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

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

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

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

Where $P = \cfrac{2}{x}$ and $Q = x\log x$

$\therefore$ $I.F. = {e^{\int {\cfrac{2}{x}dx} }} = {e^{2\log x}} = {e^{\log {x^2}}} = {x^2}$

$\therefore$ $y \cdot \left( {{\rm{I}}{\rm{.F}}{\rm{.}}} \right) = \int {Q \cdot \left( {{\rm{I}}{\rm{.F}}{\rm{.}}} \right) + C}$

$\Rightarrow$ $y \cdot {x^2} = \int {\log x \cdot {x^3}dx} + C = \log x \cdot \cfrac{{{x^4}}}{4} - \int {\cfrac{1}{x} \cdot \cfrac{{{x^4}}}{4}dx} + C$

[Integrating by parts]

$= \cfrac{{{x^4}}}{4}\log x - \cfrac{1}{4}\int {{x^3}} dx + C = \cfrac{{{x^4}}}{4}\log x - \cfrac{1}{4} \cdot \cfrac{{{x^4}}}{4} + C$

$\Rightarrow y = \cfrac{1}{4}{x^2}\log x - \cfrac{1}{{16}}{x^2} + C{x^{ - 2}} \Rightarrow y = \cfrac{{{x^2}}}{{16}}(4\log x - 1) + C{x^{ - 2}}$

which is the required solution.