Let y = y x be the solution of the differential equation cosec 2 x d y + 2 d x = 1 + y cos 2 x cosec 2 x d x , with y π 4 = 0 . Then, the value of y 0 + 1 2 is equal to:

Step-by-step solution

Given, cosec 2 x d y + 2 d x = 1 + y cos 2 x cosec 2 x d x ⇒ d y sin 2 x + 2 d x = 1 + y cos 2 x sin 2 x d x ⇒ d y d x + 2 sin 2 x = 1 + y cos 2 x ⇒ d y d x + 2 sin 2 x - 1 = y cos 2 x Using, cos 2 x = 1 - 2 sin 2 x , we get d y d x + - cos 2 x = y cos 2 x ⇒ d y d x + - cos 2 x y = cos 2 x This is a linear differential equation of the type d y d x + P y = Q , where P   &   Q are the functions of x or constants. Thus, P = - cos 2 x   &   Q = cos 2 x Now, we have i