$x\log x\cfrac{{dy}}{{dx}} + y = \cfrac{2}{x}\log x$

: The given equation is $x\log x\cfrac{{dy}}{{dx}} + y = \cfrac{2}{x}\log x$

$\Rightarrow \cfrac{{dy}}{{dx}} + \cfrac{y}{{x\log x}} = \cfrac{2}{{{x^2}}}$

…(1)
which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \cfrac{{\rm{l}}}{{x\log x}}$ and $Q = \cfrac{2}{{{x^2}}}$

$\therefore {\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\int P dx}} = {e^{\int {\cfrac{1}{{x\log x}}} dx}} = {e^{{\rm{log}}(\log x)}} = \log x$ $\therefore$
The solution is $y \cdot ({\rm{I}}.{\rm{F}}.) = \int {Q({\rm{I}}.{\rm{F}}.)dx} + C$

$\Rightarrow y \cdot \log x = 2\int {\log } x \cdot \cfrac{1}{{{x^2}}}dx + C$

$\Rightarrow y \cdot \log x = 2\left[ {\log x \cdot \cfrac{{{x^{ - 1}}}}{{ - 1}} - \int {\cfrac{1}{x}} \cdot \cfrac{{{x^{ - 1}}}}{{ - 1}}dx} \right] + C$

(Integrating by parts)
$\Rightarrow y \cdot \log x = 2\left[ { - \cfrac{{\log x}}{x} + \int {{x^{ - 2}}dx} } \right] + C$

$\Rightarrow y \cdot \log x = 2\left[ { - \cfrac{{\log x}}{x} + \cfrac{{{x^{ - 1}}}}{{ - 1}}} \right] + C$

$\Rightarrow y \cdot \log x = \cfrac{{ - 2}}{x}\left( {1 + \log x} \right) + C$,

which is the required solution.