Let f : ℝ → ℝ be a differentiable function such that f ' x + f x = ∫ 0 2 f t d t . If f 0 = e - 2 , then 2 f 0 - f 2 is equal to _____ .

Step-by-step solution

Let ∫ 0 2 f t d t = k y = f x Now, we have f x + f ' x = ∫ 0 2 f t d t ⇒ d y d x + y = k This is a linear differential equation. I . F . = e ∫ d x = e x Solution of the given differential equation is y e x = k ∫ e x d x ⇒ y e x = k e x + C Now, f 0 = e - 2 , so C = e - 2 - k Hence, y e x = k e x + e - 2 - k ⇒ y = f x = k + e - 2 - k e - x         . . . . i Now, k = ∫ 0 2 f t d t ⇒ k = ∫ 0 2 k + e - 2 - k e - t d t ⇒ k = k t - e -