Let y ( x ) be the solution of the differential equation 2 x 2 d y + e y - 2 x d x = 0 , x > 0 . If y ( e ) = 1 , then y ( 1 ) is equal to:

Step-by-step solution

d y d x = - e y 2 x 2 + 1 x e - y d y d x = e - y x + - 1 2 x 2 - e - y d y d x + e - y x = 1 2 x 2 Let e - y = t   . . . i e - y - 1 d y d x = d t d x d t d x + t x = 1 2 x 2 I . F = e ∫ 1 x d x = e l n x = x t x = ∫ 1 2 x 2 · x d x + C Using equation 1 e - y x = 1 2 ℓ n x + C Given, y e = 1 ⇒ e - 1 e = 1 2 + C ⇒ C = 1 2 e - y x = 1 2 ( 1 + ℓ n x ) Put x = 1 then y is ⇒ y = ℓ n 2   or   log e 2