If y = y x is the solution of the differential equation, d y d x + 2 y tan x = sin x , y π 3 = 0 , then the maximum value of the function y x over R is equal to :

Step-by-step solution

d y d x + 2 y tan x = sin x I . F . = e ∫ 2 tan x d x = e 2 ln sec x I . F . = sec 2 x y . sec 2 x = ∫ sin x . sec 2 x d x + C y . sec 2 x = ∫ sec x tan x d x + C y . sec 2 x = sec x + C x = π 3 ; y = 0 ⇒ C = - 2 ⇒ y = sec x - 2 sec 2 x = cos x - 2 cos 2 x Let cos x = t ,   - 1 ≤ t ≤ 1 ⇒ y = t - 2 t 2 ⇒ d y d t = 1 - 4 t = 0 ⇒ t = 1 4 Second-order derivative is negative ∴   max = 1 4 - 1 8 = 2 - 1 8 = 1 8