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    This is all to do with decay, @ and b are decay constants, etc...

    B(t) = @A_0{[e^([email protected]) - e^(-bt)]/(b - @)}
    Use L'Hopitals rule to show that the solution for b -> @ is:
    B(t) = @A_0te^([email protected])

    Help appreciated.
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    B(t) = aA_0 (e^(-at) - e^(-bt))/(b - a)

    Now:
    d/db [aA_0 (e^(-at) - e^(-bt))] = aA_0 * t e^(-bt)
    d/db (b - a) = 1

    Hence:
    lim(b->a) B(t) = lim(b->a) (aA_0 * t e^(-bt))/1 = aA_0 t e^(-at), as required.
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    One little query: We treat t as a constant here, but it's a variable (time), so why can we just drop the t down? :confused:
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    Isn't a parameter instead of a constant or something odd... I dunno (I'm only an AS mathmo!)
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    your limiting b -> infinity. All your doing is taking a limit of a 2 variable function as one goes to infinity. L'hopitals rule just shows what this limit is the diffentiating has no actual effect on the problem being considered its just a method for finding the limit.
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    It's just kind of weird thinking about it lol.

    Anyways, heres some more...

    A decays into B at rate alpha, B decays into C at rate Beta.
    I got my DE's as: dA/dt = -alpha.A, dB/dt = alpha.A - Beta.B, dC/dt = Beta.B

    Well so far I get (which are correct):

    A(t) = A_0.e^(-alpha.t)
    B(t) = [alpha.A_0/(Beta - alpha)]{e^(-alpha.t) - e^(-Beta.t)}

    Then it says find the solution for C using the solution for B, both for Beta =/= Alpha and Beta = Alpha.

    But I get a solution:
    C(t) = A_0/[Beta(Beta - Alpha]{Alpha[e^(-Beta.t) - 1] + Beta[1 - e^(-alpha.t)]}

    So, is this correct, and if so, surely I don't get a solution for C in the case alpha = beta?

    Help appreciated.
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    I'm not familiar with the physics involved here, but maybe you should also use the limiting case for B(t).
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    (Original post by dvs)
    I'm not familiar with the physics involved here, but maybe you should also use the limiting case for B(t).
    Yes I questioned that, but surely it would just say to solve for Beta -> alpha and not Beta = Alpha, which gives no solution. Meh, thanks anyways.
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    (Original post by Nima)
    C(t) = A_0/[Beta(Beta - Alpha]{Alpha[e^(-Beta.t) - 1] + Beta[1 - e^(-alpha.t)]}
    I agree with your expression for C(t) (actually Mathematica agrees with it ). After correcting the typo,

    C(t) = [A_0/(Beta - Alpha)] {Alpha[e^(-Beta.t) - 1] + Beta[1 - e^(-alpha.t)]}

    [Check: C(t) -> A0 as t -> infinity]

    To deal with the Alpha = Beta case, replace Beta by Alpha in the differential equations and solve to find B(t) and then C(t).

    Using L'Hopital's rule should give the same answers.
 
 
 
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Updated: April 12, 2006
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