Question

Let $x=x(y)$ be the solution of the differential equation $2y{e}^{x/y^2}dx+\left(y^2-4x{e}^{x/y^2}\right)dy=0$ such that $x(1)=0$. Then, $x(e)$ is equal to

• A. ${\mathrm{elog}}_e(2)$
• B. $-e{\log }_e(2)$
• C. $e^2{\log }_e(2)$
• D. $-e^2{\log }_e(2)$

Answer 1

Answer: (D)

$2y{e}^{x/y}dx+\left(y^2-4x{e}^{x/y^2}\right)dy=0$
$2e^{x/y^2}[ydx-2xdy]+y^{2}dy=0$
$2e^{x/y^2}\left[\frac{y^{2}dx-x\cdot (2y)dy}{y}\right]+y^{2}dy=0$
Divide by $y^3$
$2e^{x/y^2}\left[\frac{y^{2}dx-x\cdot (2y)dy}{y^4}\right]+\frac{1}{y}dy=0$
$2e^{x/y^2}d\left(\frac{x}{y^2}\right)+\frac{1}{y}dy=0$
Integrating
$\int 2e^{x/y^2}d\left(\frac{x}{y^2}\right)+\int \frac{1}{y}dy=0$
$2e^{x/y^2}+\ln y+c=0$
$(0,1)$ lies on it.
$2e^0+\ell n1+c=0$
$\Rightarrow c=-2$
Required curve : $2e^{x/y^2}+\ell$ ny $-2=0$
For $x(e)$
$2e^{x/e^2}+\ell \text{ ne }-2=0$
$\Rightarrow x=-e^2{\log }_{e}2$