The particular solution of the differential equation \(y(1+\log x)=\left(\log x^x\right) \frac{d y}{d x}\), when \(y(e)=e^2\) is

\(
\begin{aligned}
& y(1+\log x)=\left(\log x^x\right) \frac{d y}{d x} \\
& \therefore \int \frac{(1+\log x)}{\left(\log x^x\right)}=\int \frac{d y}{y}
\end{aligned}
\)
We know that \(\frac{d}{d x}\left(\log x^x\right)=(1+\log x) d x\)
\(
\begin{aligned}
& \therefore \log \left|\log \mathrm{x}^{\mathrm{x}}\right|=\log |\mathrm{y}|+\log \mathrm{c} \\
& \text { We have } \mathrm{y}(\mathrm{e})=\mathrm{e}^2 \\
& \therefore \log \left|\log \mathrm{e}^{\mathrm{e}}\right|=\log \left|\mathrm{e}^2\right|+\log \mathrm{c} \\
& \therefore \log |\mathrm{e} \log \mathrm{e}|=2 \log |\mathrm{e}|+\log \mathrm{c} \\
& \therefore 1=2+\log \mathrm{c} \Rightarrow \log \mathrm{c}=-1 \\
& \therefore \log \left|\log \mathrm{x}^{\mathrm{x}}\right|=\log |\mathrm{y}|-\log \mathrm{e} \\
& \therefore \log |\mathrm{x} \log \mathrm{x}|+\log \mathrm{e}=\log |\mathrm{y}| \\
& \therefore \mathrm{ex} \log \mathrm{x}-\mathrm{y}=0
\end{aligned}
\)