For differential equation y e^x/y dx = ≤ft(x e^x/y+y^2)dy, y(0) = 1,…

$y e^{\frac{x}{y}} dx = \left[x e^{\frac{x}{y}}+y^2\right]dy$

so  $\frac{dx}{dy} = \left[\frac{x}{y} + \frac{y}{e^{\frac{x}{y}}} \right]$

so let x = vy

$v + \frac{y ~dv}{dy} = v + \frac{y}{e^v}$

⇒ $\frac{y dv}{dy} = \frac{y}{e^v}$

so $\int e^v dv = \int dy$  (integrating both sides)

so e^v = y + C

so  $e^{\frac{x}{y}} = y + C$

so  x = y log(y + C)

at x = 0   y = 1

so 0 = 1 log(1 + C)

⇒ log(1 + C) = 0

⇒ log(1 + C) = log 1

⇒ C = 0

so $x = y \log y$

at  $x = e, y = e$