The solution curve of the differential equation $y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0$ passing through the point $(e, 1)$ is

<p>$$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}}{\mathrm{y}}\left(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1\right)$$</p> <p>Let $\frac{x}{y}=t \Rightarrow x=t y$</p> <p>$$\begin{aligned} & \frac{d x}{d y}=t+y \frac{d t}{d y} \\ & t+y \frac{d t}{d y}=t(\ln (t)+1) \end{aligned}$$</p> <p>$$\mathrm{y} \frac{\mathrm{dt}}{\mathrm{dy}}=\mathrm{t} \ln (\mathrm{t}) \Rightarrow \frac{\mathrm{dt}}{\mathrm{t} \ln (\mathrm{t})}=\frac{\mathrm{dy}}{\mathrm{y}}$$</p> <p>$$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t} \cdot \ln (\mathrm{t})}=\int \frac{\mathrm{dy}}{\mathrm{y}}$$</p> <p>$\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \quad$ let $\ln t=p$</p> <p>$\frac{1}{\mathrm{t}} \mathrm{dt}=\mathrm{dp}$</p> <p>$$\begin{aligned} & \Rightarrow \ln p=\ln y+c \\ & \ln (\ln t)=\ln y+c \\ & \ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c \\ & \text { at } x=e, y=1 \\ & \ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0 \end{aligned}$$</p> <p>$$\begin{aligned} & \ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y \\ & \left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y} \\ & \left|\ln \left(\frac{x}{y}\right)\right|=y \end{aligned}$$</p>