Given information:

The differential equation is x3+3y−xy′=0 .

Explanation:

The given differential equation can be written as,

  xdydx−3y=x3dydx−3xy=x3xdydx−3xy=x2

The above equation is a linear first order differential equation and its solution is calculated as,

  ye−3∫ 1 x dx=∫ x 2 e − ∫ 3 x dx dxye−3lnx=∫ x 2 e −3lnxdxyelnx −3=∫ x 2 e ln x −3 dxyx−3=∫ x 2 x −3dx

Simplify the above equation.

  yx−3=∫ 1 xdxyx−3=C+lnxy=x3(C+lnx)

Conclusion:

Thus, the solution of the differential equation x3+3y−xy′=0 is y=x3(C+lnx) .