Let x = x y be the solution of the differential equation 2 y e x y 2 d x + y 2 - 4 x e x y 2 d y = 0 such…

Given,

$2y{e}^{\frac{x}{y^2}}dx+\left(y^2-4x{e}^{\frac{x}{y^2}}\right)dy=0$

$2e^{\frac{x}{y^2}}\left[ydx-2xdy\right]+y^{2}dy=0$

$2e^{\frac{x}{y^2}}\left[\frac{y^{2}dx-x·\left(2y\right)dy}{y}\right]+y^{2}dy=0$

Divide by $y^3$

$2e^{\frac{x}{y^2}}\left[\frac{y^{2}dx-x·\left(2y\right)dy}{y^4}\right]+\frac{1}{y}dy=0$

$2e^{\frac{x}{y^2}}d\left(\frac{x}{y^2}\right)+\frac{1}{y}dy=0$

Now integrating both side we get,

$\int 2e^{\frac{x}{y^2}}d\left(\frac{x}{y^2}\right)+\int \frac{1}{y}dy=0$

$2e^{\frac{x}{y^2}}+\ln y+c=0$

Given, $\left(0,1\right)$ lies on it, 

So, $2{\mathrm{e}}^0+l\mathrm{n}1+\mathrm{c}=0\Rightarrow \mathrm{c}=-2$ 

Hence required curve: $2e^{\frac{x}{y^2}}+lny-2=0$

For $x\left(e\right)$

$2e^{\frac{x}{e^2}}+\ln e-2=0 \Rightarrow x=-e^2{\log }_{e}2$