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Linear first-order diffeq system for radioactive decay chain

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the followin[Sg decay chain- X→Y→Z
    Solve for Nx(t), Ny(t), Nz(t) for the case of Rx(t)=[itex]\alpha[/itex]t and assuming Ny(t)=Nz(t)=0

    2. Relevant equations
    dNx(t)/dt = -[itex]\lambda[/itex]xNx(t) + Rx(t)
    dNy(t)/dt = -[itex]\lambda[/itex]yNy(t) +[itex]\lambda[/itex]xNx(t)
    dNz(t)/dt = -[itex]\lambda[/itex]zNz(t) +[itex]\lambda[/itex]yNy(t)

    3. The attempt at a solution
    I know these would be solved with bateman equations and without the Rx(t)=[itex]\alpha[/itex]t term I could do these. The production term throws me off and I'm not sure exactly how to go about this.
    I have this for Nx(t) = Nx(0)e-[itex]\lambda[/itex]xt + ∫t0 dt'Rx(t')e[itex]\lambda[/itex]x(t'-t) (the integral is from 0 to t, but the itex wasn't working for me to do that)
    So how does Rx(t)=[itex]\alpha[/itex]t integrate and where does it go in the other two equations? Thanks!
     
    Last edited: Jan 26, 2012
  2. jcsd
  3. Jan 26, 2012 #2

    HallsofIvy

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    Excuse me but are you saying that you are trying to solve a system of differential equations but do not know how to integrate [itex]\alpha t[/itex]? The integral of [itex]\alpha t[/itex] with respect to t is [itex]\alpha t^2/2[/itex]. That is usually one of the first integrals you learn.
     
    Last edited: Jan 26, 2012
  4. Jan 26, 2012 #3
    Haha- no, that's cake LOL

    It's really more this term ∫dt'Rx(t')eλx(t'-t) (from 0 to t) that confuses me- I'm not sure where the primes came from and what it is indicating. It was the first step given for a solution. It seemed odd since the αt should be, like you said, extremely simple.
     
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