Find the general integral of the equation…

Find the general integral of the equation $y'=2\left(\dfrac{y+2}{x+y-1}\right)^2$.

Step 1: Eliminate the constants

Requiring $k+2=0$ and $h+k-1=0$ gives $k=-2,\ h=3$, so

\[\frac{dY}{dX}=2\left(\frac{Y}{X+Y}\right)^2.\]

Step 2: Solve the homogeneous equation

With $Y=VX$, $X\dfrac{dV}{dX}=\dfrac{2V^2}{(1+V)^2}-V=-\dfrac{V(1+V^2)}{(1+V)^2}$, and expanding $\dfrac{(1+V)^2}{V(1+V^2)}=\dfrac1V+\dfrac{2}{1+V^2}$ (the $\pm V/(1+V^2)$ terms cancel), integration gives

\[\ln|V|+2\arctan V=-\ln|X|+C \quad\Longrightarrow\quad \ln|VX|+2\arctan V=C \quad\Longrightarrow\quad \ln|Y|+2\arctan\frac{Y}{X}=C.\]

Step 3: Undo the translation

With $X=x-3,\ Y=y+2$:

\[\boxed{\,\ln|y+2|+2\arctan\frac{y+2}{x-3}=C.}\]