QJEEM23Ma74f5: JEE Main 2023 Mathematics - Class 12 JEEM23Ma74f5 Solution: Let be the solution of the differential equation . If , then is …

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

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

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),

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