Let y(x) be the solution of the differential equation (x log x) + y = 2x log x, (x 1). Then y(e) is equal to
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Integrating factor = ℓn x
General solution is
y (ℓn x) = 2(x ℓn x – x) + c
Put x = 1, we get c = 2
y(e) = 2