If I'm given a system of equation of the form $$\begin{cases} \frac{dx}{dt}= ax+by \\ \frac{dx}{dt}= cx+ey\end{cases}$$
I get the general solution finding the eigenvalues and eigenvectors of the matrix $\left( \begin{array}{ccc} a & b\\ c & e \\ \end{array} \right)$, let's say that I found the eigenvalues $\lambda_1,\lambda_2$ and eigenvectors $v_1,v_2$. Then the general solution have the form $(x,y)=c_1e^{t\lambda_1}+c_2e^{t\lambda_2}$.
What happens if I don't have "enough" eigenvectors to write a solution like the former?
This is what happened when I tried to find the solution for the system
$$\begin{cases} \frac{dx}{dt}= 2x-y \\ \frac{dx}{dt}= x+4y\end{cases}$$
Here the eigenvalues will be given by the roots of $p(\lambda)=(2-\lambda)(4-\lambda)+1=(\lambda-3)^2$. Now if I can only find one eigen vector associated to the double eigenvalue: $$\left( \begin{array}{ccc} 2-3 & -1\\ 1 & 4-1 \\ \end{array} \right)\to\left( \begin{array}{ccc} 1 & 1\\ 0 & 0 \\ \end{array} \right)\implies \left( \begin{array}{ccc} x \\ y \\ \end{array} \right)=\operatorname{gen}\Bigg\{\left( \begin{array}{ccc} 1\\ -1 \end{array} \right) \Bigg\}$$
Then the solution will be simply $\left( \begin{array}{ccc} x \\ y \\ \end{array} \right)= c_1e^{3t}\left( \begin{array}{ccc} 1 \\ -1 \\ \end{array} \right)$ ? or do I need another eigenvector to write the general solution?