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Homework Help: Diff Equations, Mixing

  1. Apr 19, 2006 #1
    I'm having a bit of difficulty with these. I'm sort of new to this so I'm somewhat confused, maybe someone could lighten a bit of my confusion.


    Here's the first problem:
    http://www.synthdriven.com/images/deletable/img01.jpg [Broken]


    This is what I've done, please let me know if I'm doing this correctly...

    [tex]\frac{dy}{dt}=[/tex]rate in - rate out
    rate in =[tex](0kg/G)(5G/hr)=0[/tex]
    rate out =[tex](\frac{y(t)}{10G})(5G/hr)=\frac{5y(t)}{10}=\frac{1}{2}y(t)[/tex]
    [tex]\frac{dy}{dt}=-\frac{1}{2}y(t)[/tex]
    [tex]\int \frac{1}{y}=-\int \frac{1}{2}dt[/tex]
    [tex]\ln{y}=-\frac{1}{2}t[/tex]

    This is where I've ended up... Am I to find y and then solve for y=(1/2)10=5?? If so, that'd give me the following:

    [tex]\ln{5}=-\frac{1}{2}t[/tex]
    [tex]t=-2\ln{5}[/tex]

    Is this correct??



    The second problem is the following:
    http://www.synthdriven.com/images/deletable/img02.jpg [Broken]

    For this, I got the following:
    [tex]\frac{dy}{dt}=[/tex]rate in - rate out
    rate in=[tex](0.15\frac{ft^3}{min})(0.06\frac{C0}{ft^3})[/tex]
    rate out=[tex](\frac{y(t)CO}{1800ft^3})(\frac{0.15ft^3}{min})=\frac{0.15y}{1800}CO/min[/tex]
    [tex]\frac{dy}{dt}=0.009-\frac{0.15y}{1800}[/tex]
    [tex]\frac{dy}{dt}=-(\frac{0.15}{1800}-0.009)[/tex]
    [tex]\int{\frac{1800}{0.15y}-\frac{1}{0.009}dy}=-\int{dt}[/tex]
    [tex]\frac{1800}{0.15}\ln{y}-\frac{y}{0.009}=-t[/tex]

    I would then plug in 0.00018 for y, right?
    When I do that, I get 51735.34 min... Is this correct...?


    Any pointers?

    Thanks
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 19, 2006 #2

    AKG

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    Homework Helper

    For number 1, you're forgetting your constant of integration, so you should have:

    ln(y) = C-t/2
    y = ce-t/2 where c = eC

    You need to solve for c, using the fact that y(0) = 10. This is easy, c=10. Now you want to find when y(t) = (1/2)10 = 5, so you set:

    5 = 10e-t/2
    -2ln(0.5) = t
    t = ln4

    For your second problem, its correct where you say dy/dt = 0.009 - 0.15y/1800. After that, it appears you've made some mistakes. Observe that you have dy/dt = a + by for some number as and b. So:

    dy/(a+by) = dt
    ln(a+by)/b = t + C

    You should be able to get the rest from here. Note that in the end, you want to find the smallest t such that y(t)/1800 > 0.00018, i.e. you want to solve y(t)/1800 = 0.00018, not y(t) = 0.00018. In case you get stuck, the solution is below, but only use it if you get stuck.

    ln(a+by) = tb + D
    a+by = cetb
    y = (cetb-a)/b
    y = detb - a/b

    where a = 0.009, b = -0.15/1800, and d has to be determined. Again, you know y(0) = 0, so d = a/b, giving:

    y = (a/b)(etb-1)

    When is y/1800 > 0.00018?

    y/1800 > 0.00018
    y > 0.324
    etb-1[/sup] < -0.003
    etb < 0.997
    tb < ln(0.997)
    t > 36.054108243584661759010672195441...
     
  4. Apr 20, 2006 #3
    I don't get it. What is y(t) represent? Gallons of water in the first problem and cubic feet of CO in the second problem at time t?
     
  5. Apr 20, 2006 #4

    AKG

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    Science Advisor
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    Yes, that's what it represents.
     
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