1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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 by a moderator: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook