The solution of the differential equation xdy = tan ⁡ y + e 1 / x 2 x sec ⁡ y dx is (where C is the constant of integration)

Question

The solution of the differential equation xdy = tan ⁡ y + e 1 / x 2 x sec ⁡ y dx is (where C is the constant of integration), sin ⁡ y = e 1 x 2 + C, 2 sin ⁡ y x + e 1 x 2 = C, sin ⁡ y x - e 1 x 2 = C, sin ⁡ y - x e 1 x 2 = C

MARKS App · Accepted Answer

d y d x = tan ⁡ y x + e 1 x 2 x 2 sec ⁡ y cos ⁡ y d y d x - sin ⁡ y x = e 1 x 2 x 2 Let, sin ⁡ y = t cos ⁡ y d y d x = d t d x ⇒ d t d x - t x = e 1 x 2 x 2 I.F. = e - ∫ 1 x d x = e ln ⁡ 1 x = 1 x ⇒ t x = ∫ e 1 x 2 x 3 d x ⇒ t x = - 1 2 ∫ - 2 x 3 e 1 x 2 d x ⇒ sin ⁡ y x = - 1 2 e 1 x 2 + C

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