Solution of differential equation dzdx+zxlog⁡z=zx2(log⁡z)2 is

Dividing the complete equation by  zlogz2  the transformed equation   obtained is1z(log⁡z)2dzdx+1xlog⁡z=1x2; Now substituting 1log⁡z=y⇒−12(log⁡z)2dzdx=dydx………..(i) Then (i) reduced to linear differential equation as below −dydx+1xy=1x2⇒dydx−1xy=−1x2 and I.F=e−1∫1xdx=1x Its solution is given by yx=−∫1x3dx+c⇒yx=12x2+c⇒yx=12x2+c⇒xlog⁡z=12+cx2 is the required general solution