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Math Help - Decay Chains

  1. #1
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    Decay Chains

    There is a decay chain with three substances, and I need to solve for each (F, G, and H)

    dF/dt = -BF
    dG/dt = BF - CG
    dH/dt = CG

    F(0) = 100
    G(0) = 30
    H(0) = 15

    The half-life of F is 13 hours
    The half-life of G is 22 hours

    B and C are constants

    I solved the first part as:

    B = 0.053319
    F(t) = 100e^(-0.053319t)

    However, I am not sure what to do next, should I just plug the new B and F into the dG/dt equation? Is C = to ln(2)/22 or does this change since the DE is different for G?

    Any help would be appreciated.
    Last edited by machi4velli; April 11th 2007 at 11:58 AM.
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  2. #2
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    Yeah, it seems that you solved correctly for B, F, and C. I got the same results. But I also stuck on the same step solving for G, and separation of variables doesn't help much either.... I believe that you work out next equation with G the same way you did solving for F. You must have dG/G on the one side to solve for G and I couldn't figure out how. Will try again...
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  3. #3
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    If my C is correct, I think I can work out dG/dt with an integrating factor and then I should get a simple dH/dt equal to a function of t.
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  4. #4
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    Quote Originally Posted by machi4velli View Post
    There is a decay chain with three substances, and I need to solve for each (F, G, and H)

    dF/dt = -BF
    dG/dt = BF - CG
    dH/dt = CG

    F(0) = 100
    G(0) = 30
    H(0) = 15

    The half-life of F is 13 hours
    The half-life of G is 22 hours

    B and C are constants

    I solved the first part as:

    B = 0.053319
    F(t) = 100e^(-0.053319t)

    However, I am not sure what to do next, should I just plug the new B and F into the dG/dt equation? Is C = to ln(2)/22 or does this change since the DE is different for G?

    Any help would be appreciated.
    It is long but straightforward.

    You solved for F(t) correctly.

    Now you have:
    G' = BF - CG

    G' + CG = BF = 100Be^{-Bt}

    Typically there would be little problem in solving this. The difficulty arises when we consider the half-life of G. How do we deal with this?

    If we weren't adding any G then we'd simply have the equation
    G' + CG = 0
    just like the first part of the problem. But we are constantly adding an amount of G from the decay of F, represented by the 100Be^{-Bt} term. This means that at 22 hours we DON'T have merely 15 (units? grams?) of G left, we should have more. So we should have the condition:
    30*e^{-Dt} < G(t), where D = (1/22)*ln(2)

    This is the only restriction I am going to put on the solution to the G equation.

    G' + CG = 100Be^{-Bt}

    The solution to the homogeneous equation is
    Gh(t) = (beta)e^{-Ct}

    The particular solution will be of the form:
    Gp' + CGp = 100Be^{-Bt}
    and the homogeneous solution is (beta)e^{-Ct}, so we try a form like:

    Gp(t) = (gamma)te^{-Bt}
    Gp'(t) = (gamma)e^{-Bt} - (gamma)Bte^{-Bt}

    Thus:
    (gamma)e^{-Bt} - (gamma)Bte^{-Bt} + (gamma)Cte^{-Bt} = 100Be^{-Bt}

    The only way to have this true for arbitrary t is for like terms to cancel. So:
    (gamma) = 100B
    -B + C = 0 ==> C = B.

    Thus
    G(t) = Gh(t) + Gp(t) = (beta)e^{-Bt} + 100Bte^{-Bt}
    (recall that C = B!)

    Now G(0) = 30, so
    G(0) = (beta) = 30, thus

    G(t) = 30e^{-Bt} + 100Bte^{-Bt} > 30e^{-Bt}
    as we needed.

    Now to solve the H equation:
    H' = CG

    H' = 30Be^{-Bt} + 100B^2te^{-Bt} <-- C = B

    Just integrate this from 0 to t:

    H(t) = 30B(-1/B)(e^{-Bt} - 1) + 100B^2[-*(t/B)*e^{-Bt} - (t/B^2)*e^{-Bt} + 1/B^2] + (delta)
    (Typically we don't need a constant of integration for a definite integral, but in this we are solving a differential equation, not simply integrating.)

    H(t) = -30(e^{-Bt} - 1) + 100B[-te^{-Bt} - (t/B)*e^{-Bt} + 1/B] + (delta)

    So at t = 0:
    H(0) = 15 = -30(1 - 1) + 100B[-0 - 0 + 1/B] + (delta)

    15 = 100 + (delta)

    (delta) = -85

    Thus
    F(t) = 100e^(-Bt)
    G(t) = 30e^{-Bt} + 100Bte^{-Bt} > 30e^{-Bt}
    H(t) = -30(e^{-Bt} - 1) + 100B[-te^{-Bt} - (t/B)*e^{-Bt} + 1/B] - 85
    where
    B = (1/13)*ln(2)

    -Dan
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