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Rate-in rate-out Differential Equation

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Skjermbilde_2012_04_22_kl_10_55_54_AM.png

    3. The attempt at a solution
    I let Q be the concentration of salt in the tank at any time t. Thus, the differential equation should be [itex]\frac{dQ}{dt} = (\frac{1}{10})(2) - \frac{Q(1)}{100 + t} = \frac{1}{5} - \frac{Q}{100 + t}[/itex]. Rearranging terms,

    [itex]\frac{dQ}{dt} + \frac{Q}{100 + t} = \frac{1}{5}[/itex]. We construct an integrating factor: [itex]θ(t) = e^{\int \frac{1}{100 + t} dt} = e^{ln(100 + t)} = 100 + t[/itex]. We introduce the integrating factor,

    [itex](Q(100 + t))' = 20 + \frac{t}{5}[/itex] and integrate both sides to find [itex]Q(100 + t) = 20t + \frac{t^2}{10} + C[/itex]. Thus,

    [itex]Q = \frac{20t + \frac{t^2}{10} + C}{100 + t}[/itex]. Given that [itex]Q(0) = 0[/itex], we can solve for C: [itex]Q(0) = 0 = \frac{20(0) + \frac{(0)^2}{10} + C}{100 + (0)} = \frac{C}{100}[/itex] which implies that C must equal zero. Thus,

    [itex]Q(100) = \frac{20t + \frac{t^2}{10}}{100 + t} = \frac{2000 + 1000}{200}[/itex]

    Have I done this correctly? I suppose my answer is more like the concentration after 100 minutes, but I'm worried because it seems like I don't have much simplifying to do, as the problem hints that I should. Suggestions?
     
  2. jcsd
  3. Apr 22, 2012 #2
    Looks good :smile:
    And that is the amount of salt in the tank after 100 minutes, not the concentration. As for not simplifying it, that should be okay, unless you just want to plug in 100 for t and do absolutely no simplifying after that.
     
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