The solution of the D.Exx2+1dydx=y1−x2+x3log⁡x is

dydx+x2−1xx2+1y=x2log⁡xx2+1I.F.=e∫x2−1xx2+1dx=e∫2xx2+1−1x dx=elog⁡x2+1x=x2+1x Solution of the D.E yx2+1x=∫xlog⁡xdxyx2+1x=logx ∫xdx -∫1x∫xdx dxyx2+1x=x22log⁡x-∫1x·x22dxyx2+1x=x22log⁡x-x24+c