The solution of differential equation ydx+x+x2ydy=0 is

Given, ydx+x+x2ydy=0⇒ dydx=−yx+x2y⇒ dxdy=−x+x2yy⇒ dxdy+xy=−x2⇒−1x2⋅dxdy−1xy=1 Put 1x=z⇒ −1x2⋅dxdy=dzdy⇒ dzdy−zy=1which is a linear differential equation of the formdzdy+P(y)z=Q(y)where ,P=−1y and Q=1IF=e∫−1ydy=e−log⁡y=1yHence, the solution of given differential equation is zy=∫1ydy+C⇒  zy=log⁡y+C⇒  1xy=log⁡y+C ∵z=1x⇒  −1xy+log⁡y=C