Let y = y x be the solution curve of the differential equation sin 2 x 2 log e tan x 2 d y + 4 x y - 4 2 x sin x 2 - π 4 d x = 0 , 0 < x < π 2 , which passes through the point π 6 , 1 . Then y π 3 is equal to _______.

Step-by-step solution

Given differential equation, sin 2 x 2 ln tan x 2 d y + 4 x y - 4 2 x sin x 2 - π 4 d x = 0 ⇒ ln tan x 2 d y + 4 x y d x sin 2 x 2 - 4 2 x sin x 2 - π 4 sin 2 x 2 d x = 0 ⇒ d y · ln tan x 2 - 4 2 x sin x 2 - cos x 2 2 2 sin x 2 cos x 2 d x = 0 ⇒ d y ln tan x 2 - 4 x sin x 2 - cos x 2 sin x 2 + cos 2 x 2 - 1 d x = 0 Now integrating both side, ⇒ ∫ d y ln tan x 2 - ∫ 4 x sin x 2 - cos x 2 sin x 2 + cos 2 x 2 - 1 d x = ∫ 0 Now let sin x 2 + cos 2 x = t ⇒ -