Question:

Step 1: Identify and simplify the given differential equation. The given differential equation is $f'(x)+f(x)=\displaystyle\int_0^2 f(t)\,dt$. Let the constant value of the integral be $C$. $$C = \int_0^2 f(t)\,dt$$ Then the differential equation becomes: $$f'(x)+f(x)=C$$ This is a first-order linear differential equation of the form $y' + P(x)y = Q(x)$, where $P(x)=1$ and $Q(x)=C$. Step 2: Solve the differential equation for $f(x)$. The integrating factor (I.F.) for $f'(x)+f(x)=C$ is $e^{\int 1\,dx} = e^x$. Multiply the differential equation by the integrating factor: $$e^x f'(x) + e^x f(x) = C e^x$$ The left side is the derivative of the product $e^x f(x)$: $$\frac{d}{dx}(e^x f(x)) = C e^x$$ Integrate both sides with respect to $x$: $$\int \frac{d}{dx}(e^x f(x))\,dx = \int C e^x\,dx$$ $$e^x f(x) = C e^x + K$$ where $K$ is the constant of integration. Divide by $e^x$ to find $f(x)$: $$f(x) = C + K e^{-x}$$ Step 3: Use the initial condition to find a relation between $C$ and $K$. We are given $f(0)=e^{-2}$. Substitute $x=0$ into the expression for $f(x)$: $$f(0) = C + K e^{-0} = C+K$$ Since $f(0)=e^{-2}$, we have: $$C+K = e^{-2} \quad \text{(Equation 1)}$$ Step 4: Use the definition of $C$ to form another equation. Recall that $C = \displaystyle\int_0^2 f(t)\,dt$. Substitute $f(t) = C + K e^{-t}$ into this integral: $$C = \int_0^2 (C + K e^{-t})\,dt$$ Evaluate the integral: $$C = \left[ Ct - K e^{-t} \right]_0^2$$ $$C = (2C - K e^{-2}) - (0 \cdot C - K e^{-0})$$ $$C = 2C - K e^{-2} - (-K)$$ $$C = 2C - K e^{-2} + K$$ Rearrange the terms to get an equation relating $C$ and $K$: $$0 = C - K e^{-2} + K$$ $$C + K - K e^{-2} = 0 \quad \text{(Equation 2)}$$ Step 5: Solve the system of equations for $C$ and $K$. We have the system of equations: 1) $C+K = e^{-2}$ 2) $C+K - K e^{-2} = 0$ Substitute Equation 1 into Equation 2: $$e^{-2} - K e^{-2} = 0$$ Factor out $e^{-2}$: $$e^{-2}(1 - K) = 0$$ Since $e^{-2} \neq 0$, we must have $1-K=0$, which implies $K=1$. Now substitute $K=1$ back into Equation 1: $$C+1 = e^{-2}$$ $$C = e^{-2}-1$$ Step 6: Write down the explicit form of $f(x)$. Substitute the values of $C$ and $K$ into $f(x) = C + K e^{-x}$: $$f(x) = (e^{-2}-1) + 1 \cdot e^{-x}$$ $$f(x) = e^{-x} + e^{-2} - 1$$ Step 7: Calculate the required expression $2f(0)-f(2)$. We are given $f(0)=e^{-2}$. Now calculate $f(2)$ using the explicit form of $f(x)$: $$f(2) = e^{-2} + e^{-2} - 1$$ $$f(2) = 2e^{-2} - 1$$ Finally, calculate $2f(0)-f(2)$: $$2f(0)-f(2) = 2(e^{-2}) - (2e^{-2}-1)$$ $$2f(0)-f(2) = 2e^{-2} - 2e^{-2} + 1$$ $$2f(0)-f(2) = 1$$ The final answer is $\boxed{1}$.