Mathematics | TS EAMCET 2019 | Discussion Page

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

Option D

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

Discussion Forum