Let f : ℝ → ℝ be a differentiable function with f 0 = 0 . If y = f x satisfies the differential equation d y d x = 2 + 5 y 5 y - 2 , then the value of lim x → - ∞ f x is

Step-by-step solution

d y d x = 25 y 2 - 4 So d y 25 y 2 - 4 = d x Integrating 1 25 × 1 2 × 2 5 ln ⁡ y - 2 5 y + 2 5 = x + c ⇒ ln ⁡ 5 y - 2 5 y + 2 = 20 x + c Now, c = 0 as f 0 = 0 Hence, 5 y - 2 5 y + 2 = e 20 x lim x → - ∞ ⁡ 5 f x - 2 5 f x + 2 = lim x → - ∞ ⁡ e 20 x Now, R H S = 0 ⇒ lim x → − ∞ ( 5 f ( x ) − 2 ) = 0 ⇒ lim x → − ∞ f ( x ) = 2 5