Let y=y(x) be the solution of the differential equation \[x{\log _e}x{{dy} \over {dx}} + y = {x^2}{\log _e}x,(x > 1)\]. If y(2)=2, then y(e) is equal to
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Official solution
\[x\ln x{{dy} \over {dx}} + y = {x^2}\ln x\]
\[{{dy} \over {dx}} + {1 \over {x\ln x}}\,.\,y = x\]
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),
\[y = {{{e^2}} \over 4} + 1 = {{4 + {e^2}} \over 4}\]