Example 7. In a certain chemical reaction, substance $B$ decomposes into two substances $X$ and $Y$, and the rate of formation of each is proportional to the current amount $b$ of substance $B$. Find the laws governing the changes in the amounts $x$ and $y$ of substances $X$ and $Y$ as functions of time $t$, if at the beginning of the process $b=b_{0}, x=0, y=0$, and after one hour $b=\frac{1}{2} b_{0}, x=\frac{b_{0}}{8}, y=\frac{3 b_{0}}{8}$.
Solution. If $x$ and $y$ denote the quantities of substances $X$ and $Y$ respectively, then the derivative $\frac{d x}{d t}$ represents the rate of formation of substance $X$, and the derivative $\frac{d y}{d t}$ represents the rate of formation of substance $Y$. At any moment of time $t$, the quantity $b$ of decomposed substance $B$ is determined by the formula $b=b_{0}-x-y$, where $b_{0}$ is the initial quantity of substance $B$. According to the problem, we obtain a system of two first-order differential equations:

$$
\frac{d x}{d t}=k\left(b_{0}-x-y\right), \frac{d y}{d t}=l\left(b_{0}-x-y\right)
$$

where $k$ and $l$ are proportionality coefficients.

This system can be reduced to a single first-order differential equation. Indeed, dividing both sides of the second equation by the corresponding parts of the first equation, we get

$$
\frac{d y}{d x}=\frac{l}{k}
$$

Solving this equation, we have

$$
y=\frac{l}{k} x+C_{1}
$$

Using the initial conditions ($x=0, y=0$ when $t=0$), we find that $C_{1}=0$, so $y=\frac{l}{k} x$. Substituting the obtained expression for $y$ into the first equation of the original system, we get a first-order linear equation

$$
\frac{d x}{d t}+(k+l) x=k b_{0}
$$

The general solution of which is

$$
x=\frac{k b_{0}}{k+l}+C_{2} e^{-(k+l) t}
$$

Since $x=0$ when $t=0$, we have $C_{2}=-\frac{k b_{0}}{k+l}$ and

$$
x=\frac{k b_{0}}{k+l}\left(1-e^{-(k+l) t}\right)
$$

Substituting the obtained expression for $x$ into the equation $y=\frac{l}{k} x$, we find that

$$
y=\frac{l b_{0}}{k+l}\left(1-e^{-(k+l) t}\right)
$$

Now let's determine the values of the coefficients $k$ and $l$. Since $x=\frac{b_{0}}{8}$ and $y=\frac{3 b_{0}}{8}$ when $t=1$, substituting these values into the expressions for $x$ and $y$ leads to the system of equations:

$$
\frac{k}{k+l}\left(1-e^{-(k+l)}\right)=\frac{1}{8} ; \frac{l}{k+l}\left(1-e^{-(k+l)}\right)=\frac{3}{8}
$$

Adding these equations term by term, we get

$$
1-e^{-(k+l)}=\frac{1}{2}
$$

from which

$$
e^{-(k+l)}=2^{-2} \text { and } k+l=\ln 2
$$

Dividing the second equation by the first term by term, we find $l=3 k$.

Therefore, $k=\frac{1}{4} \ln 2, l=\frac{3}{4} \ln 2$.

Thus, the desired solution is given by the formulas:

$$
x=\frac{b_{0}}{4}\left(1-\frac{1}{2^{\prime}}\right) ; y=\frac{3 b_{0}}{4}\left(1-\frac{1}{2^{\prime}}\right)
$$

## Problems

Formulate differential equations from the conditions of the following problems and solve them.

1. Find the curve passing through the point $M(1,3)$ and for which the segment of the tangent between the point of tangency and the $O x$ axis is bisected at the point of intersection with the $O y$ axis.
2. Find a curve passing through the point $M(0,3)$ such that the slope of the tangent at any point is equal to the ordinate of that point decreased by two units.
3. A vessel with a capacity of $a$ liters, filled with a water solution of salt, is continuously passed through by a liquid, with $b$ liters of pure water being poured in per unit time and the same amount of solution flowing out. Find the law according to which the amount of salt in the vessel changes depending on the time the liquid flows through the vessel.
4. A boat slows down under the action of water resistance, which is proportional to the boat's speed. The initial speed of the boat is 1.5 m/s, and the speed after 4 seconds is 1 m/s. When will the speed decrease to 0.01 m/s? What distance can the boat travel before stopping?
5. A body cooled from 100 to $60^{\circ}$ in 20 minutes. The temperature of the surrounding air is maintained at $20^{\circ}$. When will the body cool to $30^{\circ} ?$
6. A tank contains 100 liters of solution, containing 10 kg of salt. Water is continuously supplied to the tank at a rate of $5 \pi$ per minute. The incoming water is mixed with the existing solution, and the mixture flows out at the same rate. How much salt will remain in the tank after 1 hour?
7. Find the curve for which all normals pass through a constant point.
8. Water flows out through a hole at the bottom of a cylindrical vessel. According to what law does the water level decrease over time, if it is known that the speed $v$ at which the liquid flows out of the hole depends on the height $h$ of the liquid column as follows: $v=$ $=0.6 \sqrt{2 g h}$.
9. A ship with a displacement of $12000 \mathrm{~m}$ moves in a straight line at a speed of $v_{0}=20 \mathrm{~m} /$ sec. The water resistance is proportional to the square of the ship's speed and is equal to $36 m$ at a speed of 1 m/s. What distance will the ship travel after the engine stops, before the speed becomes equal to 5 m/s? In what time will the ship travel this distance?

## Answers

1. Differential equation $y=2 x y^{\prime}, \quad y^{2}=9 x$. 2. $y^{\prime}=y-2$, $y=e^{x}+2$. 3. $\frac{d x}{d t}=-\frac{b}{a} x, x=C_{0} e^{-\frac{b}{a} t}$, where $C_{0}$ is the initial amount of salt. 4. 50 sec; 15 m. 5. $\frac{d U}{d t}=k(U-20)$, where $U-$ is the temperature of the body. General solution $U=20+C e^{k t}$. The solution corresponding to the initial conditions of the problem: $U=20+80\left(\frac{1}{2}\right)^{\frac{1}{20}}$. The body will cool to $30^{\circ}$ in 1 hour. The cooling rate of the body in the air is proportional to the difference between the temperature of the body and the temperature of the air. 6. 0.5 kg. 7. Circle. 8.