$\begin{aligned} & (2 x-y+1) d x+(2 y-x+1) d y=0 \\ & \frac{d y}{d x}=\frac{2 x-y+1}{x-2 y-1}, \text { put } x=X+h, y=Y+k \\ & \frac{d Y}{d X}=\frac{2 X-Y+2 h-k+1}{X-2 Y+h-2 k-1} \\ & 2 h-k+1=0 \Rightarrow h-2 k-1=0\end{aligned}$
On solving $h=-1, k=-1 ; \therefore \frac{d Y}{d X}=\frac{2 X-Y}{X-2 Y}$
$\begin{aligned} & \text { Put } Y=v X ; \therefore \frac{d Y}{d X}=v+X \frac{d v}{d X} \\ & v+X \frac{d v}{d X}=\frac{2 X-v X}{X-2 v X}=\frac{2-v}{1-2 v} \\ & X \frac{d v}{d X}=\frac{2-2 v+2 v^2}{1-2 v}=\frac{2\left(v^2-v+1\right)}{1-2 v} \\ & \quad \frac{d X}{X}=\frac{(1-2 v)}{2\left(v^2-v+1\right)} d v \\ & \therefore \quad\end{aligned}$
Put $v^2-v+1=t \Rightarrow(2 v-1) d v=d t$
$\begin{aligned} & \therefore \frac{d X}{X}=-\frac{d t}{2 t} \therefore \log X=\log t^{-1 / 2}+\log c \\ & \therefore X=t^{-1 / 2} c \Rightarrow X=\left(v^2-v+1\right)^{-1 / 2} \cdot c \\ & X^2\left(v^2-v+1\right)=\text { constant } \\ & (x+1)^2\left(\frac{(y+1)^2}{(x+1)^2}-\frac{(y+1)}{x+1}+1\right)= \\ & (y+1)^2-(y+1)(x+1)+(x+1)^2=c \\ & y^2+x^2-x y+x+y=c .\end{aligned}$