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

Step-by-step solution

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 &#82