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 _____.

We need to find $$2f(0) - f(2)$$ given that $$f'(x) + f(x) = \int_0^2 f(t)\,dt$$ and $$f(0) = e^{-2}\,.$$

Let $$c = \int_0^2 f(t)\,dt$$, which is a constant. Then the differential equation becomes $$f'(x) + f(x) = c\,.$$

Using the integrating factor $$e^x$$, we have $$\frac{d}{dx}(e^x f) = ce^x\,, $$ so integrating gives $$e^x f = ce^x + K\,, $$ and hence $$f(x) = c + Ke^{-x}\,.$$

Since $$f(0)=e^{-2}$$, substituting $$x=0$$ yields $$e^{-2} = c + K\,, $$ so $$K = e^{-2} - c\,, $$ and therefore $$f(x) = c + (e^{-2} - c)e^{-x}\,.$$

To determine $$c$$, observe that $$ c = \int_0^2\bigl[c + (e^{-2}-c)e^{-t}\bigr]\,dt = 2c + (e^{-2}-c)\bigl[-e^{-t}\bigr]_0^2 = 2c + (e^{-2}-c)(1-e^{-2}) = c + ce^{-2} + e^{-2} - e^{-4}\,. $$ Hence $$ c = c(1+e^{-2}) + e^{-2} - e^{-4}\quad\Longrightarrow\quad -ce^{-2} = e^{-2}(1-e^{-2}) \quad\Longrightarrow\quad c = -(1-e^{-2}) = e^{-2}-1\,. $$

It follows that $$K = e^{-2} - c = 1$$ and so $$f(x) = (e^{-2}-1) + e^{-x}\,. $$ Checking gives $$f(0)=e^{-2}-1+1=e^{-2}$$ and $$f(2)=(e^{-2}-1)+e^{-2}=2e^{-2}-1\,. $$

Finally, $$2f(0) - f(2) = 2e^{-2} - (2e^{-2}-1) = 1\,. $$

Therefore, the answer is 1.