Let y ( x ) be a solution of the differential equation 1 + e x y ′ + y e x = 1 . If y 0 = 2 , then which of the following statements is (are) true?

Step-by-step solution

1 + e x d y + y e x d x = d x d y + d e x . y = d x y e x = x - y + c Now, x = 0 , y = 2 2 . e 0 = 0 - 2 + c ⇒ c = 4 Now, y e x = x - y + 4 y = x + 4 e x + 1 Now, d y d x = e x + 1 - x + 4 e x e x + 1 = e x + 1 - x e x - 4 e x e x + 1 = 1 - x e x - 3 e x e x + 1 f ′ 0 = - v e f ′ - 1 = 1 + e - 3 e - 1 e - 1 + 1 = + v e Critical point in - 1 , 0