The solution of the differential equation dydx=(1+y)2x(1+y)−x2 is given as ln⁡(1+y)=

consider dxdy=x1+y−x2(1+y)2⇒−1x2dydx+1x(1+y)=1(1+y)2 Substituting 1x=t, we get dtdy+t1+y=1(1+y)2 ;  Which is linear I.F =e∫11+ydy=(1+y)⇒ solution of differential equation is t1+y=∫11+ydy=c  ⇒  c+1+yx=ln1+y