A tank contains 1000 gallons of fresh water Then water conta

A tank contains 1000 gallons of fresh water. Then water containing 1/3 a pound of salt per gallon is poured into the tank at a rate of 6 gallons per minute, and the mixture is leaves the tank at the same rate. After 10 minutes the process is stopped and fresh water is poured into the tank at a rate of 3 gallons per minute, with the mixture leaving the tank at the same rate its added. Find the amount of salt in the tank at the end of 30 minutes.

Solution

dQ(t)/dt = 1/3 * 6 - Q(t)*6/1000 for 10 minutes

dQ(t)/dt = - Q(t)*6/1000 for 20 minutes

1000Q\' +6 Q(t) =2

e^{3/500 t } Q(t) = 1/500 e^{3/500 t } *500/3 + C

Q(t) = 1/3 +Ce^{-3/500t}

Q(0) = 0 => C = -1/3

Q(t) = 1/3 (1- e^{-3/500t})

Similarly can be done for second time interval