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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.
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dy/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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Shouldn't t be in the equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Using an integrating factor may be the easiest route. Don't forget to use the initial condition: y(0) = 15kg to obtain the particular solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if \[y \prime(t) = y(t)/100\] then \[y(t)=1/100t + C\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No... y(t) = ce^(t/100) So when t = 10, The particular solution will be: y = 15e^(t/100)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, okay. Gotcha. Thank you very much.
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