Example 3. Integrate the homogeneous equation

$$
\left(y^{4}-2 x^{3} y\right) d x+\left(x^{4}-2 x y^{3}\right) d y=0
$$
Solution. The functions

$$
P(x, y)=y^{4}-2 x^{3} y, Q(x, y)=x^{4}-2 x y^{3}
$$

are homogeneous functions of the fourth degree. Let $y=u x$, then $d y=u d x+x d u$ and the equation (after canceling by $x^{4} \neq 0$) will take the form

$$
\left(u^{4}+u\right) d x-\left(1-2 u^{3}\right) x d u=0
$$

or

$$
\frac{d x}{x}=\frac{1-2 u^{3}}{u+u^{4}} d u, \frac{d x}{x}=\left(\frac{1}{u}-\frac{3 u^{2}}{1+u^{3}}\right) d u(u \neq 0, u \neq-1)
$$

Integrating, we get

$$
\ln |x|=\ln |u|-\ln \left|1+u^{3}\right|+\ln |C|
$$

from which

$$
x\left(1+u^{3}\right)=C u
$$

Substituting the expression for $u$ into the last equation, we find the general integral

$$
x^{3}+y^{3}=C x y
$$

which can be rewritten as $x^{3}+y^{3}-3 a x y=0$, where $3 a=C$. The integral curves are Cartesian leaves.

Remark. When $u=-1$, we get $y=-x$ - a solution to the equation. This is a particular solution, it is obtained from the general integral $x^{3}+y^{3}=C x y$ when $C=0$. When $u=0$, we get the solution $y=0$. In addition, $x=0$ is also a solution. These solutions are particular, they are obtained from the general integral $C_{1}\left(x^{3}+y^{3}\right)=x y$ when $C_{1}=0$.