Find the general integral of the equation y'=(x-y+1)/(x+y-3)

Find the general integral of the equation $y'=\dfrac{x-y+1}{x+y-3}$.

Step 1: Eliminate the constants by translation

Setting $x=X+h,\ y=Y+k$, the numerator and denominator become $X-Y+(h-k+1)$ and $X+Y+(h+k-3)$. Requiring both constants to vanish gives $h-k+1=0,\ h+k-3=0$, so $h=1,\ k=2$. The equation becomes

\[\frac{dY}{dX}=\frac{X-Y}{X+Y}.\]

Step 2: Solve the homogeneous equation

With $Y=VX$, $X\dfrac{dV}{dX}=\dfrac{1-V}{1+V}-V=\dfrac{1-2V-V^2}{1+V}$. Since $\frac{d}{dV}(V^2+2V-1)=2(V+1)$, separating and integrating gives

\[\frac{1+V}{V^2+2V-1}\,dV=-\frac{dX}{X} \quad\Longrightarrow\quad V^2+2V-1=\frac{K}{X^2}\quad\Longrightarrow\quad Y^2+2XY-X^2=K.\]

Step 3: Undo the translation

Substituting $X=x-1,\ Y=y-2$ and simplifying,

\[\boxed{\,y^2+2xy-x^2-6y-2x=C.}\]