Is this a vector field? One of the example questions we've been given in lectures is to plot the vector field given by the first order equation $y' = y - y^{3}$. 
The solution we were given looks like this:

However, I was expecting it to be of the form

Is the first picture actually a vector field? If so, what is shown in the second picture?
Thanks
 A: The canonical answer (and in fact only answer) that would be expected in a good differential equations course is the following picture:

You can add a few more arrows in each reagion but that's it. Really one would never expected to "plot" a $2$-dimensional vector field on $\mathbb R^4$, right? So we should also not really plot a $1$-dimensional vector field on $\mathbb R^2$, neither it is really much helpful (see the following paragraph).
There is in fact a more important reason, besides being the canonical answer, for not liking much an other alternative to the former drawing: the idea is that one should understand how the solutions behave qualitatively when we look at the plot, or at what is usually called a phase portrait. This is really the main theme of a large part of the "modern" theory of differential equations: we would like to learn something about the properties of a differential equation without solving it, simply because it is usually complicated or even impossible to do it explicitly.
In the present case the drawing says it all, with the exception of whether the unbounded solutions are global or not (that is, whether they are defined for all time), but that is already a matter of how the solutions behave quantitatively.
A: They are two different visual representations of the same thing. 
The vector field corresponding to a differential equation $y'= f(t,y)$ assigns to each point $(t,y)$ of the plane a vector $[1, f(t,y)]$.
The second picture is more direct: it shows a region of the $t-y$ plane.  The little lines represent the directions of these vectors at a sample
of points (it might be a little better if they were arrows rather than lines: in this case the vectors are all supposed to point to the right).
EDIT: Unfortunately the vector field shown in your picture is not quite the one for this differential equation.  You'd need to rescale $y$ by a factor of $1/3$ so that what is shown as $y=3$ becomes $y=1$.
The first picture shows the same information schematically.
Note that in this case $f(t,y) = y - y^3$ does not depend on $t$.
Also notice that in this picture the $y$ axis is horizontal rather than vertical.
  The curve is the graph of $y - y^3$ as a function of $y$.  When this is positive (above the horizontal axis), the little lines will be sloping up, indicating that solutions of the differential equation are increasing.
When it is negative (below the horizontal axis), the little lines will be sloping down, indicating that solutions are decreasing.
A: A vector field is a vector valued function. 
The second image shows a vector field $f : \mathbb{R}^2 \to \mathbb{R^2}$, where $v = (v_x, v_y) = f(t, y)$.
The first image shows a scalar valued function $f : \mathbb{R} \to \mathbb{R}$, where $v = f(y)$.
If you consider the vectors of one dimensional vector spaces as vectors, which in the view of algebra they are, then it is a vector field as well. 
