Suppose we have a system of differential equations as follows:
$$dx/dt = x^2 \\ dy/dt = -y$$
I am certain on how to draw the phase diagram including equilibrium points, isoclines and the general direction of curvature which is marked in black in the figure below:
However, I am unsure on how to determine the "convexity" of the curves specifically in quadrants I and IV which is marked in blue. I use the word "convexity "within quotes because I do not know the formal mathematical term in this context. Note how option A (blue solid) and option B (Blue dotted) are both generally correct because they follow the correct general direction, but how would I know which one curves the correct way?
You might ask why am I not asking the same for quadrants II and III but there I do know that the curve can never cross the isoclines and the only way it can curve in the general directions without crossing the isoclines is as shown below. So I can sometimes get a clue on the curvature and "convexity" but for cases of quadrant I and IV I have no clue.
One approach (some people are able to do this using a few points only) is to calculate the slopes for a fixed value of $y$, while varying $x$.
We have
$$\dfrac{dy}{dx} = \dfrac{y'}{x'} = -\dfrac{y}{x^2}$$
If we fix $y = \dfrac{1}{2}$ and vary $-3 < x < -0.1$ in steps of $.1$ (we certainly do not need to do this many points), we get the following data:
$$\{-0.0555556,-0.059453,-0.0637755,-0.0685871,-0.0739645,-0.08,-0.0868056,-0.094518,-0.103306,-0.113379,-0.125,-0.138504,-0.154321,-0.17301,-0.195313,-0.222222,-0.255102,-0.295858,-0.347222,-0.413223,-0.5,-0.617284,-0.78125,-1.02041,-1.38889,-2.,-3.125,-5.55556,-12.5,-50.\}$$
If we plot these slopes, we get
The slope is getting more negative as we get closer to the origin and this tells us that option $A$ is the shape. Also note that you can look at the value as $x$ is approaching the origin because the slope is increasing and that tells you the shape is $A$, that is, the slope is approaching infinity. Some people can look at the values of $x'$ and $y'$ at a few points and also figure this out, but I have a hard time with that.
Compare this to the phase portrait