The solution of the differential equation \(\frac{dy}{dx}-\frac{y}{x}=x^3\) is

\(3y=x(x^3+c)\) \(y=x(x^3+3)\) \(t^2=x(x^3+c)\) \(y=x^2(x^3+c)\)

\(3y=x(x^3+c)\)

\(\frac{dy}{dx}-\frac{y}{x}=x^3\) so $\int \frac{1}{x}\frac{dy}{dx}-\frac{ydx}{x^2}=\int x^2dx$ $\frac{y}{x}=\frac{x^3}{3}+\frac{C}{3}⇒3y=x(x^3+c)$