classify stable and unstable equilibrium points for differential equation $\frac{dx}{dt} = x(\lambda -x)(\lambda + x)$ I am doing exercise to find equilibrium points and classify them as stable/unstable for the following differential equation:
$\frac{dx}{dt} = x(\lambda -x)(\lambda + x)$
With $\frac{dx}{dt} = 0$, I can find the equilibrium points $x = 0, x = \lambda, x = -\lambda$.
However, I have no idea how to infer the sign of $\frac{dx}{dt}\Bigr|_{x=0}, \frac{dx}{dt}\Bigr|_{x=\lambda}, \frac{dx}{dt}\Bigr|_{x=-\lambda}$ in order to classify them as stable or unstable.
Could you please provide some directions so that I can continue to work on it?
 A: It is easy to see that the equilibria of
$\dot x = x(\lambda - x)(\lambda + x) \tag 1$
occur at
$x = -\lambda, 0, \lambda, \tag 2$
since these are the zeroes of the cubic polynomial
$x(\lambda - x)(\lambda + x) = x(\lambda^2 - x ^2) = \lambda^2x - x^3.  \tag 3$
Note that we cannot "infer the sign" of $\dot x$ at any of these three $x$-values, since $\dot x =0$ has no sign; we can, however, find the sign of $d\dot x /dx$ at each of the critical values $0, \pm \lambda$.  Indeed, (3) yields
$\dfrac{d\dot x}{dx}(x) = \lambda^2 - 3x^2; \tag 4$
we see that, for $\lambda \ne 0$,
$\dfrac{d\dot x}{dx}(\pm \lambda) = -2\lambda^2 < 0, \tag 5$
but
$\dfrac{d\dot x}{dx}(0) = \lambda^2 > 0; \tag 6$
thus $x = \pm \lambda$ are stable, attractors, and $x = 0$ is an unstable zero, a repellor.
In the event that $\lambda = 0$, (1) reduces to
$\dot x = -x^3, \tag 7$
and (4) becomes
$\dfrac{d\dot x}{dx}(x) = -3x^2; \tag 8$
the equation now has one critical point at $0$, where
$\dfrac{d\dot x}{dx}(0) = 0; \tag 9$
the stability thus cannot be decided via the derivative of $\dot x$; we can nevertheless infer that $0$ is stable since $\dot x > 0$ to its left, and $\dot x < 0$ to its right, so points on either side flow towards $0$ under the action of $\dot x$.
A: Assume $\lambda >0$.
Solving $x(\lambda - x)(\lambda + x)\geq 0$ you get $x\leq -\lambda$ or $0\leq x \leq \lambda$. Therefore $\pm \lambda$ are stable, while $0$ is unstable.
For $\lambda <0$ things are more or less the same.
On the other hand, if $\lambda =0$, the unique stationary point $0$ is stable.
