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Thread: Isotop Modeling

  1. #1
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    Isotop Modeling



    I have:

    $\displaystyle A=kte^{-at}$

    since at any time there will be kt of A but its also decaying

    $\displaystyle B = kt -kte^{-at} $

    since at any moment in time kt will be the amount of substance so kt - (amount of A) = amount of B?

    I don't know how to set up the differential equations at the start and I don't think your supposed to start with the solution!
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  2. #2
    MHF Contributor
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    Re: differential equations,

    You are producing A at rate k and it is decaying at a rate $\displaystyle \alpha{A}$, so $\displaystyle \frac{dA}{dt} = k - \alpha{A}$. Your equation $\displaystyle A=kte^{-\alpha{t}}$ is incorrect since some of the A produced has decayed for a long time and some for a short time.

    To solve this equation, substitute $\displaystyle A=Y+\frac{k}{\alpha}$. Everything else is pretty straightforward. Let us know if you have any trouble.

    - Hollywood
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  3. #3
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    Re: differential equations,

    Thanks,

    This is my solution:

    $\displaystyle \frac{dA}{dt} = k-A\alpha$

    $\displaystyle \frac{dB}{dt} = A\alpha$

    $\displaystyle A(t) = \frac{k}{\alpha} $

    I got the general solution using the integrating factor method

    for part v)

    $\displaystyle A(t) = \frac{k}{\alpha} - \frac{k}{\alpha}{e}^{-\alpha t}$

    $\displaystyle A(0) > C$ C is some constant greater than k/a

    $\displaystyle A(t) = \frac{k}{\alpha} + (C - \frac{k}{\alpha}{e}^{- \alpha t} $

    for vi I subbed in the given B(t) into the differential equation I wrote for dB/dt

    $\displaystyle {c}_{1} = k$ and $\displaystyle {c}_{2} = -{A}_{0}$ (not sure about this)

    then for B(t) I get

    $\displaystyle B(t) = kt - {A}_{0}{e}^{- \alpha t} + {A}_{0}$
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