$$ \frac{dx}{dt}=x-x^3 \\ \frac{dy}{dt}=-y$$ Show that the linearization of points $(-1,0)$ and $(1,0)$ are the same.
I found that for $(1,0)$, the linearization is $-2x+2=\frac{dx}{dt}, -y = \frac{dy}{dt}$,
but for $(-1,0)$, I found $4x+4= \frac{dx}{d}, -y = \frac{dy}{dt}$.
Further, I believe the phase lines to be $\leftarrow -1 \rightarrow 0 \leftarrow 1 \rightarrow$ and $\rightarrow 0 \leftarrow$.
I'm not sure how to draw the phase portrait of the system, however, and any help there would be great.
We are given the system
$$ \begin{align} f(x, y) = \frac{dx}{dt}&=x-x^3 \\ g(x, y) = \frac{dy}{dt}&=-y \end{align}$$
We find the three critical points by simultaneously solving $f(x, y) = g(x, y) = 0$
$$(x, y) = (-1, 0), (0,0), (1, 0)$$
We find the Jacobian as
$$J(x,y)=\begin{pmatrix} \dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y} \\ \dfrac{\partial g}{\partial x} & \dfrac{\partial g}{\partial y} \end{pmatrix} = \begin{pmatrix} 1-3x^2 & 0 \\ 0 & -1 \\ \end{pmatrix}$$
Now we evaluate the Jacobian at each critical point
$$J(-1,0)=\begin{pmatrix} -2 & 0 \\ 0 & -1 \\ \end{pmatrix}\\J(0,0)=\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}\\J(1,0)=\begin{pmatrix} -2 & 0 \\ 0 & -1 \\ \end{pmatrix}$$
We see that the linearization of the first and third points are identical as asked for in the problem and that those are asymptotically stable.
We also see that the origin is an unstable saddle point.
What does this linearization tell you about those points? Hint, they are not the marginal cases.
Putting all of this together and using these notes, with critical points and nullclines, on manually plotting the phase portrait, we arrive at