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I got confused with a seemingly simple problem.

Consider a system of linear ODEs $\dot{x}=Ax+bu$, $u(\cdot)\in R$, $x(0)=0$. I want to solve this system for an exponential input $u=e^{st}$, $s\in R$.

The solution is given by $x(t)=\int^t_0 e^{A(t-\tau)}be^{s\tau}d\tau$. I solve this integral using integration by parts to get $x(t)=(sI-A)^{-1}(I e^{st}-e^{At})b$.

It appears that the solution becomes singular when $s$ is an eigenvalue of $A$. How can it be?

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This is called resonance and quite normal. If you look closely you can interpret $$ (sI−A)^{-1}(I·e^{st}−e^{At}) $$ as some kind of difference quotient, which it actually becomes when evaluated at an eigenvector of $A$.

To see more easily that the expression has no poles, write it as $$ e^{At}(sI−A)^{-1}(e^{(sI-A)t}-I) $$ and note that $$X^{-1}(e^{Xt}-I)=t\sum_{k=0}^\infty\frac{(Xt)^k}{(k+1)!} $$ converges everywhere.

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    Dear @LutzL, thank you for the reply. Indeed, I thought about resonance, but there are some things that I cannot comprehend. What kind of difference quotient is this? I mean, quotient of what w.r.t. what? Also, when I consider a simple case $\dot{x}=-x+u$ and choose $u=e^{-t}$, I get $x(t)=te^{-t}$, not a singularity. – 2017-01-21
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    I see that $Ie^{st}$ and $e^{At}$ share a common eigenvector when $s$ is an eigenvalue, but I do not see yet how is this related to the point :( – 2017-01-21
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    If $b=\sum c_kv_k$ is a representation in an eigenbasis, then $$(sI-A)^{-1}(I e^{st}-e^{At})b=\sum c_k\frac{e^{st}-e^{λ_kt}}{s-λ_k}v_k$$ which obviously has a smooth continuation to the definition gaps $s= λ_k$. – 2017-01-21
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    The for resonance typical unlimited amplitude growth you only get for oscillating excitations $A\sin(\omega t+\phi)$, i.e., purely imaginary eigenvalues. If the system has friction, i.e., the eigenvalues of the system near $\pm\omega i$ have (small) negative real parts, then the amplitude some saturation. – 2017-01-21