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Math Help - Integrating Factor error

  1. #1
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    Australia
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    Integrating Factor error

    Hello,
    Could someone point out where I'm going wrong here - I'm making a mistake somewhere, and it's not clear to me what I'm doing wrong...

    \frac{dy}{dt}(t) - 3y +2 = 0

    = \frac {dy}{dt} - \frac {3y}{t} = \frac {-2}{t}

    I(t)=e^{\int \frac {-3}{t}}

    =I(t) = e^{-3lnt} = t^{-3}

    = \frac {dy}{dt}\cdot t^{-3} \frac {-3y}{t}\cdot t^{-3} = \frac {-2}{t} \cdot t^{-3}

    = \int \frac {d}{dt}t^{-3}y = \int \frac {-2}{t^4}

    = yt^{-3} = -2 \int t^{-4}

    = yt^{-3} = -2 \frac {t^{-3}}{-3}+C

    = yt^{-3} = \frac {2}{3} t^{-3}+C = \frac {2}{3t^3}+C

    = y = \frac {2}{3} + C

    Now, I end up cancelling the t-value out, which can't be right. What am I doing wrong exactly?
    Last edited by astuart; November 9th 2012 at 04:20 PM.
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  2. #2
    Member
    Joined
    Jun 2012
    From
    Australia
    Posts
    86

    Re: Integrating Factor error

    Quote Originally Posted by astuart View Post
    Hello,
    Could someone point out where I'm going wrong here - I'm making a mistake somewhere, and it's not clear to me what I'm doing wrong...

    \frac{dy}{dt}(t) - 3y +2 = 0

    = \frac {dy}{dt} - \frac {3y}{t} = \frac {-2}{t}

    I(t)=e^{\int \frac {-3}{t}}

    =I(t) = e^{-3lnt} = t^{-3}

    = \frac {dy}{dt}\cdot t^{-3} \frac {-3y}{t}\cdot t^{-3} = \frac {-2}{t} \cdot t^{-3}

    = \int \frac {d}{dt}t^{-3}y = \int \frac {-2}{t^4}

    = yt^{-3} = -2 \int t^{-4}

    = yt^{-3} = -2 \frac {t^{-3}}{-3}+C

    = yt^{-3} = \frac {2}{3} t^{-3}+C = \frac {2}{3t^3}+C

    = y = \frac {2}{3} + C

    Now, I end up cancelling the t-value out, which can't be right. What am I doing wrong exactly?
    Ahh, god, I think I can already see where I went wrong....

     y = \frac {2}{3} + Ct^3
    Last edited by astuart; November 9th 2012 at 04:25 PM.
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