Let y = y x be the solution of the differential equation, x d y d x + y = x log e x , x > 1 . If 2 y 2 = log e 4 - 1 , then y e is equal to

Step-by-step solution

The given differential equation can be written as d y d x + y x = log e x , which is a linear differential equation. Now, integrating factor I . F . = e ∫ 1 x d x = e l n x = x Hence, the solution of the given differential equation is y · x = ∫ x · log e x   d x ⇒ y · x = log e x ∫ x d x - ∫ 1 x ∫ x d x d x (using integration by parts) ⇒ y · x = log e x x 2 2 - ∫ 1 x · x 2 2 d x ⇒ y · x = log e x x 2 2 - ∫ x 2 d x ⇒