Question

Suppose that we have a system consisting of two interconnected tanks, each containing a brine solution. Tank A contains

x(t) pounds of salt in 200 gallons of brine, and tank B contains
y(t) pounds of salt in 300 gallons of brine. The mixture in each tank is kept uniform by constant stirring. When t = 0, brine is pumped from tank A to tank B at 20 gallons/minute and from tank B to tank A at 20 gallons/minute. Find the amount of salt in each tank at time t if x(0) = 10 and y(0) = 40.

Answer 1

First notice that neither tank will change in total gallons at any time.
The pump is just mixing the water among the 2 tanks continuously. So eventually the concentration in each tank will be the same as concentration of both tanks put together.
Total gallons = 500, Total Salt = 50
Concentration = 50/500 = 1/10
If tank A has this concentration , then salt = 20
For tank B the salt amount is then 30.
This gives us 2 points, an initial amount and a finishing amount as t -> infinity.

To find functions for x(t) and y(t) , we need to look at the rate of change.

`Rate = R_(in) - R_(out)`
For tank A, 20/300 or 1/15 of tank B salt is coming in. While 20/200 or 1/10 of its own salt is leaving.
For tank B its just the other way around.
This will give a system of 2 differential equations:

`(dx)/(dt) = (y)/(15) - (x)/(10)`

`(dy)/(dt) = (x)/(10) - (y)/(15)`
From this we can say:

`(dy)/(dt) = -(dx)/(dt)`

`y = -x+C`
Plug this in for y in the dx/dt equation:

`(dx)/(dt) = (-x+C)/(15) -(x)/(10) = -(x+C)/(6)`
Solving this differential equation gives:

`\int (dx)/(x+C) = \int -(dt)/(6)`

`\ln (x+C) = -(t)/(6) +k`

`x = k e^(-t/6) +C`
The dy/dt equation is solved exactly the same way, so you have:

`y = k e^(-t/6) +C`
All thats left is to apply initial and final conditions to get constants.
The final solution is:

`x(t) = -10e^(-t/6)+20`

`y(t) = 10e^(-t/6)+30`