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Mixture DE

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A tank contains 200 liters of fluid which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

    2. Relevant equations

    initial amount of liquid = 200L
    A(t=0) = 30g
    rate in = 4L/min
    rate out = 4L/min
    concentration of the inflow liquid = 1 g/L

    3. The attempt at a solution

    [tex]\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}[/tex]

    [tex]\frac{dA}{dt} = (4) - \frac{4A}{200}[/tex]

    [tex]\frac{dA}{dt} + \frac{4A}{200} = (4)[/tex]

    [tex]\frac{dA}{dt} + \frac{A}{50} = (4)[/tex]

    [tex]I(t) = e^{\int\frac{A}{50}dt}[/tex]

    [tex]I(t) = e^{\frac{At}{50}}[/tex]

    [tex]f(t) = \frac{1}{e^{\frac{At}{50}}}[\int (e^{\frac{At}{50}})(4) dt + c][/tex]

    [tex]f(t) = \frac{1}{e^{\frac{At}{50}}}[\frac{200e^{\frac{At}{50}}}{A} + c][/tex]

    [tex]f(t) = \frac{200e^{\frac{At}{50}}}{Ae^{\frac{At}{50}}} + \frac{c}{e^{\frac{At}{50}}}][/tex]
     
  2. jcsd
  3. Oct 9, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, no, no! "A" is your dependent variable, not a constant. You find the integrating factor by integrating the coefficient of A. Your integrating factor is
    [tex]e^{\int\frac{dt}{50}}= e^{\frac{t}{50}}[/tex]

    And now you've completely lost me! Where did "f(t)" come from? There was no "f" before this line. You are supposed to be finding A(t).

    Since the integrating factor is
    [tex]e^{\int\frac{dt}{50}}= e^{\frac{t}{50}}[/tex]
    you have
    [tex]e^{\frac{t}{50}}\frac{dA}{dt}-e^{\frac{t}{50}}\frac{A}{50}= 4e^{\frac{t}{50}}[/tex]
    [tex]\frac{d}{dt}\left(e^{\frac{t}{50}}A\right)= 4e^{\frac{t}{50}}[/itex]
    Integrate that to find A. The left side is, of course, just
    [tex]e^{\frac{t}{50}}A(t)[/tex]


    Frankly, I think it would be easier not to use the "integrating factor" method at all. This is a separable equation.

    [tex]\frac{dA}{dt}= 4-\frac{A}{50}= \frac{200- A}{50}[/tex]
    [tex]\frac{dA}{200- A}= \frac{dt}{50}[/tex]
    Integrate both sides of that.
     
  4. Oct 9, 2009 #3
    opps I made some typos and I don't know what I was thinking doing the integrating factor :confused: .

    Here is my next attempt:

    [tex]\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}[/tex]

    [tex]\frac{dA}{dt} = (4) - \frac{4A}{200}[/tex]

    [tex]\frac{dA}{dt} + \frac{4A}{200} = (4)[/tex]

    [tex]\frac{dA}{dt} + \frac{A}{50} = (4)[/tex]

    [tex]I(t) = e^{\int\frac{1}{50}dt}[/tex]

    [tex]I(t) = e^{\frac{t}{50}}[/tex]

    [tex]A(t) = \frac{1}{e^{\frac{t}{50}}}[\int (e^{\frac{t}{50}})(4) dt + c][/tex]

    [tex]A(t) = \frac{1}{e^{\frac{t}{50}}}[200e^{\frac{t}{50}} + c][/tex]

    [tex]A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}[/tex]
     
  5. Oct 9, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I hope you understand that this can be simplified! :biggrin:

    And, of course, use the fact that A(0)= 30 to find c.
     
  6. Oct 9, 2009 #5
    [tex]A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}[/tex]

    [tex]A(t) = 200 + \frac{c}{e^{\frac{t}{50}}}[/tex]

    [tex]A(0) = 30 = 200 + \frac{c}{e^{\frac{0}{50}}}[/tex]

    [tex]c = 30 - 200 = -170[/tex]

    [tex]A(t) = 200 - \frac{170}{e^{\frac{t}{50}}}[/tex]
     
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