Given the differential equation $$\dot{x}=x-x^{3}$$ sketch the graph of $x(t)$ for different initial conditions.
Instead of drawing the graph of $x-x^{3}$ in a plot of $\dot{x}$ vs $x$. I drew the graph of $y = x$ and $y = -x^{3}$ in the plot of $\dot{x}$ vs $x$ and found that $\dot{x} < 0$ between the interval $(0,-1)$. Then taking the derivative of $\dot{x} = x-x^{3}$, I found that $\ddot{x} = 1-3x^{2}$. Drawing the graph of $1-3x^{2}$ in a plot of $\ddot{x}$ vs $x$, I find that $\ddot{x}$ is positive in the interval $\bigg(0,-\dfrac{1}{\sqrt{3}}\bigg)$ and $\ddot{x}$ is negative in the interval $\bigg(-\dfrac{1}{\sqrt{3}}, \infty\bigg)$. So when I sketch the graph of $x(t)$ for different initial conditions, if I start at $x_{0}=-\dfrac{1}{\sqrt{3}}-0.001$ for example I get a curve that is concave down from $\bigg(-\dfrac{1}{\sqrt{3}}, -1 \biggr)$. I was wondering why this is happening? I thought that the curve would eventually asymptote to $-1$, instead of actually touching $-1$.
What about first solving the ODE an then discuss the solution curves?
The ODE is
$$\dot x(t) = x(t) - x(t)^3\tag{1}$$
Let $z(t) = x(t)^2$, then
$$\dot{z}= 2 x \dot{x}= 2 x (x-x^3) = 2 (x^2-x^4) = 2 z(1-z)\tag{2}$$
This is a separable ODE
$$2 dt = \frac{dz}{z(1-z)} =dz\left(\frac{1}{z}+\frac{1}{1-z}\right) = d\left(\log\left(\frac{z}{1-z}\right) \right)$$
which is immediately integrated with the result
$$z(t) = \frac{1}{1+a e^{-2 t}}\tag{3}$$
At $t=0$ we have $z(0) = \frac{1}{1+a }$ or $a=\frac{1}{z(0)}-1$.
Finally we have
$$x(t,x_0) = \pm \frac{1}{\sqrt{1+(\frac{1}{x_{0}^2}-1) e^{-2 t}}}\tag{4}$$
It can be easily shown using $(1)$ that the sign is that of $x(0)$.
For given initial values $x_0$ we can now discuss the curves.
There are four regions for $x(0)$ to be distinguished
region 1: $x_0<-1$: $x$ goes to $-1$ from below
region 2: $-1 < x_0<0$: $x$ goes to $-1$ from above
region 3: $0<x_0<1$: $x$ goes to $+1$ from below
region 4: $1<x_0$: $x$ goes to $+1$ from above
The pictures shows all 4 cases: