"Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int_\limits{0}^{2} f(t) d t$. If $f(0)=e^{-2}$, then $2 f(0)-f(2)$ is equal to ____________."

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Accepted Answer

"$f^{\prime}(x)+f(x)=k$

$$ \begin{aligned} & \Rightarrow e^{x} f(x)=k e^{x}+c \\\\ & f(x)=k+c e^{-x} \\\\ & k=\int_{0}^{2}\left(k+c e^{-t}\right) d t \\\\ & k=2 k+\left.c \cdot \frac{e^{-t}}{-1}\right|_{0} ^{2} \\\\ & k=2 k+c\left(\frac{e^{-2}}{-1}+1\right) \\\\ & -k=c\left(1-\frac{1}{e^{2}}\right) \\\\ & f(x)=c e^{-x}-c\left(1-\frac{1}{e^{2}}\right) \\\\ & f(0)=c-c+\frac{c}{e^{2}}=\frac{1}{e^{2}} \Rightarrow c=1 \\\\ & f(2)=e^{-2}-r\left(1-e^{-2}\right) \\\\ & =2 e^{-2}-1 \\\\ & 2f(0)-f(2)=1 \end{aligned} $$ "

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

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

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