Let y = y x be the solution of the differential equation d y d x = 2 x x + y 3 − x x + y − 1 , y 0 = 1 . Then, 1 2 + y 1 2 2 equals:

Step-by-step solution

Given: d y d x = 2 x x + y 3 − x x + y − 1 Let, x + y = t ⇒ 1 + d y d x = d t d x ⇒ d y d x = d t d x - 1 ⇒ d t d x − 1 = 2 x t 3 − x t − 1 ⇒ d t d x = 2 x t 3 − x t ⇒ d t d x + x t = 2 x t 3 ⇒ 1 t 3 d t d x + x t 2 = 2 x . . . i Putting, 1 t 2 = u ⇒ - 2 t 3 d t d x = d u d x ⇒ 1 t 3 d t d x = - 1 2 d u d x Putting this value in equation i ⇒ - 1 2 d u d x + x u = 2 x ⇒ d u d x - 2 x u = - 4 x Integrating factor, I F = e ∫ - 2 x d x ⇒ I F = e - x 2 So, solution of the given differential equation is, u × e - x 2 = ∫