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Better substitution?

  1. Jun 22, 2008 #1
    Given y' = y / (x + y^2), the substitution u = y^2 will give a homogeneous DE which can then be easily solved. Is there a substitution which would make things easier?
     
  2. jcsd
  3. Jun 22, 2008 #2

    rock.freak667

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    Try V=y/x

    But it is kinda long in my opinion.


    EDIT: The easiest way is your substitution of [itex]u=y^{-2}[/itex], anything else, is just harder.
     
    Last edited: Jun 22, 2008
  4. Jun 23, 2008 #3
    I think the substitution u = y^2 + x is better. I haven't tried it though.
     
  5. Jun 28, 2008 #4
    There is a solution that does not involve a substitution... if that's any help...

    First, multiply through by [tex]x + y^2[/tex], to get

    [tex]x y^{\prime} + y^2 y^{\prime} = y [/tex]

    rearrange to get

    [tex]x y^{\prime} - y = -y^2 y^{\prime} [/tex]

    but

    [tex] x y^{\prime} - y = y^2 ( \phi - \frac{x}{y})^{\prime}[/tex]

    (where [tex]\phi[/tex] is a constant.) So,

    [tex]( \phi - \frac{x}{y})^{\prime} = -y^{\prime} [/tex]

    which you can integrate to get

    [tex]\phi - \frac{x}{y} = - y [/tex]

    which you can turn into a quadratic by multiplying through by [tex]y[/tex], leaving you with.

    [tex]y(x) = \frac{-\phi \pm \sqrt{\phi^2 + 4x}}{2} [/tex]
     
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