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Thread: 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:

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

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

    to

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

    Given the differential equation:

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

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

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

    $\displaystyle y^2+y = t + k$

    to

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

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