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Math Help - Change of variable

  1. #1
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    Change of variable

    Help Please

    Thank you

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  2. #2
    Forum Admin topsquark's Avatar
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    There's probably a more elegant way, but this does work:

    x^2\frac{dy}{dx}=y^2+4xy

    This is essentially a Bernoulli equation, so if we go ahead and let v = y^{1-2}=\frac{1}{y} we obtain a linear equation:

    y=\frac{1}{v}
    \frac{dy}{dx}=-\frac{1}{v^2}\frac{dv}{dx}

    So
    -x^2\frac{1}{v^2}\frac{dv}{dx}=\frac{1}{v^2}+4x \frac{1}{v}

    x^2\frac{dv}{dx}+4xv=-1

    (4xv+1)dx+x^2dv=0

    Now, \frac{1}{x^2}(4x-2x) depends only on x, so the integrating factor will be:
    exp \left ( \int dx \frac{1}{x^2}(4x-2x) \right )=e^{2ln|x|}=x^2

    So (4x^3v+x^2)dx+x^4dv=0 is an exact differential equation.

    Thus x^4v+\frac{1}{3}x^3=C

    or
    v=\frac{C-\frac{1}{3}x^3}{x^4}

    Thus y=\frac{3x^4}{3C-x^3}

    -Dan
    Last edited by topsquark; April 22nd 2006 at 06:04 AM. Reason: Oops! I guess it IS a Bernoulli equation. I need coffee.
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  3. #3
    Super Member Rebesques's Avatar
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    Note that the right hand side of

    <br />
\frac{dy}{dx}=\frac{y^2+4xy}{x^2}<br />

    is homogeneous in x and y, and change variables to z=y/x.
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