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Thread: Differential equation for a tank with a salt solution and a changing volume

  1. #1
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    Question Differential equation for a tank with a salt solution and a changing volume

    On my most recent test the last question completely stumped me. The problem is the following:
    A tank contains 500 liters of salt solution with 50 kg of salt to start with. Another solution with the concentrate of 0,015 kg/liter is pumped in at a speed of 6 liters/min. At the same time 4 liter/min is being pumped out of the tank. The solution is always assumed to be evenly distributed and the tank has a maximum capacity of 1000 liters. How much salt does the tank contain right before it spills over the edge?

    This is what I managed to do:

    t= time in minutes
    It adds 2 liter/min therefor:
    t=(1000-500)/2=250

    I got the follow differential equation which I don't know how to solve.

    S(t)= the amount of salt in kg

    S'(t)=0,09-4((S(t)/(500+2t))

    With the condition S(0)=50

    How do you solve this type of equation? I could do it if wasn't for the "t" mixed in to it. Any help would be greatly appreciated.
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  2. #2
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    Re: Differential equation for a tank with a salt solution and a changing volume

    Separate the variables.
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  3. #3
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    Re: Differential equation for a tank with a salt solution and a changing volume

    That is a "linear, first order differential equation" and there is a general formula for its solution. Check your text book.

    However, I would use a little different method, derived from solving higher order equations: First solve the associated homogeneous equation: S'(t)= dS/dt= -4((S(t)/(500+2t)). Then you can "separate variables" as Archie suggests- dS/S= -4dt/(500+ 2t). Integrate both sides.

    Now, find the general solution to the entire equation by adding any single solution to the entire equation to that.
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