Let $y(x)$ be the solution of the differential equation $(x \log x) \frac{d y}{d x}+y=2 x \log x,(x \geq 1)$. Then $y(e)$ is equal to

2

We have,

$(x \log x) \frac{d y}{d x}+y=2 x \log x$

$\Rightarrow \frac{d y}{d x}+\frac{1}{x \log x} y=2$                   .....(i)

This is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$, where $P=\frac{1}{x \log x}$ and $Q=2$

Integrating factor = $e^{\int P d x}=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x$

Multiplying both sides of (i) by integrating factor $=\log x$, we obtain

$\log x \frac{d y}{d x}+\frac{1}{x} y=2 \log x$

Integrating both sides with respect to $x$, we get

$y \log x =\int 2 \log x d x+C$

$\Rightarrow y \log x =2 x(\log x-1)+C$                  ......(ii)

It is given that $x \geq 1$. Putting $x=1$ in (ii), we obtain

$y \times 0=2(0-1)+C \Rightarrow C=2$

Putting $C=2$ in (ii), we get

$y \log x=2 x(\log x-1)+2$

Putting $x=e$, we get $y(e)=2$.