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(C) Given the differential equation $f^{\prime}(x)+f(x)=k$,where $k = \int_0^2 f(t) dt$ is a constant.The general solution of the linear differential

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(C) Given the differential equation $f^{\prime}(x)+f(x)=k$,where $k = \int_0^2 f(t) dt$ is a constant.The general solution of the linear differential equation $\frac{dy}{dx} + y = k$ is given by the integrating factor $e^{\int 1 dx} = e^x$.Multiplying by the integrating factor: $e^x f(x) = \int k e^x dx = k e^x + c$.Thus,$f(x) = k + c e^{-x}$.Using the condition $f(0) = e^{-2}$,we get $e^{-2} = k + c$,so $c = e^{-2} - k$.Substituting $c$ back: $f(x) = k + (e^{-2} - k)e^{-x}$.Now,calculate $k = \int_0^2 f(t) dt = \int_0^2 (k + (e^{-2} - k)e^{-t}) dt$.$k = [kt - (e^{-2} - k)e^{-t}]_0^2 = (2k - (e^{-2} - k)e^{-2}) - (0 - (e^{-2} - k)) = 2k - (e^{-2} - k)e^{-2} + e^{-2} - k$.$k = k - (e^{-2} - k)e^{-2} + e^{-2} \implies (e^{-2} - k)e^{-2} = e^{-2}$.Dividing by $e^{-2}$ (since $e^{-2} 
eq 0$): $e^{-2} - k = 1$,so $k = e^{-2} - 1$.Then $f(x) = (e^{-2} - 1) + 1 \cdot e^{-x} = e^{-2} - 1 + e^{-x}$.$f(0) = e^{-2} - 1 + 1 = e^{-2}$.$f(2) = e^{-2} - 1 + e^{-2} = 2e^{-2} - 1$.$2f(0) - f(2) = 2(e^{-2}) - (2e^{-2} - 1) = 1$.

Let $f : R \rightarrow R$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int \limits_0^2 f(t) dt$. If $f(0)=e^{-2}$,then $2f(0)-f(2)$ is equal to $.........$.

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