# Sketching the phase portrait for an autonomous vector field

Consider the following autonomous vector field on the plane:

\begin{align*} \dot x & = x \\ \dot y & = 2x^2, \qquad (x,y) \in \mathbb{R}^2 \end{align*}

I'm trying to sketch the phase portrait for this system. I know that the equilbria are $\{(x,y)\colon x=0 \}$ but I'm unsure of what to do next.

• Well, you are suposed to show that on the halfplane $x>0$ the vector field points to the NE and that on the halfplane $x<0$ the vector field points to the NW. (And the next question of the exercise might be to show that every solution stays on a curve $y=x^2+c$.) – Did Nov 2 '16 at 20:45

First, note that $\dot{y} = 2x^2 \ge 0$ for all $(x,y)$, that means that $y-$ component of the phase flow is non-negative, or in other words, the solutions will always grow in $y$.
The $x-$ component is actually $x = \dot{x}$. And so, for $x>0$ the flow points NE, while in for $x<0$ the flow points NW. Here's an sketch
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2x^2}{x} = 2x$$
$$y(x) = x^2 + c$$