Example 2. Solve the equation

$$
\frac{d y}{d x}=-\frac{y}{x}(x \neq 0)
$$

Find the particular solution that satisfies the condition: $y=3$ when $x=2$.
Solution. Separating the variables, we get

$$
\frac{d y}{y}+\frac{d x}{x}=0
$$

Integrating, we find

$$
\ln |y|+\ln |x|=C_{1}
$$

The constant $C_{1}$ can be written as

$$
C_{1}=\ln |C|(C \neq 0)
$$

(since any positive or negative number $C_{1}$ can be represented as the natural logarithm of another, positive number $|C|$).

Substituting the expression for $C_{1}$ into equation (A) and transforming it, we get

$$
\ln |x y|=\ln |C|(C \neq 0)
$$

Exponentiating the last equation, we find the general integral $x y=C$ $(C \neq 0)$, which geometrically represents a family of hyperbolas with the coordinate axes as asymptotes.

Let's solve the Cauchy problem for this equation with the given initial conditions: $y_{0}=3$ when $x_{0}=2$. Substituting these values into the equation $x y=C$, we get $2 \cdot 3=C$ or $C=6$.

Thus, the particular solution is $x y=6$. This is a hyperbola passing through the point $M_{0}(2,3)$.

Remark. When separating the variables, we assumed that $y \neq 0$, and thus could have lost the solution $y=0$. Indeed, $y=0$ is a solution to the equation (which can be verified directly), and it can be obtained from the general integral when $C=0$. Therefore, the general integral is given by the formula

$$
x y=C
$$

where $C$ can take any real values (including $C=0$).