Example 4. Solve the equation $y^{\prime}-y^{2}=4 x^{2}+4 x y+2$.
Solution. Rewriting the given equation in the form $y^{\prime}=(2 x+y)^{2}+2$, we notice the appropriateness of the substitution (change of variable) $2 x+y=v$. Differentiating this equality (with respect to the variable $x$), we get $2+y^{\prime}=v^{\prime}$, or $y^{\prime}=v^{\prime}-2$. The original equation takes the form $(y=v-2 x): v^{\prime}-2=v^{2}+2$, or $v^{\prime}=v^{2}+4$. This is a separable equation with respect to the unknown function $v$: $\frac{d v}{v^{2}+4}=d x$. Integrating: $\frac{1}{2} \operatorname{arctg} \frac{v}{2}=x+\frac{C}{2}, \frac{v}{2}=\operatorname{tg}(2 x+C)$ - this is the general solution of the auxiliary equation. Returning to the original unknown: $v=2 x+y$.

## Answer. $y=2 \operatorname{tg}(2 x+C)-2 x$