The general solution of the differential equation $(2x-y+1)dx+(2y-x+1)dy=0$ is:

$(2x-y+1)dx+(2y-x+1)dy=0$

$\frac{dy}{dx}=\frac{2x-y+1}{x-2y-1},$put $x=X+h,y=Y+k$

$\begin{array}{l} \frac{dY}{dX}=\frac{2X-Y+2h-k+1}{X-2Y+h-2k-1}\\ 2h-k+1=0\\ \Rightarrow h-2k-1=0\end{array}$

On solving $h=-1, k=-1$

$\begin{array}{l}\therefore \frac{dY}{dX}=\frac{2X-Y}{X-2Y}\text{ Put }Y=vX\\ \therefore \frac{dY}{dX}=v+X\frac{dv}{dX}\end{array}$

$\begin{array}{l} v+X\frac{dv}{dX}=\frac{2X-vX}{X-2vX}=\frac{2-v}{1-2v}\\ X\frac{dv}{dX}=\frac{2-2v+2v^2}{1-2v}=\frac{2\left(v^2-v+1\right)}{1-2v}\\ \therefore \frac{dX}{X}=\frac{(1-2v)}{2\left(v^2-v+1\right)}dv\end{array}$

Put $v^2-v+1=t$

$\begin{array}{l}\Rightarrow (2v-1)dv=dt\\ \therefore \frac{dX}{X}=-\frac{dt}{2t}\\ \therefore \log X=\log t^{-1/2}+\log c\\ \therefore X=t^{-1/2}c\\ \Rightarrow X={\left(v^2-v+1\right)}^{-1/2}\cdot c\end{array}$

$X^2\left(v^2-v+1\right)=$Constant

$(x+1{)}^2\left(\frac{(y+1{)}^2}{(x+1{)}^2}-\frac{(y+1)}{x+1}+1\right)=$Constant

$\begin{array}{l}(y+1{)}^2-(y+1\left)\right(x+1)+(x+1{)}^2=c\\ y^2+x^2-xy+x+y=c.\end{array}$