$$x\ln x{{dy} \over {dx}} + y = {x^2}\ln x$$

$${{dy} \over {dx}} + {1 \over {x\ln x}}\,.\,y = x$$

If $$ = {e^{\int {{1 \over {x\ln x}}dx} }} = {e^{\int {{1 \over t}dt} }}$$, where $$t = \ln x$$

$$ = {e^{\ln t}} = t = \ln x$$

$$y.\ln x = \int {x\ln x = {{{x^2}} \over 2}\ln x - \int {{{{x^2}} \over 2}\,.\,{1 \over x}} } $$

$$y\ln x = {{{x^2}} \over 2}\ln x - {{{x^2}} \over 4} + C$$ ..... (i)

$$y(2) = 2 \Rightarrow C = 1$$

Putting $$x = e$$ in (i),

$$y = {{{e^2}} \over 4} + 1 = {{4 + {e^2}} \over 4}$$