Let x = x(y) be the solution of the differential equation 2ye x y ^...

Given,

$2ye\dfrac{x}{{y}^{2}}dx + \left({y}^{2} - 4xe\dfrac{x}{{y}^{2}}\right)dy = 0$

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

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

Divide by ${y}^{3}$

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

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

Now integrating both side we get,

$∫ 2e\dfrac{x}{{y}^{2}}d$\left(\dfrac{x}{{y}^{2}}\right) + ∫ \dfrac{1}{y}dy = 0$</p><p>$2e\dfrac{x}{{y}^{2}} + \ln y + c = 0$</p><p>Given, \left(0, 1\right)$ lies on it,

So, $2{e}^{0} + \ln 1 + c = 0 ⇒ c = - 2$

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

For $x\left(e\right)$

$2e\dfrac{x}{{e}^{2}} + lne - 2 = 0 ⇒ x = - {e}^{2}\log_{e}2$