Example 4. A vessel containing $10 \pi$ of water is continuously supplied with a solution at a rate of $2 \pi$ per minute, with each liter of the solution containing 0.3 kg of salt. The incoming solution is mixed with the water, and the mixture flows out of the vessel at the same rate. How much salt will be in the vessel after 5 minutes?
Solution. Let the independent variable be time $t$, and the function we are looking for be $y(t)$ - the amount of salt in the container $t$ minutes after the start of the experiment. We will determine how the amount of salt changes over the time interval from moment $t$ to moment $t+\Delta t$. Into the container, $2 \pi$ liters of solution flow in one minute, and in $\Delta t$ minutes, $2 \Delta t$ liters flow in, containing $0.3 \cdot 2 \Delta t=0.6 \Delta t$ kg of salt. On the other hand, over the same time interval $\Delta t$, $2 \Delta t$ liters of solution flow out of the container. At moment $t$, the entire container (10 liters) contains $y(t)$ kg of salt, so in the $2 \Delta t$ liters of the outflowing solution, there would be $0.2 \Delta t \cdot y(t)$ kg of salt if the salt concentration in the container remained unchanged over the time $\Delta t$. However, since the concentration changes by an infinitesimally small amount as $\Delta t \rightarrow 0$, the $2 \Delta t$ liters of outflowing solution contain $0.2 \Delta t \cdot [y(t)+\alpha]$ kg of salt, where $\alpha \rightarrow 0$ as $\Delta t \rightarrow 0$.

Therefore, the amount of salt in the inflowing solution over the time interval $\Delta t$ is $0.6 \Delta t$ kg, and in the outflowing solution, it is $-0.2 \Delta t \cdot [y(t)+\alpha]$ kg. The change in the amount of salt over this time is the difference between these two expressions, i.e.,

$$
y(t+\Delta t)-y(t)=0.6 \Delta t-0.2 \Delta t \cdot [y(t)+\alpha]
$$

where $\alpha \rightarrow 0$ as $\Delta t \rightarrow 0$. Dividing both sides of the obtained equation by $\Delta t$ and taking the limit as $\Delta t \rightarrow 0$, we get

$$
y'(t)=0.6-0.2 y(t)
$$

This is a first-order differential equation with separable variables, which can be written as:

$$
\frac{d y}{0.6-0.2 y}=d t
$$

Integrating this equation, we obtain

$$
y=3-C e^{-0.2 t}.
$$

The value of the arbitrary constant is determined from the initial condition
$y=0$ when $t=0$ (at the initial moment, there was no salt in the container). Substituting the values $t=0$ and $y=0$ into the general solution, we find:

$$
0=3-C, C=3
$$

Thus, the change in the amount of salt in the container is given by the formula

$$
y=3-3 e^{-0.2 t}
$$

At $t=5$, we get

$$
y(5)=3-3 e^{-0.2 \cdot 5}=3-3 e^{-1} \approx 1.9(\text{kg})
$$