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