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How to solve this differential equation?

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Assume NA = NAo exp(-t/a)

    Solve the differential equation:

    dNB/dt + NB/b = NA/a

    2. Relevant equations

    differential equations

    3. The attempt at a solution

    trial function: NB = C exp(-t/a) + D exp (-t/b)

    with initial condition: C + D = NBo

    I tried plugging in this and NA into the original equation, but was not able to solve for C or D...
     
  2. jcsd
  3. Sep 27, 2011 #2

    vela

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    Show us what you got because what you described should work.

    Note that the D term will vanish when you plug it in because it's the solution to the homogeneous differential equation. This allows you to solve for C.
     
  4. Sep 27, 2011 #3
    Unfortunately I don't really have much experience with this kind of differential equation. After I substituted the trial function and C = NB - D into the equation, I got:

    [ -(NB-D)/a*exp(-t/a) - D/b*exp(-t/b) ] + [ NB-D/b*exp(-t/a) - D/b*exp(-t/b) ] = NA/a*exp(-t/a)

    how would you solve 'D' from this?
     
  5. Sep 27, 2011 #4

    vela

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    You want to plug the trial function into the differential equation:
    [tex]\frac{d}{dt}(C e^{-t/a} + D e^{-t/b}) +\frac{1}{b}(C e^{-t/a} + D e^{-t/b}) = \frac{N_{A_0}}{a} e^{-t/a}[/tex]
    When you differentiate the first term, you'll see the e-t/b terms cancel out, which leaves only the e-t/a terms, allowing you to solve for C. Once you know C, you can solve for D.
     
  6. Sep 27, 2011 #5
    Ok, thanks for the insightful help, vela. This is what I got for NB, with D=NBo-C :

    NB = [NAo / (a/b-1)] *exp(-t/a) + [NBo - NAo / (a/b - 1)] *exp(-t/b)

    assuming this is correct.
     
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