x2dy=(2xy+y2) dx
dydx=2xy+y2x2=2yx+(yx)2
which is homogeneous ....(i)
Put y=vx
⇒ v+xdvdx=2vx2+v2x2x2=2v+v2
⇒ xdvdx=v+v2
⇒ 1v(v+1)dv=1xdx
⇒ 1v−1v+1dv=1xdx
log (v)−log(v+1)=log x+log c
log∣∣vv+1∣∣=log cx
yy+x=cx ...(ii)
When y=1, x=1 then
11+1=c
⇒ c=12
Put the value of c in eq. (ii)
yy+x=12x
⇒ 2y=xy+x2
⇒ x2+xy−2y=0