non-homogenous differential equation eigenvalues? I am confused by the fact that the following system is not homogenous:



How can we speak of the eigenvectors of a system like this? without the $2$'s, we simply have the equation $Av=\lambda v\to (A-\lambda I)v=0$. So we know that the determinant of $A-\lambda I$ is zero. 
But I don't know what I can conclude from the equation $(A-\lambda I)v=\overrightarrow 2$
So how do I answer this question?
 A: You can change coordinates by calling
$$
x=u+2\text{ and }y=v+2
$$
from which you can easily differentiate with respect to $t$:
$$
x'=u'\text{ and }y'=v'
$$
and the system becomes
$$
\left\{
\begin{array}{l}
u'=-2u+v\\
v'=u-2v
\end{array}\right.
$$
The change of coordinates just preserves the direction of the coordinate axis and changes the origin from $(0,0)$ to $(-2,-2)$. 
A: If we write the given system
$\dot x = -2x + y + 2, \tag 1$
$\dot y = x - 2y + 2, \tag 2$
in matrix-vector form, setting
$\vec r = \begin{pmatrix} x \\ y \end{pmatrix} \tag 3$
and
$A = \begin{bmatrix} -2 & 1 \\ 1 & -2 \end{bmatrix}, \tag 4$
we obtain
$\dot{\vec r} = A \vec r + \begin{pmatrix} 2 \\ 2 \end{pmatrix}. \tag 5$
Since 
$\det A = (-2)^2 - 1 = 3, \tag 6$
$A$ is invertible and hence we may find a vector $\vec b$ such that
$A\vec b = \begin{pmatrix} 2 \\ 2 \end{pmatrix}; \tag 7$
in fact, an easy calculation shows that
$\vec b = \begin{pmatrix} -2 \\ -2 \end{pmatrix}; \tag 8$
now (5) may be written
$\dot{\vec r} = A \vec r + A\vec b = A(\vec r + \vec b); \tag 9$
if we define new variables $x_1$, $y_1$, $\vec r_1$ with
$\vec r_1 = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \vec r + \vec b, \tag {10}$
i.e., $x_1 = x -2$, $y_1 = y - 2$, 
then since $\vec b$ is constant, 
$\dot {\vec r_1} = \dot {\vec r}; \tag{11}$
combining (9), (10), and (11) yields
$\dot {\vec r_1} = \dot{\vec r} = A(\vec r + \vec b) = A \vec r_1, \tag{12}$
which is homogeneous, and so may be solved by ordinary eigenvalue methods.  The eigenvalues of $A$ are easily seen to satisfy
$\lambda^2 + 4\lambda + 3 = 0 \tag{13}$
and hence are
$\lambda = \dfrac{1}{2}(-4 \pm \sqrt{4}) = -1, -3, \tag{14}$
and the system has an attractive node at the point $\vec r_1 = 0$, i.e. a the point $\vec r = (2, 2)^T$.  With this information, the eigevectors of (12) may easily be found and the phase portrait sketched; I leave these operations to my readers.
The point here is, of course, that the ordinary eigenvalue/eigenvector computations may be used on the coefficient matrix $A$ of the system (5) once it is transformed into the homogeneous equation (12) by a mere translation of coordinates, which shifts the origin to the point $(2, 2)^T$ (in the original $x$-$y$ coordinate system).
