Example 2. Solve the equation $y^{\prime}=\frac{x+y-1}{2 x-y+4}$.
Solution. This equation would be homogeneous if there were no free terms in the numerator and denominator of the right-hand side. It has the form indicated in point $2^{\circ}$, and can be reduced to a homogeneous equation after additional searching.

From the linear system $\left\{\begin{array}{r}x+y-1=0, \\ 2 x-y+4=0\end{array}\right.$ we find $\left\{\begin{array}{l}x=-1, \\ y=2 .\end{array}\right.$

These numbers are used in subsequent linear substitutions of the argument $x$ and the function $y$.

Let $x=u-1, y=v+2, y^{\prime}=v^{\prime}$. We obtain (verify this) the equation $v^{\prime}=\frac{u+v}{2 u-v}$. This new equation is homogeneous (it involves the new unknown function $v$ and the new argument $u$).
By substituting $v=t u, v^{\prime}=t^{\prime} u+t$ (here $u$ is the independent variable, $t=t(u)$ is the unknown function of $u$), we arrive at a separable equation. We have $t^{\prime} u+t=\frac{1+t}{2-t}$, or $t^{\prime} u=\frac{t^{2}-t+1}{2-t}$. Replacing $t^{\prime}=\frac{d t}{d u}$, separating variables, and integrating: $\int \frac{(t-2) d t}{t^{2}-t+1}=$ $=-\int \frac{d u}{u}$, i.e., $\frac{1}{2} \ln \left(t^{2}-t+1\right)-\sqrt{3} \operatorname{arctg} \frac{2 t-1}{\sqrt{3}}=-\ln |u|+C$.

Returning to $x$ and $y: t=\frac{v}{u}=\frac{y-2}{x+1}$. Under the logarithm, we bring to a common denominator, replace the logarithm of the quotient with the difference of logarithms, and cancel equal terms in the right and left parts. The general solution is obtained in implicit form.

Answer.
$\ln \left((y-2)^{2}-(y-2)(x+1)+(x+1)^{2}\right)-2 \sqrt{3} \operatorname{arctg} \frac{2 y-x-5}{\sqrt{3}(x+1)}=C_{\mathrm{I}}$.