rearranging,
dy/dx=(-x-y-1)/(-x+y-3)=(x+y+1)/(x-y+3)
This is to find what substitution we need:
let a + b +1 = 0
a - b +3 =0
solving you get a= -2, b=1
now let x=z-2, y=w+1
dy/dx=(z-2+w+1+1)/(z-2-w-1+3)=(z+w)/(z-w)
dy/dx=dw/dx=(dw/dx)*(dz/dx)=dw/dz
so dw/dz=(z+w)/(z-w)
Now let w=vz
dw/dz=v+ (dv/dz)*z
Thus DE becomes
v+ (dv/dz)*z=(1+v)/(1-v)
Solving this, you get tan-1(w/z)=1/2 ln(z^2 + w^2) + c
so answer: tan-1[(y-1)/(x+2)]=1/2ln[(x+2)^2 + (y-1)^2] +c
