Sketch phase portray of $ x'' + 3x^2 + - 3 = 0 $ I am sketching the phase portray of $ x'' + 3x^2 + - 3 = 0 $. First of all I found the corresponding ODE system: \begin{align*} x' = y \\\ y' = -3x^2 + 3 \end{align*} The critical points are (-1,0) and (1,0). However I am going to center it in the origin so I will ignore the +3, the solution is the same anyway. I tried to use the method of linearization but I an eigenvalue of real part equals to 0 and this does not give information. Then I tried using Liaupnov method, so I need to find a function $V(x,y)$ such that $V(0,0)=0$, $V(x,y) \geq 0$ and $V'(x,y) \leq 0$, this would mean that the origin is stable. However I am struggling to find such a function here. Any hints?

Write the Jacobian at the critical points ${\bf x}^*$

$$ J({\bf x}^*) = \pmatrix{0 & 1 \\\ -6x^* & 0} $$

with eigenvalues $\lambda^2 = -6x^*$. That means that for

> ${\bf x}_1^* = (+1,0)$ the eigenvalues are $\lambda_1^{\pm} = \pm i \sqrt{6}$

Solution is a cycle

> ${\bf x}_2^* = (1,0)$ the eigenvalues are $\lambda_2^{\pm} = \pm \sqrt{6}$

Solution is a saddle point. You can find the directions of the stable and unstable manifolds by calculating the eigenvectors. I will leave that part to you

Here's a sketch

![enter image description here](