Discussion Forum



1)

The solution of the differential equation  $ydx-xdy+3x^{2}y^{2}e^{x^{3}}dx=0$ satisfying y=1 when x=1, is 


A) $y\left( e^{x^{3}}-(1+2e)\right)-x=0$

B) $y\left( e^{x^{3}}+(1-e)\right)+x=0$

C) $y\left( e^{x^{3}}+(1+e)\right)-x=0$

D) $y\left( e^{x^{3}}-(1+e)\right)+x=0$

Answer:

Option D

Explanation:

 We have,

$ydx-xdy+3x^{2}y^{2}e^{x^{3}}dx=0$

 

$\frac{ydx-xdy}{y^{2}}+3x^{2}e^{x^{3}}dx=0$

$\Rightarrow$     $d\left(\frac{x}{y}\right)+d(e^{x^{3}})=0$

 On integrating  , we get

 $\frac{x}{y}+e^{x^{3}}$=c

 putting x=y=1, we get

                         c=1+e

$\therefore$       $\frac{x}{y}+e^{x^{3}}$ =1+e

 x+$y\left( e^{x^{3}}-(1+e)\right)+x=0$