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Math Help - Differential constants

  1. #1
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    Differential constants

    Could somebody please check I have got the value of the constant correct for this D.E

     \frac{dy}{dt}= - \frac{8t}{1+t^2}

    I get

    y = \frac{1}{(1+t^2)^4}

    where C = c2 - c1 = 0
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  2. #2
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    Quote Originally Posted by davefulton View Post
    Could somebody please check I have got the value of the constant correct for this D.E

     \frac{dy}{dt}= - \frac{8t}{1+t^2}

    I get

    y = \frac{1}{(1+t^2)^4}

    where C = c2 - c1 = 0
    Does your answer work when you substitute it into the DE?

    (You will need to show all your work if you want your error(s) corrected. And you will need to include whatever boundary condition came with the question).
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  3. #3
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    Thank you, sorry I missed off the y term

    So if

    \frac{dy}{dt}=-\frac{8ty}{(1+t^2}

    then

    \frac{1}{y}dy=-\frac{8t}{(1+t^2}dt

    integrating

    \ln(y)+c1=-4\ln(1+t^2) + c2

    therefore

    for y(0)=1

    y=\frac{1}{(1+t^2)^4}

    sorry about the sloppiness
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