I want to solve the following system of ODE $$ \begin{cases} y_1' = 3y_1 - y_2 +1\\ y_2'= 4 y_1 +2 y_2 +x\\[.5em] y_1(0)= 1, y_2(0)= 0 \end{cases} $$ For this, I calculate the eigenvalues of the matrix $A$, and I found $\lambda_1= \dfrac{5-\sqrt{15}i}{2}$ and $\lambda_2= \dfrac{5+\sqrt{15}i}{2}$.
Now I search the eigenvectors associated to $\lambda_1$ and $\lambda_2$.
Let $v_1$ the eigenvector associated to $\lambda_1$, then $(A-\lambda_1 Id) v_1=0$.
We put $v_1= (x,y)$ and we resolve the system $$ \begin{cases} (3-\lambda_1) x - y =0\\ 4 x + (2-\lambda_1)y =0 \end{cases} $$ My problem is that I obtain $x=y=0$, and it's not normally to found $v_1=0$.
How do we find the eigenvector associate to $\lambda_1$?
By the first equation of the linear system for the first eigenvalue the eigenvector is (a multiple of) $v_1=\pmatrix{1\\3-λ_1}$. This inserted into the second equation $$4\cdot 1+(2-λ_1)\cdot(3-λ_1)=λ_1^2-5λ_1+10=0$$ just reproduces the characteristic equation. Thus if you insert the correct root into the linear system, you should get non-trivial solutions.
The linear system for the first eigenvector $v_1=\pmatrix{v_{11}\\v_{21}}$ is (after multiplying everything by $2$) \begin{cases} (1+\sqrt{15}i)v_{11}-2v_{21}=0\\ 8v_{11}+(-1+\sqrt{15}i)v_{21}=0 \end{cases} and indeed the second equation is a multiple by the factor of $\frac{1-\sqrt{15}i}2$ of the first equation.
Note that by the same reasoning, $v_2=\pmatrix{1\\3-λ_2}$ is one variant for the second eigenvector.