The solution of the differential equation d y d x + y 2 sec ⁡ x = tan ⁡ x 2 y , where 0 ≤ x π 2 and y 0 = 1 , is given by

Question

The solution of the differential equation d y d x + y 2 sec ⁡ x = tan ⁡ x 2 y , where 0 ≤ x π 2 and y 0 = 1 , is given by, y 2 = 1 + x sec ⁡ x + tan ⁡ x, y = 1 + x sec ⁡ x + tan ⁡ x, y = 1 - x sec ⁡ x + tan ⁡ x, y 2 = 1 - x sec ⁡ x + tan ⁡ x

MARKS App · Accepted Answer

d y d x + y 2 sec ⁡ x = tan ⁡ x 2 y 2 y d y d x + y 2 sec ⁡ x = tan ⁡ x Put y 2 = t ⇒ 2 y d y d x = d t d x d t d x + t sec ⁡ x = tan ⁡ x I . F = e ∫ sec ⁡ x d x = e ln ( sec ⁡ x + t a n x ) = sec ⁡ x + tan ⁡ x ⇒ t sec ⁡ x + tan ⁡ x = ∫ sec ⁡ x + tan ⁡ x tan ⁡ x d x ⇒ t sec ⁡ x + tan ⁡ x = ∫ sec ⁡ x tan ⁡ x d x + ∫ tan 2 ⁡ x d x ⇒ y 2 sec ⁡ x + tan ⁡ x = sec ⁡ x + tan ⁡ x - x + c y 0 = 1 ⇒ c = 0 ⇒ y 2 = 1 - x sec ⁡ x + tan ⁡ x

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