Equilibrium point linear differential equation as far as I know, there is a comfortable criterium to find out whether the y=0 solution of $y'=Ay$ is stable if the maximum real part of the eigenvalues of A is negative and unstable if it is positive. now I was wondering whether this is also true if we can find a different solution $y=c$ where c is not necessarily 0. or is this really only applicable to the situation $y=0$?  
 A: First of all, there is indeed a "comfortable criterium" for stability of the zero solution of
$y' = Ay$, and that indeed is that the real parts of the eigenvalues of $A$ be negative; and of course there is the companion criterion (I am now reverting to standard American English usage, with which I am natively familiar) that the zero solution is unstable when there exists an eigenvalue of $A$ with positive real part.  These are standard, well-known results in the event that $A$ is a constant (i.e. non-time-varying) matrix of finite size; see for example the book Ordinary Differential Equations by Jack K. Hale, Dover Press (2009), chapter III, especially sections III.4 and III.5.
The reason I have somewhat belabored the points about $A$ being constant and of finite size in the above paragraph is that it is only under these conditions that I can positively affirm the accuracy of what I have said, and of what I am about to say.  For example, if $A = A(t)$ is time-dependent, then the eigenvalue criterion for stability fails.  A famous example, first published in 1960 by Markus and Yamabe (see Markus, L. and H. Yamabe, Global stability criteria for differential systems. in Osaka Math. J. 12 (1960), pp. 305-317), well illustrates this point.  Suppose for the moment we let
$A$ be the matrix
$A(t) = \begin{bmatrix}
-1 + \frac{3}{2} {cos^2 t} & 1 - \frac{3}{2} ({cos \: t})({sin \: t}) \\
-1 - \frac {3}{2}(cos \: t)(sin \: t) & -1 + \frac{3}{2} sin^2 t
\end{bmatrix}$;
then $A(t)$ has constant eigenvalues $-1 \pm i \frac{\sqrt {7}}{4}$, which have a common negative real part; nevertheless the vector $y(t) = exp{(\frac {t}{2})}(- cos \: t, sin \: t)$ is a solution of  $y' = A(t)y$ which becomes unbounded as $t \to \infty$; indeed, we have $\| y \| = exp{(\frac {t}{2})}$.  This example shows that the standard eigenvalue criteria break down if $A$ is time-dependent.  Likewise, if $A$ is allowed to be of infinite size, for example an operator on an infinite dimensional Hilbert space, it is not clear, to me at least at the present moment, that there are not possible phenomena which invalidate the typical, finite dimensional criteria for stability stated above, or at least require additional assumptions to ensure they apply.
Having noted these facts, let me return to the OP's question as stated.  Lipschitz asks whether the standard eigenvalue criterion for stability/instability applies to non-zero solutions of the form "$y = c$", with $c \ne 0$ (necessarily).  Suppose then that there existed a non-zero constant solution $y = c$.  Then we would have $Ac = c' = 0$, which is tantamount to asserting that $c \ne 0$ is an eigenvector of $A$, and that the corresponding eigenvalue is $0$.  But this possibility is ruled out if it is assumed that
the maximum of the real part of the eigenvalues is negative, since then $A$ must be nonsingular.  Similar remarks apply if the maximum of the real part of the eigenvalues is positive, provided the possibility of a zero eigenvalue is ruled out; as long as the eigenvalues are all nonzero, the matrix $A$ has exactly one zero.  This fact is of course well-known.  It should also be observed here that, in the event that a constant solution
$c \ne 0$ exists, so that $Ac = 0$, then in fact any vector collinear with $c$ is also such a solution; indeed, the entire eigenspace corresponding to eigenvalue $0$ consists of
