Two 1000 liter tanks are containing salt water. Tank 1 has 800 liters of water initially having 20 grams of salt dissolved in this and tank 2 has 1000 liters of water and initially has 80 grams of salt dissolved into this. Salt water along with a concentration of ½ gram/liter of salt enters tank 1 at a rate of 4 liters/hour. Fresh water enters into tank 2 at a rate of 7 liters/hour. With a connecting pipe water flows from tank 2 in tank 1 at a rate of 10 liters/hour. By a different connecting pipe 14 liters/hour flows out of tank 1 and 11 liters/hour are  drained out of the pipe and thus out of the system totally and only 3 liters/hour flows back in tank 2. Set up the system which will provide the amount of salt in each tank at any specified time.

Solution:

Okay, assume that Q1 (t) and Q2 (t) be the amount of salt into tank 1 and in tank 2 at any time t correspondingly.

 This time all we want to do is set up a differential equation for both tanks just as we did back while we had a particular tank. The only difference is that we now require dealing along with the fact that we've found a second inflow to both tank and the concentration of the second inflow will be the concentration of the other tank.

Recall that the basic differential equation is the rate of change of salt (Q′) equals the rate at that salt enters minus the rate at salt leaves. All entering/leaving rates are found through multiplying the flow rate times the concentration.

Now there is the differential equation for tank 1.

Q₁' = (4) (1/2) + (10) (Q₂/1000) - (14) (Q₁/800)                                 Q₁(0) = 20

= 2 + (Q₂/1000) - (7Q₁/400)

Under this case of differential equation the initial pair of numbers is the salt entering from the external inflow. The second set of numbers is the salt which entering in the tank from the water flowing in from tank 2. The third set is the salt leaving tank as water flows out.

Now there is the second differential equation.

Q₂' = (7) (0) + (3) (Q₁/800) - (10) (Q₂/1000)                          Q₂(0) = 80

= (3Q₁/800) - (Q₂/100)

Note that since the external inflow in tank 2 is fresh water the concentration of salt in it is zero.

Summarized here that the system we'd require to solve,

Q₁' = 2 + (Q₂/1000) - (7Q₁/400)                                 Q₁(0) = 20

Q₁' =(3Q₁/800) - (Q₂/100)                                          Q₂(0) = 80

This is a non-homogeneous system due to the first term in the first differential equation. If we had clean and fresh water flowing in both of these we would actually have a homogeneous system.