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What is the integral of dydx=y^2/x^2

  1. Apr 5, 2010 #1

    math_trouble

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    im having problem integrating the equation

    dydx=y^2/x^2

    and also

    dydx=3*y^2/x
     
    math_trouble, Apr 5, 2010
  2. jcsd
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  3. Apr 5, 2010 #2

    g_edgar

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    I'm assuming they should be dy/dx=y^2/x^2
    and dy/dx=3*y^2/x .

    Your differential equations textbook should discuss "separation of variables" near the very beginning.
     
    g_edgar, Apr 5, 2010
  4. Apr 9, 2010 #3

    StalkerM

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    This can be rewritten in this way:
    y'=y^2/x^2 with x different from zero.
    y'/y^2=1/x^2
    using chain rule:
    d/dx[-1/y]=d/dx[-(1/x)+C]
    consequentely:
    1/y=(1/x)-C
    y=1/[(1/x)+C]
     
    StalkerM, Apr 9, 2010
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