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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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Give me a minute
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
Thats correct

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Other answers:

Shouldn't t be in the equation?
dy/dt = -y(t)/100
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.
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.
So if \[y \prime(t) = -y(t)/100\] then \[y(t)=-1/100t + C\]?
No... y(t) = ce^(-t/100) So when t = 10, The particular solution will be: y = 15e^(-t/100)
Sorry, when t = 0
Ah, okay. Gotcha. Thank you very much.

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