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Math Help - Decay of a Decay. Is the diff eqn correct

  1. #1
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    Decay of a Decay. Is the diff eqn correct

    You have a decaying isotope x=x_0e^{-kt} which decays in isotope y. Now y decays in z which is stable. Is the following correct

    \dfrac{dy}{dt}=kx-cy

    Also what is the equation of z?

    Edit
    I think z=z_0+y_0-y+x_0-x which means \dfrac{dz}{dt}=cy what you would expect
    Last edited by fobos3; July 26th 2010 at 11:39 AM.
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  2. #2
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    Here is what I have

    x=x_0e^{-kt}

    \dfrac{dy}{dt}=kx-cy

    \dfrac{dy}{dt}+cy=kx_0e^{-kt}

    \dfrac{d}{dt}(e^{ct}y)=kx_0e^{(c-k)t}

    e^{ct}y=\dfrac{kx_0}{c-k}e^{(c-k)t}+A

    y=\dfrac{kx_0}{c-k}e^{-kt}+Ae^{-ct}

    Assuming y_0=0=\dfrac{kx_0}{c-k}+A\therefore A=-\dfrac{kx_0}{c-k}

    y=\dfrac{kx_0}{c-k}(e^{-kt}-e^{-ct})
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  3. #3
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    Quote Originally Posted by fobos3 View Post
    You have a decaying isotope x=x_0e^{-kt} which decays in isotope y. Now y decays in z which is stable. Is the following correct

    \dfrac{dy}{dt}=kx-cy

    Also what is the equation of z?

    Edit
    I think z=z_0+y_0-y+x_0-x which means \dfrac{dz}{dt}=cy what you would expect
    That looks OK

    CB
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