Equilibrium of a system of 2nd-order ODEs Given 
$$
\ddot{x} = \begin{cases}
-x + c\cdot \operatorname{sgn}(x)& |x| > c\\
 0 & |x|\leq c
\end{cases}
$$ 
where $c > 0$ are some given constants. 
(a) Find all equilibria
(b) Graph the phase portrait of that system
(c) Make the conjecture, and prove it, about the nature of the non-equilibrium solutions.
My attempt: For part (a), From the given system, we see that the only case where the RHS is $0$ is when $|x|< c$ (the equation, $- x + c\cdot \operatorname{sgn}(x) = 0$, only has roots $x=\pm c$, which contradicts to the condition $|x| > c$). Thus all equilibria $:= \left\{x\in R\ |\  |x| < c\right\}$.
For part (b), it seems weird to me that the only possible solution is $x(t) = At+B$, where $A$ and $B$ are some constants such that $|x| < c$ for all $t\in R$. This means $A=0$ is the only choice, so $x(t) = B$ ($B < c$), is the only solution. So the phase portrait is just the horizontal line $x=B$. For the other equation, I couldn't see how a specific solution $x(t)$ that satisfies the constraint $|x| > c$ AND the equation: $\ddot{x} + x = c\cdot \operatorname{sgn}(x)$ exists.
For part (c), all the non-equilibrium solutions don't exist, based on what I found in part (b).
My question: I'm skeptical with my answers above. So I hope someone can try this problem out and/or give me some thoughts about my work.
 A: This solution is inspired by the approach used in the amazing book "Theory of Oscillators" by A.A. Andronov, A.A. Vitt and S.E. Khaikin (See Section $III-I$ and $VII-9$).
We are given 
$$ \tag 1
\ddot{x} = \begin{cases}
-x + c\cdot \operatorname{sgn}(x)& |x| > c\\
 0 & |x|\leq c
\end{cases}
$$ 
where $c > 0$ is a constant. 
We can re-write $(1)$ as a system of first order equations:
$$ \tag 2 \begin{align} x' &= y \\ y' &= -x + c ~\mbox{sgn}(x) \end{align}$$
From $(2)$, we have that all points in the interval $| x | \le c, y = 0$ are equilibrium points.
Now, to see what is going on with the phase portrait, lets consider three cases.
Case 1: $~x \lt -c$, from $(2)$, we get the DEQ
$$\dfrac{dy}{dx} = \dfrac{-x-c}{y} \implies y^2 + (x + c)^2 = C_1$$
This gives us semicircles centered at $(-c, 0)$.
Case 2: $~-c \le x \le c$, from $(2)$, we get the DEQ
$$\dfrac{dy}{dx} = 0 \implies y = C_2$$
This gives us straight lines.
Case 3: $~x \gt c$, from $(2)$, we get the DEQ
$$\dfrac{dy}{dx} = \dfrac{-x+c}{y} \implies y^2 + (x - c)^2 = C_3$$
This gives us semicircles centered at $(c, 0)$.
From these three cases, we see that we have a phase portrait with semicircles, where the variable $c$ is elongating those. We can now draw the phase portrait with direction fields by hand as:

Of course, I wanted to try and verify this using numerical methods and needed help with Mathematica to get it to work. Here are three cases for $c = 1, 5, 10$ (sorry I didn't add the direction arrows, but already spent too much time with it). If you were to overlay them, you would have the hand drawn graphic shown above.



Do you see how the phase portrait is being elongated by the choice of the variable $c$, since those are constant phase lines? The other areas give us the two sets of semicircles as described by the three cases.
With a bunch of time, it should be possible to draw a general phase portrait or vector field showing many initial conditions and it will look like the items already shown, but with many elongated semicircles and direction fields.
A: For $x>c$ set $y=x-c$ to obtain the ODE $\ddot y+y=0$ which gives upper half circles in the $(x,\dot x)$ phase space.
Similarly lower half circles for $x<c$.
For the solution continuation in $(x,\dot x)=(c,-r)$, $r\ne 0$ you get the line $x(t)=c-rt$, $\dot x(t)=-r$ which at time $t=2c/r$ reaches the point $(x,\dot x)=(-c,r)$ and continues in a half circle towards $(-c,r)$, then in a line straight up to $(c,r)$ and an upper half circle closing the stadium oval.
This leaves the consideration of the points $(x,\dot x)\in[-c,c]×\{0\}$
