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Q.
Let $x=x(y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x+\left(y^{2}-4 x e^{x / y^{2}}\right) d y=0$ such that $x(1)=0$. Then, $x(e)$ is equal to
$2 y e^{x / y} d x+\left(y^{2}-4 x e^{x / y^{2}}\right) d y=0$
$2 e^{x / y^{2}}[y d x-2 x d y]+y^{2} d y=0 $
$2 e^{x / y^{2}}\left[\frac{y^{2} d x-x \cdot(2 y) d y}{y}\right]+y^{2} d y=0$
Divide by $y ^{3}$
$2 e^{x / y^{2}}\left[\frac{y^{2} d x-x \cdot(2 y) d y}{y^{4}}\right]+\frac{1}{y} d y=0$
$2 e^{x / y^{2}} d\left(\frac{x}{y^{2}}\right)+\frac{1}{y} d y=0$
Integrating
$\int 2 e^{x / y^{2}} d\left(\frac{x}{y^{2}}\right)+\int \frac{1}{y} d y=0$
$2 e ^{ x / y ^{2}}+\ln y + c =0$
$(0,1)$ lies on it.
$2 e ^{0}+\ell n 1+ c =0$
$ \Rightarrow c =-2$
Required curve : $2 e^{x / y^{2}}+\ell$ ny $-2=0$
For $x ( e )$
$2 e ^{ x / e ^{2}}+\ell \text { ne }-2=0 $
$\Rightarrow x =- e ^{2} \log _{ e } 2$