1)

 The solution of the differential equation  $(x+2y^{3})\frac{dy}{dx}=y$ is 


A) $x=y^{3}+c$

B) $x=y^{3}+cy$

C) $y=x^{3}+c$

D) $y=x^{3}+cx+d$

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$