non-vanishing, constant solutions to $y' = Ay$.  Are these solutions stable?  Well, in the event that $A$ has an eigenvalue with positive real part, obviously not; we need merely add to any $c \in ker(A)$ an arbitrarily small linear combination of eigenvectors whose eigenvalues have positive real part to obtain a point, as close to $c$ as we please, whose orbit becomes unbounded as $t \to \infty$, where I am denoting the independent variable by $t$, so that $y' = \frac{dy}{dt}$.  And in fact, constant solutions in $ker(A)$ may be unstable even if $A$ has no eigenvalue of positive real part.  Such instability occurs
when $A$ is possessed of so-called generalized eigevectors corresponding to eigenvalue $0$; that is, when there are vectors $v$ such that $Av \ne 0$ but $A^kv = 0$ for some $k \ge 2$.  The solution of $y' = Ay$ with $y(0) = v$ for such a $v$ is of course given by
the formula $y(t) = e^{At}v$ and since $A^kv = 0$ the power series in $t$ which $e^{At}v$ defines is truncated to a finite number of terms, i.e. is a polynomial in $t$, albeit with vector coefficients.  Such a solution clearly grows without bound as $t \to \infty$, but its rate of growth is polynomial, not exponential as in the case of eigenvalues with positive real part.  If such a solution $e^{At}v$ is added to a constant solution in $ker(A)$, the result is again an unstable solution, though there is no eigenvalue with positive real part involved.  These assertions for the most part follow from the Jordan normal form of the matrix $A$; the theory is fully explained in the book by Hale cited above.
One point to be made here is that the general eigenvalue criteria for stability don't really seem to apply in the simple, obvious sense when $ker(A) \ne \{0\}$, i.e. when $A$ has a zero eigenvalue or, equivalently, when there is a non-vanishing, constant solution to
the differential equation $y' = Ay$.  I would be very interested in hearing (well, reading) the answer Lipschitz found (referred to in his answer to this question).  I suspect it must have something to do with the more general approach I outlined the preceding paragraph, in which the matrix $A$ is allowed to have eigenvalues with other than negative real parts.  
In closing, I think it is worth observing that the stability of any solution of a linear system, time dependent or not, may be analyzed as follows:  let $y(t)$ be a solution of $y' = Ay$, and let $y + \delta y$ be a "neighboring" solution; then we have $(y + \delta y)' = A(y + \delta y)$ which leads, after some elementary manipulations, to $(\delta y)' = A(\delta y)$.  Thus, once we understand the stability of the origin (in, say, the case of $A$ nonsingular), we have a handle on the stability of the non-zero solutions as well.  Similar remarks, appropriately modified, apply to more general $A$ as well.
OK, I've said enough.  I'll shut up now.
For the time being at least.
A: Actually I found the answer and it is yes, there is no difference between the case where the constant is not zero to the one where it is zero
A: The idea of studying the equation $x'=Ax$ near $x=0$ is to study the linearization of an equation of motion such as
$$
y'=u(y)\tag{1}
$$
near a point $c$ where $u(c)=0$ (an equilibrium point). We linearize $(1)$ near $y=c$ as
$$
\begin{align}
y'
&=\color{#C00000}{u(c)}+
\color{#00A000}{
\begin{bmatrix}
\frac{\partial u_1}{\partial y_1}&\frac{\partial u_1}{\partial y_2}\\
\frac{\partial u_2}{\partial y_1}&\frac{\partial u_2}{\partial y_2}
\end{bmatrix}}
\color{#0000FF}{
\begin{bmatrix}y_1-c_1\\y_2-c_2\end{bmatrix}}\\[6pt]
&=\color{#C00000}{0}+\color{#00A000}{\mathscr{J}u}\color{#0000FF}{(y-c)}\tag{2}
\end{align}
$$
where $\mathscr{J}u$ is the Jacobian of $u$. There is no need for $c$ to be $0$, if that was your question.
If, however, you are asking about a case where $y'=Ay=0$ for some $y=c\ne0$, then $Ac=0$ means that $0$ is an eigenvalue. This means the equilibrium at $y=c$ is astable at best.
