Example 3. Solve the equation $y^{\prime}=\frac{2 x+3 y-5}{4 x+6 y+8}$.
Solution. Obviously, the right-hand side of the given equation is a function of $2 x+3 y$. Let $v=2 x+3 y$. This equality is differentiable with respect to the variable $x$ ( $x$ is an argument for both the function $y$ and the function $v$ ): $2+3 y^{\prime}=v^{\prime}$.

From this, we find $y^{\prime}=\frac{1}{3}\left(v^{\prime}-2\right)$ and substitute this expression into the original equation:

$$
\frac{1}{3}\left(v^{\prime}-2\right)=\frac{v-5}{2 v+8}, \quad v^{\prime}=\frac{3(v-5)}{2 v+8}+2, \quad \frac{d v}{d x}=\frac{7 v+1}{2 v+8}, \quad \frac{2 v+8}{7 v+1} d v=d x
$$

Integrating: $\int \frac{2 v+8}{7 v+1} d v=\int d x, \frac{2}{7} \int\left(1+\frac{27}{7 v+1}\right) d v=x+C_{1}, v+$ $+\frac{27}{7} \ln |7 v+1|=\frac{7}{2} x+C$. We have obtained the general solution of the auxiliary equation. Returning to $x$ and $y$, we write the general solution of the original equation in implicit form.

Answer. $3 y+\frac{27}{7} \ln |7(2 x+3 y)+1|=\frac{3}{2} x+C$.