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Diff Eq's Mixing problem

  1. Feb 3, 2014 #1
    1. The problem statement, all variables and given/known data

    A tank contains 1160 L of pure water. A solution that contains 0.03 kg of sugar per liter enters a tank at the rate 7 L/min The solution is mixed and drains from the tank at the same rate.

    a.) How much sugar is in the tank initially?
    b.) Find the amount of sugar in the tank after t minutes.
    c.) Find the concentration of sugar in the solution in the tank after 72 minutes.

    2. Relevant equations



    3. The attempt at a solution

    I made s(t) be the amount of salt at time 't'

    ds/dt = rate in - rate out

    Well the rate in would be the concentration times the rate... so

    ds/dt = (.03kg)(7L/min) - (s/1160)(7L/min)

    ds/dt = .21 - (7s/1160)

    Kind of get confused from there...I figure I'll have to get the 's' on one side right? separable?


    Am I even on the right track here? Just starting out in the class.
     
  2. jcsd
  3. Feb 3, 2014 #2

    LCKurtz

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    Yes, you are on the right track. So far so good.

    If you call ##a = \frac 7 {1160}## your equation can be rewritten$$
    \frac{ds}{dt} +as = .21$$It is constant coefficient, linear, and separable, so any of these methods would work. I would suggest the constant coefficient method first, linear (integrating factor) second, and separation of variables last in order of preference.
     
  4. Feb 4, 2014 #3
    Constant coefficient? I wouldn't just integrate here? I'm looking all over the web for an explanation of "constant coefficient" but not seeing too much
     
  5. Feb 4, 2014 #4

    LCKurtz

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    That's "constant coefficient differential equation". If you haven't studied those yet use the integrating factor method for first order linear DE's.
     
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