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Insane differential equation help?

  1. Jan 28, 2009 #1
    The equation is as follows:

    2x * (dy/dx) + y = y2 * sqrt(x - (x2 * y2))

    Any ideas how to solve it or begin solving it?
  2. jcsd
  3. Jan 28, 2009 #2
    Hmm maple gives a fairly straightforward answer maybe a subst. u = xy, just a guess
  4. Apr 9, 2009 #3
    Hello itsthewoo,

    I finally found the solution. It kept running in my head and I couldn't put it aside, so here it is. First of all I used an advanced technique called Lie point transformations to find the answer. It comes down for this equation to find an integrating factor so the equation becomes exact. Writing the equation as:

    [tex]\left(2x\right)\cdot dy-\left(-y+y^2\sqrt{x-x^2y^2}\right)\cdot dx=0[/tex]

    The integrating factor is the following:


    The new (exact) equation is now:

    [tex]\left(\frac{2x}{xy^2\sqrt{x-x^2y^2}}\right)\cdot dy+
    \left(\frac{1}{xy\sqrt{x-x^2y^2}}-\frac{1}{x}\right)\cdot dx=0[/tex]

    Which has as solution:


    This is the same solution as maxima (like maple or mathematica) gives. If any more info is necessary, I would recommend the following book:

    "Ordinary differential equations, an elementary text-book with an introduction
    to Lie's theory of the group of one parameter." written by James Morris Page

    Hope this helps,

  5. Apr 18, 2009 #4
    [tex]u = x y^2[/tex] will separate your equation.
  6. Apr 18, 2009 #5
    Hello Mathwebster,

    I considered the following one-parameter group:

    [tex]Uf=ax\frac{\partial f}{\partial x}+by\frac{\partial f}{\partial y}+(b-a)y'\frac{\partial f}{\partial y'}[/tex]

    And found the unknown numbers to bea=-1 and b=1/2. From this it is possible to find the integrating factor, but also the transformation you gave. It works indeed. I did not think of doing it like that although I should have :blushing:. Many thanks for the hint.

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