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Math Help - differential equation

  1. #1
    Member akhayoon's Avatar
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    differential equation

    y\prime+\frac{2}{t}y=\frac{arctant}{t^{2}}

    here they want the y(t) equation

    so first what I tried doing was integrating t/2

    so I got the factor was e^{lnt^{2}} which equals to 2

    so now I have to integrate

    \frac{d}{dt}2y=2\frac{arctant}{t^2}

    which in the end would turn out to be

    y(t)=\frac{tarctant-0.5ln(t^{2}+1)}{t^{2}}+C/2

    so I can't help but feel I did something wrong???
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akhayoon View Post
    y\prime+\frac{2}{t}y=\frac{arctant}{t^{2}}

    here they want the y(t) equation

    so first what I tried doing was integrating t/2

    so I got the factor was e^{lnt^{2}} which equals to 2

    so now I have to integrate

    \frac{d}{dt}2y=2\frac{arctant}{t^2}+C/2

    which in the end would turn out to be

    y(t)=\frac{tarctant-0.5ln(t^{2}+1)}{t^{2}}

    so I can't help but feel I did something wrong???
    are you sure it isnt e^{\int\frac{2}{t}dt}?
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  3. #3
    Member akhayoon's Avatar
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    yeah but wouldn't that be e^{2lnt}

    which means that it would equal e^{lnt^{2}} which equals 2???
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akhayoon View Post
    yeah but wouldn't that be e^{2lnt}

    which means that it would equal e^{lnt^{2}} which equals 2???
    e^{2\ln(t)}=e^{\ln(t^2)}=t^2...remember that since e^{u(x)} and \ln(u(x)) are inverse operations e^{\ln(u(x))}=u(x)
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