The solution of $y^{\mathrm{\prime }}(y-x-4)=x+y-2$is given by 

$\begin{array}{l}{y}^{\mathrm{\prime }}=\frac{x+y-2}{y-x-4}.\text{ Put }x=X+h,y=Y+K\\ \frac{dY}{dX}=\frac{X+Y+(h+K-2)}{Y-X+(-h+K-4)}\end{array}$

Select h, K such that $h+K-2=0,-h+K-4$

$\begin{array}{l}=0\\ \Rightarrow K=3,h=-1\\ \frac{dY}{dX}=\frac{X+Y}{Y-X}\end{array}$

Putting  $Y=vX$ , we have 

$\begin{array}{l}v+X\frac{dV}{dX}=\frac{1+V}{V-1}\\ \Rightarrow X\frac{dV}{dX}=\frac{1+V}{V-1}-V=\frac{1+2V-V^2}{V-1}\\ \frac{dX}{X}=\frac{-1}{2}\frac{-2V+2}{1+2V-V^2}dV\end{array}$

Integrating, we have $X^2\left(1+2V-V^2\right)=\text{ Const }$

$X^2-Y^2+2XY=\text{ Const }$ 

$(x+1{)}^2-(y-3{)}^2+2(x+1)(y-3)=\text{ Const }$

$x^2+2xy-y^2-4x+8y=\text{ Const. }$