Drawing direction fields on phase portraits by hand I'm looking for an idiot's guide to drawing the direction fields included on phase portraits by hand.  All I can find online (and in my course notes) - is the instruction that you should do it by finding dy/dx - and then they show a computer generated plot. I cannot figure out HOW to do it by hand as I need to.
Consider x' = x; y' = xy - y
I know I need to find dy/dx = (xy - y) / x = y - y/x
I can substitute in a bunch of x and y values to calculate the gradients
x = 0 -> infinity.       y = 0 -> 0.  (1,1) -> 0.    (2,1) -> 0.5.  (-1,1) -> 2    (-2,1) -> 1.5 etc
So - at (1,1) I draw an arrow with gradient zero, at (2,1) gradient 0.5 etc.  What length should the arrow be? How do I figure out the direction of each arrow?
 A: Your system has the general form
\begin{align*}
  x' &= f(x, y) \bigl[= x\bigr], \\
  y' &= g(x, y) \bigl[= xy - y\bigr].
\end{align*}
There are two common ways to draw this information:

*

*Vector field: Draw the arrow from $(x, y)$ to $(x + f(x, y), y + g(x, y))$ for as many points as you like.


*Direction field: Draw a short segment of fixed length with midpoint $(x, y)$ and slope $m(x, y) := \frac{g(x, y)}{f(x, y)}$, or a vertical segment if $f(x, y) = 0$ and $g(x, y) \neq 0$, for as many points as you like. The direction of flow can be indicated by arrows, but books often draw only the slope segments.
The qualitative aim for either technique is to sketch curves "everywhere tangent" to the field. Only the direction of the field matters for this, not its magnitude from point to point. Precisely, if we multiply the right-hand side of our system by a smooth, positive function, the solutions to the new system are reparametrizations of the original solutions, and therefore trace the same curves in the $(x, y)$-plane. (This observation can be surprisingly useful.)

In sketching a direction field, we may parallelize the computations by using isoclines, curves where $m(x, y)$ is constant.
In your example, we have $m(x, y) = (xy - y)/x = y(x - 1)/x$. This is $0$ if $y = 0$ and $x \neq 0$, or if $x = 1$. Along those two curves we can immediately draw as many short horizontal segments as we want.
Similarly, $f(x, y) = 0$ where $x = 0$ and $y \neq 0$, so along the $y$-axis we have vertical segments. (At the origin, the field is $(0, 0)$, so the origin is a constant solution.)
Generally (in this example), if $m$ is a real number, we have $y(x - 1)/x = m$ along the graph $y = mx/(x - 1)$. To sketch the direction field along an isocline, pick a value of $m$ (say $m = 1$, or $-1$, or $2$, etc.) and lightly sketch the corresponding isocline. Along this curve, draw several short segments of slope $m$. With a small selection of $m$, we can often fill out the $(x, y)$-plane well enough to make a good qualitative sketch of solutions to the system of differential equations.
