v= \frac{1}{6}\left(3v- \frac{1}{v}\right)[/tex]
    [tex]x\frac{dv}{dx}= \frac{1}{6}\left(3v- \frac{1}{v}\right)- v= \frac{1}{6}\left(-3v- \frac{1}{v}\right)[/tex]
    [tex]x\frac{dv}{dx}= -\frac{1}{6}\left(\frac{3v^2+ 1}{v}\right)[/tex]
    [tex]\frac{v}{3v^2+ 1}dv= -\frac{1}{6x}dx[/tex]
     
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