Considering the linear system $Y'=AY$ What would be an equation that I can use when I compute the eigenpairs for the coefficient matrix $A.$ 
 A: 
Here is the fundamental idea for solving a system of linear ode's with constant coefficients. 

To solve the system $Y'(t)=AY(t)$, assuming for simplicity $A_{2\times 2}$ matrix, we assume the solution to have the form 

$$ Y(t) = \begin{bmatrix}
  k_1 \\
  k_2
\end{bmatrix}\rm e^{\lambda t} \implies Y'(t) = \begin{bmatrix}
  k_1 \\
  k_2
\end{bmatrix}\lambda \rm e^{\lambda t}. $$

Now, we substitute back in the Diff. eq. which gives 

$$ Y'(t) = \begin{bmatrix}
  k_1 \\
  k_2
\end{bmatrix}\lambda \rm e^{\lambda t} = A \begin{bmatrix}
  k_1 \\
  k_2
\end{bmatrix}\rm e^{\lambda t}\implies \begin{bmatrix}
 \lambda k_1 \\
 \lambda k_2
\end{bmatrix}= \begin{bmatrix}
 a & b \\
 c & d
\end{bmatrix}\begin{bmatrix}
 k_1 \\
 k_2
\end{bmatrix}. $$

Solve the last system for $k_1$ and $k_2$. In order to get a non trivial solution for the system we force the determinant of the matrix of the coefficients to be zero which results in getting $\lambda's$ the eigenvalues. 
A: In the two dimensional case, the eigenvalues of the matrix $A=\begin{pmatrix}a& b\\ c& d\end{pmatrix}$ are the roots of the characteristic polynomial
$$
\chi_A(\lambda)=\lambda^2-\mathrm{tr}(A)\lambda+\det(A),
$$
where
$$
\mathrm{tr}(A)=a+d,\qquad\det(A)=ad-bc.
$$
Let me assume you know how to compute these roots. To compute an eigenvector $u_\lambda=\begin{pmatrix}x_\lambda\\ y_\lambda\end{pmatrix}$ for the eigenvalue $\lambda$, note that $Au_\lambda=\lambda u_\lambda$ if and only if
$$
by_\lambda=(\lambda-a)x_\lambda,\qquad cx_\lambda=(\lambda-d)y_\lambda,
$$
and that these two equations are actually equivalent (this is the sign that $\lambda$ is indeed an eigenvalue), hence a first choice for $u_\lambda$ is
$$
u_\lambda=\begin{pmatrix}b\\ \lambda-a\end{pmatrix}.
$$
This choice fails if the RHS is the zero vector, that is, if $b=0$ and $\lambda=a$, then choose
$$
u_\lambda=\begin{pmatrix}\lambda-d\\ c\end{pmatrix}.
$$
The only case when both choices fail is when $A=aI$, then both eigenvalues are $a$ and every nonzero vector is an eigenvector.
