Answer
$x^2y'=xy-y^2$ and $y(e)=e$
Work Step by Step
Apply the quotient rule of differentiation:
Thus, $y'=\dfrac{ \ln x -x(\dfrac{1}{x})}{(\ln x)^2}=\dfrac{1}{\ln x} -\dfrac{1}{(\ln x)^2}$
Since, we have $y=\dfrac{ x}{\ln x}$
Therefore, $x^2y'=\dfrac{x^2}{\ln x} -\dfrac{x^2}{(\ln x)^2}$
or, $x^2y'=xy-y^2$
Apply the initial conditions:
$y(e)=\dfrac{e}{\ln e}=e$