If c is an arbitrary constant, the general solution of the differenti…

\((x+y^3)\frac{dy}{dx}=y\)

$⇒\frac{(x+y^3)}{y}=\frac{dx}{dy}$

so $\frac{dx}{dy}-\frac{x}{y}=y^2$

so $I.F.=e^{\int -\frac{1}{y}dy}=e^{ln|\frac{1}{y}|}=\frac{1}{y}$

so $\int\frac{1}{y}\frac{dx}{dy}-\frac{1}{y^2}xdy=\int ydy$

$\frac{x}{y}=\frac{y^2}{2}+C$

$x=\frac{y^3}{2}+\frac{2cy}{2}$