Example 1. $\left(y^{\prime}\right)^{2}+\left(\frac{x-1}{y}+\frac{1+2 x}{\sqrt{1+x+x^{2}}}\right) y^{\prime}+\frac{(x-1)(2 x+1)}{\sqrt{1+x+x^{2}} \cdot y}=0$.
Solution. Analysis of the left side of the equation allows us to factorize it:

$$
\left(y^{\prime}+\frac{2 x+1}{\sqrt{1+x+x^{2}}}\right)\left(y^{\prime}+\frac{x-1}{y}\right)=0
$$

This equation is equivalent to a system of two equations with separable variables:

$$
\left[\begin{array} { l } 
{ y ^ { \prime } = - \frac { 2 x + 1 } { \sqrt { 1 + x + x ^ { 2 } } } } \\
{ y ^ { \prime } = \frac { 1 - x } { y } }
\end{array} \Rightarrow \left[\begin{array}{l}
y=-2 \sqrt{1+x+x^{2}} \\
y^{2}+(x-1)^{2}=C_{2}^{2}
\end{array}\right.\right.
$$

The general solution of the original equation consists of a combination of two functions, each depending on its own arbitrary constant (the right function is defined explicitly, the second implicitly, but for uniformity, we will write the solution in implicit form).

Answer. $y+2 \sqrt{1+x+x^{2}}=C_{1}$ or $y^{2}+(x-1)^{2}=C_{2}^{2}$.

Note. In the general solution, we use two constants $C_{1}$ and $C_{2}$ to emphasize the independence of the solutions of the separate equations. However, using one constant is not an error.

Example $2 .\left(y^{\prime}\right)^{2}-(2 x+y) y^{\prime}+x^{2}+x y=0$.

Solution. The given equation must be solved as a quadratic equation in terms of the derivative $y^{\prime}$.

We find: $y^{\prime}=x+y$ or $y^{\prime}=x$.

We arrive at a system of equations, each of which can be solved by one of the known methods. The first one is solved by substitution:

$$
\begin{gathered}
y^{\prime}=x+y, \quad x+y=z, \quad 1+y^{\prime}=z^{\prime}, \quad z^{\prime}-1=z, \quad z^{\prime}=z+1 \\
\frac{d z}{z+1}=d x, \quad \ln |z+1|=x+C_{1}, \quad z+1=e^{x+C_{1}}, \quad x+y+1=e^{x+C_{1}}
\end{gathered}
$$

The second one is integrated directly:

$$
y^{\prime}=x, \quad y=\frac{x^{2}}{2}+C_{2}
$$

Answer. $y=C_{1} \cdot e^{x}-x-1 ; \quad y=\frac{x^{2}}{2}+C_{2}$.

Note. The existence and uniqueness theorem is not applicable to the original equations of examples 1 and 2 (not resolved with respect to the derivative $y^{\prime}$). As an illustration, note that in example 2, through the point $M_{0}(1,2)$, there pass two different integral curves: $y=4 e^{x-1}-x-1$ and $y=\frac{x^{2}}{2}+\frac{3}{2}$.