$X'(t)=Ax(t)+g(t)$ with $x(0)=x_0$ has the solution $x(t)=\exp(tA)x_o+\int_{0}^{t}\exp\big((t-s)A\big)g(s)ds$
I wanted to know what happens if I take $\lim_{t\to\infty}x(t)$ with $\lim_{t\to\infty} |g(t)|=0$
If all eigenvalues of $A$ satisfy $Re(\alpha_j)<0$ then $\lim_{t\to\infty}x(t)=0$, but how to prove? $\exp(tA)x_o$ goes to 0 for $t\to\infty$ but what about the integral term?
What if I suppe $\lim_{t\to\infty} g(t)=g_0$ ? Is $\lim_{t\to\infty}x(t)=g_o$ ?
In all cases $A$ satisfy $Re(\alpha_j)<0$