If d y d x + 2 y tan x = sin x , 0 < x < π 2 and y π 3 = 0 , then the maximum value of y x is

Step-by-step solution

We know that, d y d x + y 2 tan x = sin x is a linear differential equation so, I . F = e ∫ 2 tan x d x = e 2 ln sec x = sec 2 x The general solution will be y sec 2 x = ∫ sin x sec 2 x d x + C ⇒ y sec 2 x = sec x + C ∵ y π 3 = 0 ⇒ C = - 2 Hence the particular solution is y sec 2 x = sec x - 2 ⇒ y = cos x - 2 cos 2 x ⇒ y = 1 8 - 2 cos x - 1 4 2 So, the maximum value of y x is 1 8