The solution of the differential equation d y d x + x 2 x + y = x 3 2 x + y 3 - 2 is ( C being an arbitrary constant)

Step-by-step solution

Let, 2 x + y = t ⇒ d y d x + 2 = d t d x d t d x + x t = x 3 t 3 ⇒ 1 t 3 d t d x + 1 t 2 x = x 3 Let, 1 t 2 = u ⇒ - 2 t 3 d t d x = d u d x d u d x + - 2 x u = - 2 x 3 I.F. = e - ∫ 2 x d x = e - x 2 ⇒ u . e - x 2 = ∫ e - x 2 - 2 x 3 d x e - x 2 2 x + y 2 = - 2 ∫ e - x 2 . x 3 d x e - x 2 2 x + y 2 = ∫ e - x 2 . x 2 - 2 x d x Let, - x 2 = v   − 2 x d x = d v ⇒ e − x 2 ( 2 x + y ) 2 = − ∫ e v v d v e − x 2 ( 2 x + y ) 2 + v &#890