What to do to find yet another vector for ODEs system? I have the system that can be descrubed as $\dot{x} = Ax$
where $A = \begin{pmatrix}3 & 1  \\ 0 & 3\end{pmatrix}$
so characteristic polinomial is $$(3-\lambda)^2 = 0$$ and therefore $\lambda_{1,2} = 3$
so solving the system for value $3$ I got the first eigenvector $a = (1,0)$
$\textbf{My question}$ is: how should I find yet another linear-independent vector? and how will the final answer look like for this case?

$a=(1,0)$ is the only eigenvector. There is no other linearly independent eigenvector. However there is another generalized eigenvector, satisfying $(A-\lambda)^2b=0.$ In our case, $(A-3)^2=0$, so its nullspace is all of $\mathbb{R}^2.$ For our second generalized eigenvector, linearly independent of $a$, we can take $b=(0,1).$ Then the general solution to $\dot x=Ax$ will be $x=C_1te^{\lambda t}a+C_2e^{\lambda t}b.$
