Answer:
Option B
Explanation:
We have $(x+2y^{3})\frac{dy}{dx}=y$
$\Rightarrow$ $y \frac{dy}{dX}=x+2y^{3} $
$\Rightarrow$ $\frac{dx}{dy}-\frac{x}{y}=2y^{2}$
It is linear differential equation of the firm $\frac{dx}{dy}+Px=Q$
$\Rightarrow$ $ P= -\frac{1}{y}. Q= 2y^{2}$
$\therefore$ $IF= e^{\int P dy}=e^{\int \frac{1}{y} dy}=e^{-\log y}=\frac{1}{y}$
Hence, required solution is
x.(IF)= $\int (IF.Q) dy$
$\Rightarrow$ $\frac{x}{y}=\int\frac{2y^{2}}{y}dy=y^{2}+c$
$\Rightarrow$ $ x=y^{3}+cy$