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anonymous

  • 5 years ago

A 1000L brine tank is filled with 15kg of salt. The tank is constantly mixed. Water enters and exits at 10L/min. Set up a differential equation for the amount of salt at any time.

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  1. anonymous
    • 5 years ago
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    Give me a minute

  2. anonymous
    • 5 years ago
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    dy/dt = (rate in) - (rate out)... rate in = 0 (only water is entering the tank) rate out = (y(t)/1000) kg/L * 10 L/min = y(t)/100 kg/min... dy/dt = -y/100

  3. anonymous
    • 5 years ago
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    Thats correct

  4. anonymous
    • 5 years ago
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    Shouldn't t be in the equation?

  5. anonymous
    • 5 years ago
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    dy/dt = -y(t)/100

  6. anonymous
    • 5 years ago
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    It is correct how it is now... What you'll have to do now is just integrate by separation of variables or by using the integrating factor.

  7. anonymous
    • 5 years ago
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    Using an integrating factor may be the easiest route. Don't forget to use the initial condition: y(0) = 15kg to obtain the particular solution.

  8. anonymous
    • 5 years ago
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    So if \[y \prime(t) = -y(t)/100\] then \[y(t)=-1/100t + C\]?

  9. anonymous
    • 5 years ago
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    No... y(t) = ce^(-t/100) So when t = 10, The particular solution will be: y = 15e^(-t/100)

  10. anonymous
    • 5 years ago
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    Sorry, when t = 0

  11. anonymous
    • 5 years ago
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    Ah, okay. Gotcha. Thank you very much.

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