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In this problem, I tried the following: First I show that $$x(t) = e^{tA}x_0 + \int\limits_0^t e^{(t-s)A}f(s)ds$$ Then I take the norm for both sides $$\|x(t)\|\leq Ke^{-\alpha t}\|x_0\|+\left\|\int\limits_0^t e^{(t-s)A}f(s)ds\right\|$$ I have used Meiss’s Lemma ($\|e^{tA}\|\leq K e^{-\alpha t}\|$ for some $K>0$, $\alpha >0$).

Could you please help me in what is remaining, and if there is any mistake. Thanks.

Ahmed
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Following the already given hints, $$ \begin{split} \|x(t)\| &\leq Ke^{-\alpha t} \|x_0\| + \left\|\int_0^t e^{(t-s)A}f(s)\,ds\right\| \\ &\leq Ke^{-\alpha t} \|x_0\| + \int_0^t \bigl\|e^{(t-s)A}f(s)\bigr\|\,ds \\ &\leq Ke^{-\alpha t} \|x_0\| + \int_0^t \bigl\|e^{(t-s)A}\bigr\|\|f(s)\|\,ds \\ &\leq Ke^{-\alpha t} \|x_0\| + K\|f\|_\infty \int_0^t e^{\alpha(s-t)}\,ds \\ &= Ke^{-\alpha t} \|x_0\| + K\|f\|_\infty \frac{1-e^{-\alpha t}}{\alpha} \\ &\leq Ke^{-\alpha t} \|x_0\| + \frac{K\|f\|_\infty}\alpha . \end{split} $$

Federico
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  • Should we eliminate also $e^{-\alpha t}$ as $t\to\infty$ – Ahmed Dec 11 '18 at 13:49
  • What do you mean? I'm not taking any limits here. I only used $1-e^{-\alpha t}\leq 1$. – Federico Dec 11 '18 at 14:00
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    Oh, I see, you mean in the first term? Well then yes you can eliminate $Ke^{-\alpha t}|x_0|\leq K|x_0|$ because $e^{-\alpha t}\leq 1$. But you can also keep it because it provides a slightly better inequality – Federico Dec 11 '18 at 14:01