Let $f$ is a differentiable function such that $f'(x) = f(x) + \int_0^2 f(x)dx, f(0) = \frac{4 - e^2}{3}$, then:
Step-by-Step Solution
Key Concept: Recognize that $\int_0^2 f(x)dx$ is a constant; set $c = \int_0^2 f(x)dx$ and solve the differential equation $f'(x) - f(x) = c$ using integrating factor $e^{-x}$, then use the boundary condition and self-consistency equation to determine $c$.
Given $f'(x) = f(x) + \int_0^x f(t)dt$, rewrite as $f'(x) - f(x) = c$ where $c = \int_0^x f(t)dt$. Differentiating gives $\frac{f'(x)}{f(x)+c} = 1$, leading to $\log|f(x)+c| = x + \text{constant}$, so $f(x) + c = ke^x$. Using $f(0) = \frac{4-e^2}{3}$, determine that $k = 1$. Finally, integrate from 0 to 2 to find the constant, obtaining $f(x) = e^x - 1 + \frac{4-e^2}{3}$.
Correct Answer: 2,3