The solution curve of the differential equation y d x d y = x log e x - log e y + 1 , x > 0 , y > 0…

Given: $y\frac{dx}{dy}=x\left({\log }_{e}x-{\log }_{e}y+1\right), x>0, y>0$

$\Rightarrow \frac{dx}{dy}=\left(\frac{x}{y}\right)\left[{\log }_e\left(\frac{x}{y}\right)+1\right]$

Putting, $x=vy$

$\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dx}$

$\Rightarrow v+y\frac{dv}{dy}=\left(\frac{vy}{y}\right)\left[{\log }_e\left(\frac{vy}{y}\right)+1\right]$

$\Rightarrow v+y\frac{dv}{dy}=v\left[{\log }_e\left(v\right)+1\right]$

$\Rightarrow v+y\frac{dv}{dy}=v{\log }_e\left(v\right)+v$

$\Rightarrow y\frac{dv}{dy}=v{\log }_e\left(v\right)$

$\Rightarrow \frac{dv}{v{\log }_{e}v}=\frac{dy}{y}$

$\Rightarrow \int \frac{dv}{v{\log }_{e}v}=\int \frac{dy}{y}$

Putting, ${\log }_{e}v=t$

$\Rightarrow \frac{dv}{v}=dt$

$\Rightarrow \int \frac{dt}{t}=\int \frac{dy}{y}$

$\Rightarrow \log t=\log y+c$

$\Rightarrow \log \left({\log }_e\frac{x}{y}\right)=\log y+c$

Using point $\left(e,1\right)$

$\Rightarrow \log \left({\log }_{e}e\right)=\log \left(1\right)+c$

$\Rightarrow c=0$

$\Rightarrow \log \left({\log }_e\frac{x}{y}\right)=\log y$

$\Rightarrow \left|{\log }_e\frac{x}{y}\right|=y$