I have such a problem:
There are two tanks, each one of contains $100 \text{ L}$ of liquid. The first one: compound of alcohol and water ($20\%$ of alcohol), the second one: pure water. Then a pump is installed. It works in both directions and pumps $10 \text{ L/min}$ (from the first one to the second one and from the second one to the first one).
We are to find the formula which describes the concentration of alcohol in the first tank. So just the equation is needed.
The problem is simple but I would like to see a full and correct solution as I'm new to differential equations.
Let the amount of alcohol in the first tank at any given time $t$ be $x(t)$ liters. Since no alcohol is lost, the total alcohol is always $20 \text{ L}$, therefore the amount of alcohol in the second tank at the same time must be $20 - x$. It follows that the concentration of alcohol in the two tanks must be $\frac{x}{100}$ and $\frac{20-x}{100}$, respectively.
The rate of change of the alcohol in the first tank, $dx/dt$, must be equal to the inflow minus the outflow. The outflow of alcohol is equal to the volume being pumped out times the concentration of alcohol in the first tank, while the outflow is the volume times the concentration of the second tank. Thus we have the following equation $$ \frac{dx}{dt} = 10\cdot\frac{20-x}{100} - 10 \cdot \frac{x}{100} $$
with the initial condition $x(0) = 20$
Just for some intuition, the solution is $$ x(t) = 10 + 10e^{-0.2t} $$ which makes sense as after pumping back and forth, both tanks would contain equal amounts ($10 \text{ L}$) of alcohol