Answer
$x^2y'=xy-y^2$ and $e$
Work Step by Step
Need to use quotient rule of differentiation.
This implies, $y'=\dfrac{ \ln x -x(\dfrac{1}{x})}{(\ln x)^2}=(\dfrac{1}{\ln x}) -\dfrac{1}{(\ln x)^2}$
As we are given that $y=\dfrac{ x}{\ln x}$
Thus, $x^2y'=(\dfrac{x^2}{\ln x}) -(\dfrac{x^2}{(\ln x)^2})$
and $x^2y'=xy-y^2$
Now, apply the initial conditions.
$y(e)=\dfrac{e}{\ln e}=e$
