Stability of equilibrium points of system of differential equations Given $$\ddot{x}+\lambda\dot{x}=x-x^3,$$
which I rewrote to
\begin{align}
\dot{x}&=y \\
\dot{y} &= -\lambda y + x - x^3,
\end{align}
For the exercise we take $\lambda\geq 0$.
I have determined that the equilibrium points are given by $(x,y)$ equal to $(0,0)$, $(-1, 0)$ and $(1,0)$. I now need to determine what their stability type is.
From phase planes I saw that $(0,0)$ was a saddle point, thus unstable, and $(-1,0)$ and $(1,0)$ are spiral sinks, thus stable, but how do I prove this? I tried proving they were Lyaponov stable, because I think that determines stability or instability, but I don't know how to do that with a system of differential equations:
What I tried for the point $e=(-1,0)$:
Let $a$ be a point close to $e$, so $||a-e||<\delta$, with $\delta>0$. I then want to find a solution, which I'll call $z(t)$, corresponding to this point $a$ and show that for all $t$ there exists an $\epsilon>0$ such that $||z(t)-e||<\epsilon$. I don't know how to find such a solution or how to proceed.
 A: Let's utilize the Jacobian matrix $A$ at each equilibrium point and check its eigenvalues:
$$
A=
\begin{bmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} 
\end{bmatrix} 
= 
\begin{bmatrix}
0 & 1 \\
1-3x^2 & -\gamma
\end{bmatrix} 
$$
For the origin, we get 
$$
A=
\begin{bmatrix}
0 & 1 \\
1-3x^2 & -\gamma
\end{bmatrix}_{x=y=0}
= 
\begin{bmatrix}
0 & 1 \\
1 & -\gamma
\end{bmatrix}
$$
The eigenvalues of $A$ are 
$$
\lambda_{1,2} = \frac{-\gamma\pm \sqrt{\gamma^2+4}}{2}
$$
They are real and opposite signs, therefore, the origin is saddle (i.e unstable). 
For the equilibrium point (-1,0), we have 
$$
A=
\begin{bmatrix}
0 & 1 \\
1-3x^2 & -\gamma
\end{bmatrix}_{x=-1,y=0}
= 
\begin{bmatrix}
0 & 1 \\
-2 & -\gamma
\end{bmatrix}
$$
The eigenvalues of $A$ are 
$$
\lambda_{1,2} = \frac{-\gamma\pm \sqrt{\gamma^2-8}}{2}
$$
If $\gamma < \sqrt{8}$, the eigenvalues are complex with negative real part, hence, the equilibrium point is stable focus. If $\gamma \geq \sqrt{8}$, the eigenvalues are real and negative, therefore, the equilibrium point is stable node. Apply same procedure for the equilibrium point (1,0), you will end up with the same result. 
