You are given the time dependent linear system in $\mathbb{R^n}$ $\bf{\dot{x}}$=$A$$\bf{x}$ where $A$ is a fixed $n\times n$ matrix and $b : \mathbb{R} \rightarrow \mathbb{R^n}$ is continuous. $i)$ Show that $$x(t)=e^{tA}x_o +\int_{0}^{t} e^{(t-s)A} b(s) ds$$ is the unique solution satisfying the initial condition $x(0)=x_0$ ii)Suppose that $\lim_{t \rightarrow \infty} b(t)=0$ and that all eigenvalues of $A$ have negative real parts. Prove that regardless of the initial condition, the solution $x(t)$ approaches zero as $t \rightarrow \infty$
Proof
For the function $\bf{x(t)}$ defined above $$x^{'}(t)=Ae^{tA} I x_o+e^{tA} I b(t)+\int_{0}^{t} Ae^{(t-s)A} b(s) ds$$ And since $e^{tA}$ is a fundamental matrix solution, it follows that $$x^{'}(t)=A\big[e^{tA} x_o+\int_{0}^{t} e^{(t-s)A} b(s) ds\big]+ b(t)=Ax(t)+b(t)$$ Here is what I have for part i. But I am unclear how to tackle part ii.
For part i you should also note that $x(t)$ is of class $C^1$ (for that it is sufficient to observe that the derivative exists, as you showed, and that it is continuous). Moreover, the domain goes up to $+\infty$ since $x(t)$ is always defined.
For part ii, you know that there exist $\alpha,K>0$ such that $\|e^{At}\|\le K e^{-\alpha t}$ for $t>0$. Hence, $$ \|x(t)\|\le K e^{-\alpha t}\|x_0\|+\int_0^tK e^{-\alpha (t-s)}\|b(s)\|\,ds. $$ Let's say that $\|b(t)\|<\varepsilon$ for $t>T$. Then $$ \begin{align} \int_0^tK e^{-\alpha (t-s)}\|b(s)\|\,ds &\le \int_0^TK e^{-\alpha (t-s)}\|b(s)\|\,ds+\int_T^tK e^{-\alpha (t-s)}\varepsilon\,ds\\ &= e^{-\alpha t}\int_0^TK e^{\alpha s}\|b(s)\|\,ds+\frac{\varepsilon K}{\alpha}(1-e^{-\alpha (t-T)}). \end{align} $$ The right-hand side tends to $\varepsilon K/\alpha$ when $t\to+\infty$. Hence, $$ \limsup_{t\to+\infty}\|x(t)\|\le \frac{\varepsilon K}{\alpha}, $$ but since $\varepsilon$ is arbitrary you get the desired property.