Example 2. Find the general integral of the homogeneous equation

$$
\left(x^{2}-y^{2}\right) d y-2 y x d x=0
$$
Solution. In this case, we have:

$$
Q(x, y)=x^{2}-y^{2} ; P(x, y)=-2 x y .
$$

These functions are homogeneous functions of the second degree. Indeed:

$$
\begin{gathered}
Q(k x, k y)=(k x)^{2}-(k y)^{2}=k^{2}\left(x^{2}-y^{2}\right) \equiv k^{2} Q(x, y) \\
P(k x, k y)=-2(k x)(k y)=k^{2}(-2 x y) \equiv k^{2} P(x, y)
\end{gathered}
$$

Assuming $y=u x$, we find

$$
d y=x d u+u d x .
$$

Substituting the expressions for $y$ and $d y$ into the original equation and dividing by $x^{2} \neq 0$, we obtain a separable equation:

$$
\left(1-u^{2}\right)(x d u+u d x)-2 u d x=0
$$

or

$$
\frac{d x}{x}+\frac{u^{2}-1}{u\left(1+u^{2}\right)} d u=0(u \neq 0)
$$

Considering that

$$
\frac{1-u^{2}}{u\left(1+u^{2}\right)}=\frac{1}{u}-\frac{2 u}{1+u^{2}}
$$

and integrating the obtained equation, we find

$$
\ln |x|=\ln |u|-\ln \left|1+u^{2}\right|+\ln |C|
$$

from which

$$
x=\frac{C u}{1+u^{2}}
$$

Substituting the expression for $u$, defined by the equality $y=u x$, we obtain the general integral

$$
x^{2}+y^{2}=C y
$$

representing a family of circles with centers on the $O y$ axis and passing through the origin.

Remark. When $u=0$, we obtain the solution $y=0$. This solution is particular. Indeed, the general integral $x^{2}+y^{2}=C y$ can be rewritten as $C_{1}\left(x^{2}+y^{2}\right)=y$, where $C_{1}=\frac{1}{C}$. When $C_{1}=0$, we get $y=0$.