If x = x y is the solution of the differential equation y d x   d y = 2 x + y 3 y + 1 e y , x 1 = 0 ;…

Given

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

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

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

$\Rightarrow \frac{x}{y^2}=y·e^y+C$

$0=e+C\Rightarrow C=-e$

$\Rightarrow \frac{x}{y^2}=y{e}^y-e$

Putting $y=e$ in the equation we get

$\Rightarrow \frac{x}{e^2}=e.e^e-e$

  $\Rightarrow x=e^3\left(e^e-1\right)$