Let y = y ( x ) be solution of the following differential equation e y d y d x - 2 e y sin x + sin x cos 2 x = 0 , y π 2 = 0 . If y 0 = log e α + β e - 2 , then 4 ( α + β ) is equal to .

Step-by-step solution

Let e y = t ⇒ e y d y d x = d t d x So, e y d y d x - 2 e y sin x + sin x cos 2 x = 0 ⇒ d t d x - 2 sin x t = - sin x cos 2 x So, the integrating factor I . F . = e ∫ - 2 sin x d x = e 2 cos x So, the solution of differential equation is t · e 2 cos x = ∫ e 2 cos x · - sin x cos 2 x d x ⇒ e y · e 2 cos x = ∫ e 2 z · z 2   d z (Assume, z = cos x ⇒ d z = - sin x ) Using integration by parts, ⇒ e y · e 2 cos x = 1 2 · z 2 · e 2 z -