The particular solution of the differential equation y 1 + log ⁡ x d x d y - x log ⁡ x = 0 when y ( e ) = e 2 is

Question

The particular solution of the differential equation y 1 + log ⁡ x d x d y - x log ⁡ x = 0 when y ( e ) = e 2 is, y = e x log ⁡ x, e y = x log ⁡ x, x y = e log ⁡ x, y log ⁡ x = e x

MARKS App · Accepted Answer

y 1 + log ⁡ x d x d y - x log ⁡ x = 0 ⇒ 1 + log ⁡ x x log ⁡ x d x = d y y Integrating on both side ∫ 1 + log ⁡ x x log ⁡ x d x = ∫ d y y log ⁡ x log ⁡ x = log ⁡ y + log ⁡ C ⇒ log ⁡ x log ⁡ x = log ⁡ y . c ∴ x log ⁡ x = y . c ...... (i) As x = e , y = e 2 ∴ e = e 2 . c ∴ c = 1 e Putting c = 1 e in equation (i) we get x log ⁡ x = y e ⇒ y = e x log ⁡ x

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