If the solution curve $f(x, y)=0$ of the differential equation

$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$,

passes through the points $(1,0)$ and $(\alpha, 2)$, then $\alpha^{\alpha}$ is equal to :

We have, $\left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^{y}, x>0$ <br/><br/>Put $x \log _e x=t$ <br/><br/>$$ \begin{aligned} & \Rightarrow \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \Rightarrow \left(1+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \therefore \frac{d t}{d y}-t =e^{y} \end{aligned} $$ <br/><br/>Now, IF $=e^{\int-d y}=e^{-y}$ <br/><br/>$\begin{aligned} & \therefore \text { General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\\\ & \Rightarrow t e^{-y}=\int d y+c \\\\ & \Rightarrow t e^{-y}=y+c \\\\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}$ <br/><br/>Equation (i), passes through the point $(1,0)$ <br/><br/>$$ \begin{aligned} & \therefore 0=0+c \\\\ & \Rightarrow c=0 \\\\ & \therefore \left(x \log _e x\right) e^{-y}=y \\\\ & \Rightarrow x \log _e x=y e^y \end{aligned} $$ <br/><br/>Which passes through $(\alpha, 2)$ <br/><br/>$$ \begin{aligned} & \therefore \alpha \log _e \alpha=2 e^2 \\\\ & \Rightarrow \log _e \alpha^\alpha=2 e^2 \\\\ & \Rightarrow \alpha^\alpha=e^{2 e^2} \end{aligned} $$