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.

Question

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 · Answer

Give me a minute

anonymous · Answer

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

anonymous · Answer

Thats correct

anonymous · Answer

Shouldn't t be in the equation?

anonymous · Answer

dy/dt = -y(t)/100

anonymous · Answer

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.

anonymous · Answer

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.

anonymous · Answer

So if $$y \prime(t) = -y(t)/100$$ then $$y(t)=-1/100t + C$$?

anonymous · Answer

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

anonymous · Answer

Sorry, when t = 0

anonymous · Answer

Ah, okay.  Gotcha.  Thank you very much.