Let f: R R be a continuous function such that for all x R , f(x) = 1 + 3 ₀^x f(t) dt + (1 - e^ -x ) , where is a real constant. If _ x f(x) = 0 , then the value of f(- 5) is equal to

Step-by-step solution

Given integral equation is: f(x) = 1 + 3 ₀^x f(t) dt + (1 - e^ -x ) Substituting x = 0 , we get: f(0) = 1 + 0 + (1 - 1) = 1 Differentiating the given equation with respect to x using the Newton-Leibniz rule: f'(x) = 3f(x) + e^ -x dy dx - 3y = e^ -x (where y = f(x) ) This is a linear differential equation with Integrating Factor (I.F.) = e^ -3 dx = e^ -3x . The general solution is: y e^ -3x = e^ -x e^ -3x dx y e^ -3x = e^ -4x dx y e^ -3x = - 4 e^ -4x + C f(x) = - 4 e^ -x + C e^ 3x Using the initial condition f(0) =