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Integrating factor

  1. Nov 12, 2014 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    ##C_{B}## is a function of ##\tau'##, and ##k_{1}##,##k_{2}##, and ##C_{A0}## are constants. I want to solve this differential equation
    [tex] \frac {dC_{B}}{d \tau'} + k_{2}C_{B} = k_{1}C_{A0}e^{-k_{1} \tau'} [/tex]

    2. Relevant equations


    3. The attempt at a solution
    Using the integrating factor, we get
    [tex] \frac {d(C_{B}e^{k_{2} \tau'})}{d \tau'} = k_{1}C_{A0}e^{(k_{2} - k_{1})
    \tau'} [/tex]

    However, from here I am unsure how to separate this. I was thinking of using chain rule on the inside, but that seems to only get me back to where I started (that's the purpose of the integrating factor??)
     
  2. jcsd
  3. Nov 12, 2014 #2

    pasmith

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    The next step is to integrate both sides with respect to [itex]\tau'[/itex]. By the fundamental theorem of calculus, the integral of the left hand side is
    [tex]\int \frac{d}{d\tau'} (C_Be^{k_2 \tau'})\,d\tau' = C_Be^{k_2 \tau'} + \mbox{constant}.[/tex]
     
  4. Nov 12, 2014 #3

    HallsofIvy

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    I don't know what you mean by "separate" or what it is you want to "separate". There is no need to "separate" anything- just go ahead and integrate:
    [tex]\frac{d C_Be^{k_2\tau'}}{d\tau'}= k_1C_{A0}e^{(k_2- k_1)\tau'}[/tex]
    [tex]\int d C_Be^{k_2\tau'}= \int k_1C_{A0}e^{(k_2- k_1)\tau'} d\tau'[/tex]

    [tex]C_Be^{k_2\tau'}= \frac{k_1C_{A0}}{k_2- k_1}e^{(k_2- k_1)}\tau'+ C[/tex]

    That's the whole point of an "integrating factor".
     
  5. Nov 12, 2014 #4

    Maylis

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    [tex] C_{B} e^{k_{2} \tau'} = \frac {k_{1}C_{A0}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) \tau'} + C [/tex]
    My boundary condition is ##C_{B} = 0## at ##\tau' = 0##, this means ##C = -\frac {k_{1}C_{A0}}{k_{2} - k_{1}}##

    Then the solution is
    [tex] C_{B} = \frac {k_{1}C_{A0}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) \tau'} - \frac {k_{1}C_{A0}}{k_{2} - k_{1}}[/tex]

    But the solution given is
    [tex] C_{B} = k_{1}C_{A0} \Big( \frac {e^{-k_{1} \tau'} - e^{-k_{2}
    \tau'}}{k_{2} - k_{1}} \Big)[/tex]
     
  6. Nov 12, 2014 #5

    pasmith

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    No, so far you have [tex]
    C_B e^{k_2 \tau'} = \frac {k_{1}C_{A0}}{k_{2} - k_{1}} e^{(k_{2} - k_{1}) \tau'} - \frac {k_{1}C_{A0}}{k_{2} - k_{1}}.
    [/tex]
    Dividing both sides by [itex]e^{k_2 \tau'}[/itex] yields the given solution.
     
  7. Nov 12, 2014 #6

    Maylis

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    Gold Member

    Woops, I see where I made my mistake. Thank you!
     
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