Let $$f : \mathbb{R} \to \mathbb{R}$$ be a differentiable function such that $$f'(x) + f(x) = \int_0^2 f(t) dt$$. If $$f(0) = e^{-2}$$, then $$2f(0) - f(2)$$ is equal to _____.

Given: $$f'(x) + f(x) = \int_0^2 f(t)\, dt$$, $$f(0) = e^{-2}$$.

Let $$K = \int_0^2 f(t)\, dt$$ (a constant). The ODE becomes:

$$ f'(x) + f(x) = K $$

This is a first-order linear ODE with integrating factor $$e^x$$:

$$ \frac{d}{dx}(e^x f(x)) = Ke^x $$

$$ e^x f(x) = Ke^x + C $$

$$ f(x) = K + Ce^{-x} $$

Using $$f(0) = e^{-2}$$,

$$ K + C = e^{-2} \quad \ldots (1) $$

Using the integral condition,

$$ K = \int_0^2 (K + Ce^{-t})\, dt = 2K + C[-e^{-t}]_0^2 = 2K + C(1 - e^{-2}) $$

$$ -K = C(1 - e^{-2}) $$

$$ K = -C(1 - e^{-2}) = C(e^{-2} - 1) \quad \ldots (2) $$

Substituting (2) into (1):

$$ C(e^{-2} - 1) + C = e^{-2} $$

$$ Ce^{-2} = e^{-2} $$

$$ C = 1 $$

From (2): $$K = e^{-2} - 1$$

So $$f(x) = (e^{-2} - 1) + e^{-x}$$

Computing the required value,

$$ f(0) = e^{-2} - 1 + 1 = e^{-2} \quad \checkmark $$

$$ f(2) = e^{-2} - 1 + e^{-2} = 2e^{-2} - 1 $$

$$ 2f(0) - f(2) = 2e^{-2} - (2e^{-2} - 1) = 1 $$

Therefore, $$2f(0) - f(2) = 1$$.