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Question

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

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

Solution

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