Let y = y ( x ) be the solution of the differential equation d y d x = 1 + x e y - x , - 2 < x < 2 , y 0 = 0 , then the minimum value of y x , x ∈ - 2 , 2 is equal to :

Step-by-step solution

Given, d y d x = 1 + x e y - x ⇒ d y - d x e y - x = x d x ⇒ ∫ d y - x e y - x = ∫ x d x ⇒ - e x - y = x 2 2 + c At x = 0 , y = 0 ⇒ c = - 1 So, the particular solution is e x - y = 2 - x 2 2 ⇒ y = x - ln 2 - x 2 2 ⇒ d y d x = 1 + 2 x 2 - x 2 = 2 + 2 x - x 2 2 - x 2 ⇒ d y d x = x 2 - 2 x - 2 x 2 - 2 ⇒ d y d x = x 2 - 2 x - 2 x + 2 x - 2 If d y d x = 0 ⇒ x 2 - 2 x - 2 = 0 ⇒ x = 2 ± 12 2 ⇒ x = 1 ± 3 So minimum value occurs at x = 1