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Math Help - Find the general solution of the differential equation specified..

  1. #1
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    Find the general solution of the differential equation specified..

    Hey all,

    Given the differential equation:

    \frac{dy}{dt}=\frac{1}{2y+1}, where x is not -\frac{1}{2}

    I do not understand how they went from:
    \int(2y+1)dy = \int tdt
    y^2+y = t + k

    to

    y(t) = \frac{-1\pm \sqrt{4t+4k+1}}{2}
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  2. #2
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    They completed the square on the \displaystyle y terms and solved for \displaystyle y.
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  3. #3
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    Quote Originally Posted by Oiler View Post
    Hey all,

    Given the differential equation:

    \frac{dy}{dt}=\frac{1}{2y+1}, where x is not -\frac{1}{2}

    I do not understand how they went from:
    \int(2y+1)dy = \int tdt
    This should be

    \displaystyle\int(2y+1)\,dy=\int dt.

    y^2+y = t + k

    to

    y(t) = \frac{-1\pm \sqrt{4t+4k+1}}{2}
    Follow Math Help Forum on Facebook and Google+

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