Let y ( x ) be the solution of the differential equation ( x log x ) d y d x + y = 2 x log x ,   ( x…

Given, $\left(x\log x\right)\frac{dy}{dx}+y=2x\log x$

$\Rightarrow \frac{dy}{dx}+\left(\frac{1}{x\log x}\right)y=2$
Integrating factor $I.F=e^{\int Pdx}$$=e^{\int \frac{\mathit{1}}{x\log \mathit{x}}dx}=e^{\int \frac{\mathit{1}\mathit{/}x}{\log \mathit{x}}dx}=e^{\log \left(\log \mathit{x}\right)}$$=\log x$

The solution of the equation becomes 

$y·e^{\int Pdx}=\int Q·e^{\int Pdx}dx+c$, where, $c$ is the constant of integration.

$\Rightarrow y\mathrm{log\; x}=\int 2\mathrm{log\; x}dx+c$

$\Rightarrow y\mathrm{log\; x}=2\left(\mathrm{x\; log\; x}-x\right)+c$

Let, $P\left(1,y_1\right)$ be any point on the curve.

$\Rightarrow y_1·\left(0\right)=2\left(0-1\right)+c$$\Rightarrow c=2$

$\Rightarrow y\log x=2\left(x\log x-x\right)+2$

Now, putting $x=e$, we get, $y=2$.