Let $\mathrm{y}\left(\mathrm{x}\right)$ be the solution of the differential equation $\left(\mathrm{x} \log \mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=2\mathrm{x} \mathrm{logx}, (\mathrm{x} \ge 1).$ Then $\mathrm{y}\left(\mathrm{e}\right)$ is equal to :

$\begin{array}{l}Given\frac{dy}{dx}+\frac{1}{x\log x}y=2\\ I.F=e^{\int \frac{1}{x\log x}dx}=e{}^{\log \left(\log x\right)}=\log x\\ Solutionis\\ y(I.F)=\int 2(I.F)dx\\ \Rightarrow y\left(\log x\right)=2\int \log xdx\\ =2(\log x.x-\int \frac{1}{x}.xdx)\\ =2[x\log x-x]+c\\ Putx=1\end{array}$

$\begin{array}{l}\Rightarrow y\left(1\right)\log 1=2[1\log \left(1\right)-1]+c\\ \Rightarrow c=2\\ y\log x=2[x\log x-x]+2\\ putx=e\\ y\left(e\right)\log e=2[e\log e-e]+2\\ y\left(e\right)=2\end{array}$