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Homework Help: Differential equations incongruecy

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    I am going to copy-paste this text that my friend made (because we both have the same doubt and we don't know to work around it. This is a long post, so warning):

    "I'm currently unsure of how these two problems work. I've tried working at them in different ways but i don't understand why they are different (even if we are dealing with percentages):

    1.The air in a room with volume 180m3 contains 0.15% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2m3/min and the mixed air flows out at the same rate. Find the percentage of carbon dioxide in the room as a function of time. What happens in the long run?

    2.A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?

    The solutions for both problems are different:

    -> 1. The rate of change in the amount of CO2 is dy/dt = Rate(in) - Rate(out), so we have:
    Rate(in): 0.0005(2) Rate(out): (2)y/(180)
    Then, our differential equation would be dy/dt = 0.001 - y/90, and by integrating it we would reach that our final equation is y = 0.06e-t/90 + 0.09. However, note that in contrast to the second problem WE DO NOT CHANGE IT TO A PERCENTAGE (even though we did the same set up for the exercise!)

    -> 2. The rate of change in the amount of alcohol is dy/dt = Rate(in) - Rate(out), so it should just be:
    Rate(in): 0.06(5) Rate (out): y(t)/500(5)
    So, our differential equation would be dy/dt = (30-y)/100. After integrating and such (which i understand), the book arrives at the expression y = 30 - 10e-t/100, then they multiply it by 100 and divide it by 100 to obtain the percentage."

    Long story short, we are presented with two different problems, and in one they convert it to a percentage and the other one is just left as it is (but it express results as a percentage).

    2. Relevant equations

    3. The attempt at a solution

    Why would it be that in the second case we are converting the expression into a percentage, but in the first case we are completely ignoring that fact (but still getting the "supposed" correct answer?). Thank you very much for your time.
    Last edited: Nov 3, 2013
  2. jcsd
  3. Nov 4, 2013 #2
    Update: I solved the problem (the issue was with how problem 1 was interpreted. Problem 2 is perfectly done and does not need any modifications).

    The thing was that when plugging in the value to find the constant in the expression, I first had to convert the initial volume of the CO2 given as a percentage into units of volume (this means, dividing by 100 and multiplying by the volume in the room), then the constant is found to be 0.18. After that, you transform the whole expression into an expression that gives volume as a percentage with respect to time (and then you can finally solve the exercise, which its answer is 0.05%!)
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