The solution of the differential equation  dydx=x+y2x+2y−2 is

dydx=(x+y)2(x+2)(y−2)=[(x+2)+(y−2)]2(x+2)(y−2) put X=x+2 and Y=y+2dYdX=(X+Y)2XYput Y=vXv+XdvdX=v2+2v+1v=v+2v+1v⇒Xdvdx=2v+1v⇒v2v+1dv=dxX⇒∫1−11+2vdv=2∫dxXv−12ln⁡|1+2v|=2ln⁡X+ln⁡c2v-ln1+2v=4ln X+2lnc2v=ln⁡c2X4(1+2v)2YX=ln⁡c2X41+2YXe2YX=c2X3(X+2Y)⇒e2(y−2)x+2=c2(x+2)3(x+2+2y−4)=c2(x+2)3(x+2y-2)