Example 1. Integrate the homogeneous differential equation

$$
x d y-\left(y+\sqrt{x^{2}+y^{2}}\right) d x=0
$$
Solution. The coefficients of $d y$ and $d x$ are respectively:

$$
Q(x, y)=x ; P(x, y)=-\left(y+\sqrt{x^{2}+y^{2}}\right) .
$$

The functions $P(x, y)$ and $Q(x, y)$ are homogeneous functions of the first degree. Indeed,

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

Thus, the identities (9.12) are satisfied.

Let $y=u x$, then $d y=u d x+x d u$. Substituting these expressions into the original equation:

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

Expanding the brackets, combining like terms, and canceling by $x$, we obtain a separable equation

$$
x d u-\sqrt{1+u^{2}} d x=0
$$

Separating the variables and integrating, we find

$$
\ln x=\ln \left(u+\sqrt{1+u^{2}}\right)+\ln C
$$

from which

$$
x=C\left(u+\sqrt{1+u^{2}}\right)
$$

Substituting here the expression $u=\frac{y}{x}$ (obtained from the equality $y=u x$), we get

$$
x=C\left(\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^{2}}\right)
$$

or

$$
\frac{x^{2}}{C}-y=\sqrt{x^{2}+y^{2}}
$$

Squaring both sides of the equation and canceling by $x^{2}$, we obtain the general integral

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

geometrically representing a family of parabolas with vertices on the $O y$ axis.