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: $a=(1,0)$ 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 $x=C_1te^{\lambda t}a+C_2e^{\lambda t}b.$
A: Let me answer specific to this case first. The matrix that you have considered has got eigenvalue 3 whose multiplicity is 2 (Eigenvalues are not distinct). But as you have shown it has got only one eigenvector (and not two linearly independent ones that could form a basis for $\mathbb{R}^2$). This means that the matrix is not diagonalizable.
The way out is to look for a generalized eigenvector $v$ which satisfies the equation $(A-\lambda I)^2 v = 0$ such that $(A-\lambda I) v \in Ker(A-\lambda I)$.
As your computations show, $Ker(A-\lambda I) = Span\{(1,0)\}$. So we must have the vector $v$ such that $$(A-\lambda I)v = (k,0)$$ for some constant $k$. Calculations will show that $v = (0,k)$. Choosing the constant to be $1$, we get the two linearly independent generalized eigenvectors $\{(1,0),(0,1)\}$. 
Now let us denote the original eigenvector to be $u$. The matrix $P = (u|v)$ turns out to be invertible and the matrix $P^{-1}AP = B$, called the Jordan Canonical Form.
In this case $P = I$ and the given matrix is itself in the Jordan Form, where the eigenvalues are in the diagonal entries and the $1$ that is above the second diagonal element indicates that there is only one eigenvector for the eigenvalue.
Now wherever we have a non-diagonalizable matrix in the system of ODEs, we convert it into the Jordan Form by determining the matrix with generalized eigenvectors ($P$ in this case) and sole it as follows:
$x(t) = P diag(\exp B_j) P^{-1} x_0 $
where $diag(\exp B_j)$ is a block diagonal matrix where each block is a Jordan Block. 
In this case there is only one Jordan Block and that is $A$ itself. So $\exp A$ is given by 
$$e^{3t}\begin{bmatrix}
    1&t\\
    0&1
\end{bmatrix}$$
So the solution is given by $$x(t) = e^{3t}\begin{bmatrix}
    1&t\\
    0&1
\end{bmatrix} \begin{bmatrix} C_1\\C_2\end{bmatrix}$$
Please refer to the book Differential Equations, Dynamical Systems and Linear Algebra by Hirsch and Smale for detailed understanding. 
http://sgpwe.izt.uam.mx/files/users/uami/jdf/EDO_II/Hirsch-Smale.pdf
