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Mixing Problem (variable volume)

  1. Mar 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A tank contains 150L of water in which initially 30g or salt is dissolved. Brine runs in at a rate of 4 L/m and each litre contains 1.5g of dissolved salt. The mix in the tank is kept uniform. Brine runs out at 3L/m. What is that amount of salt in the tank at any time t?
    Note that the volume of brime in the tank is not constant in time


    2. Relevant equations

    [tex] y' +p(x)y = r(x) [/tex]

    General solution to first order, linear ODE

    [tex] y(t) = exp(-\intp(p(t)) dt) [\int r(t)\exp(\intp(p(t))dt) dt +C] [/tex]



    3. The attempt at a solution

    Input/Output Gives

    [tex]y(t)=(6-3y)/(150+t)
    [/tex]
    -
    [tex]y'+3y = 6/(150+t)[/tex]

    Substituting these values into the general equation gives me

    y(t) = e^(-3t)[\nt(6/(150+t)*e^(3t) dt) + C]


    And from here ... well... some sort of a clue as to where to go next would be very nice.
    Basically i am having trouble with that integral, is it possible to take [tex]\int(ln((6/(150+t)*\exp^(3t) dt)[/tex] somehow? I'm not sure if/how to make this step.
    I can take the log of the other parts of the equation...

    ln(y(t)) = ln(exp^3t)+ [this is the bit i'm not sure of] + ln(C)
     
    Last edited: Mar 17, 2008
  2. jcsd
  3. Mar 17, 2008 #2
  4. Mar 17, 2008 #3
    Yes, thanks for showing me that - perhaps it's just the tiredness setting in, or perhaps i'm just completely dense! But i am still not able to work this out.
     
  5. Mar 17, 2008 #4
    Well, i'm an idiot :P solved it now! Thanks!
     
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