$x \log x \frac{d y}{d x}+y=2 x \log x \Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=2$
$$
\begin{aligned}
& \mathrm{IF}=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x \\
& y(\log x)=\int(2)(\log x) \cdot d x+C \\
& \Rightarrow \quad y \log x=2 \int \log x d x+C \\
& \Rightarrow \quad y \log x=2[x \log x-x]+C
\end{aligned}
$$
Putting $x=1$, we get
$$
y \cdot 0=2[1 \times 0-1]+C \Rightarrow C=2
$$
Putting $C=2$ in Eq. (i), we get
$$
y \log x=2[x \log x-x]+2
$$
At
$$
\begin{aligned}
x & =e \Rightarrow y \cdot 1=2(e \times 1-e)+2 \\
y & =0+2=2 \\
y(e) & =2
\end{aligned}
$$