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$X'(t)=Ax(t)+g(t)$ 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 $Re(\alpha_j)<0$

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If real parts of eigenvalues are negative, then we need to show that $\int_0^t e^{\lambda(t-s)} g(s) ds$ converges to zero.

$|x(t)| \le \int_0^t e^{\lambda (t-s)} |g(s)| ds = e^{\lambda (t-p)} \int_0^t |g(s)|$, where $0 < p < t$. By the mean value theorem for integrals. Then it's easy to show that the r.h.s. converges to zero.

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