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Math Help - Initial condition DE

  1. #1
    Senior Member mollymcf2009's Avatar
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    Initial condition DE

    This is probably really easy, but I am not getting the right answer. Can someone check this please?
    Thanks!!

    Solve.


    \frac{dy}{dx} = \frac{x}{y}, y(0)=-8

    \int ydy = \int xdx

    y = \frac{x^2}{2} + C

    C = -8

    So I get:

    y = \frac{x^2}{2} - 8

    Where am I screwing this up?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    [snip]
    Solve.


    \frac{dy}{dx} = \frac{x}{y}, y(0)=-8

    \int ydy = \int xdx

    {\color{red}\frac{1}{2}}y^{{\color{red}2}} = \frac{x^2}{2} + C

    [/snip]
    See my correction in red.
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