Solution of the equation xdy−y+xy3(1+log⁡x)dx=0 is

We have xdy−ydx=xy3(1+log⁡x)dx⇒−ydx−xdyy2=xy(1+log⁡x)dx Observing the complete differentials, we get −dxy=xy(1+log⁡x)dx⇒−xydxy=x2(1+log⁡x)dx−xy2=(1+log⁡x)x33−∫x33⋅1xdx⇒−x22y2=x33(1+log⁡x)−x39+c2⇒−x2y2=2x3323+log⁡x+c