## 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.

1. anonymous

Give me a minute

2. anonymous

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

Thats correct

4. anonymous

Shouldn't t be in the equation?

5. anonymous

dy/dt = -y(t)/100

6. anonymous

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

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

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

9. anonymous

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

10. anonymous

Sorry, when t = 0

11. anonymous

Ah, okay. Gotcha. Thank you very much.