Phase portrait in 2 dimensions

Question

I am trying to plot the phase portrait for the system:

$\dot{x} = 1+y -e^{-x}$

$\dot{y} = x^3-y$

Now I worked out my eigenvalues to be $\lambda_1 = 2, \lambda_2 = -1$ and these correspond to 2 eigenvectors, $v_1 = (1,0)$ and $v_2 = (1,-3)$ However I am not sure how to plot this on the phase plane. I can plot $v_2$ however I'm not sure how I would plot $v_1$. I would aussume that the trajectory is just along the x axis? But I put the equations into a matlab and it does not look like this is the case. Thanks.

Answer 1

$$\begin{align}\dot{x}&=(1-e^{-x})+y&=\phantom{0}x+y+(\text{higher order terms})\\\dot{y}&= x^3-y&=0x-y+(\text{higher order terms})\end{align}$$

Therefore, the linearization is

$$\begin{pmatrix}\dot{x}\\\dot{y}\end{pmatrix}=\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$

We get the characteristic polynomial $(1-t)^2$.

Now you can use this to sketch the phase diagram.