If the solution curve f ( x ,   y ) = 0 of the differential equation 1 + log e x d x d y - x log e x =…

The given differential equation is

$\left(1+\ln \left(x\right)\right)\frac{dx}{dy}-x\ln \left(x\right)=e^y$

Put $x\ln \left(x\right)=t$

$\Rightarrow \left(1+\ln \left(x\right)\right)dx=dt$

$\Rightarrow \frac{\mathrm{dt}}{\mathrm{dy}}-t=e^y$ which is a linear differential equation.

$\mathrm{I}.\mathrm{F}=e^{-\int dy}$

$=e^{-y}$

The solution of the differential equation is

$\Rightarrow t\times e^{-y}=\int e^y\times e^{-y}dy+c$

$\Rightarrow t\times e^{-y}=y+c$

$\Rightarrow x\ln \left(x\right)=y e^y+c e^y$

Put $x=1, y=0$

$\Rightarrow c=0$

Put $x=a, y=2$

$a\ln \left(a\right)=2e^2$

$\therefore a^a=e^{2e^2}$

Hence this is the correct option.