If x = x(y) is the solution of the differential equation y(dx)/(dy)...

Given

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

Finding integrating factor $I . F . = {e}^{∫ \dfrac{- 2}{y}dx} = {e}^{- 2\ln y} = \dfrac{1}{{y}^{2}}$

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

$⇒ \dfrac{x}{{y}^{2}} = y \cdot {e}^{y} + C$

$0 = e + C ⇒ C = - e$

$⇒ \dfrac{x}{{y}^{2}} = y{e}^{y} - e$

Putting $y = e$ in the equation we get

$⇒ \dfrac{x}{{e}^{2}} = e . {e}^{e} - e$

$⇒ x = {e}^{3}\left({e}^{e} - 1\right)$