Plotting the direction field of a differential equation

Question

I have to sketch the direction field for the following differential equation:

$$\frac{dy}{dx}=\frac{-0.02 y +0.00002 xy}{0.08 x-0.001xy}$$

This is the code I used, which gives an incorrect plot:

StreamPlot[Normalize[{1, (y (-0.016 + 0.00008 x))/(x (0.12 - 0.006 y))}], 
    {x, -200, 200}, {y, -200, 200}, Axes -> True]

The following picture shows what I need to get:

Answer 1

I believe this is getting close to what you want:

f[x_, y_] := (-0.02 y + 0.00002 x y)/(0.08 x - 0.001 x y)

VectorPlot[{1, f[x, y]}, {x, 0, 3100}, {y, 0, 150}, VectorStyle -> Arrowheads[{}]]

Answer 2

I think the following is a bit closer

f[x_, y_] := (-0.02 y + 0.00002 x y)/(0.08 x - 0.001 x y)
{X1, X2} = {0, 3100};
{Y1, Y2} = {0, 150};
AR = 0.5;
length = 0.04;
VectorPlot[{1, f[x, y]}, {x, X1, X2}, {y, Y1, Y2}, AspectRatio -> AR, 
 PlotRange -> {{X1, X2}, {Y1, Y2}}, VectorStyle -> Arrowheads[{}], 
 VectorScale -> {length, Automatic, If[#5 > 0, #5/Sqrt[#3^2 + (AR(X1-X2)#4/(Y1-Y2))^2],0] &}]

This solution gives the equal lengths of the arrows with taking into account the aspect ratio.

Answer 3

I'll assume that what you really want is a StreamPlot and not a vector plot, because that's in your code.

The equation you're plotting in the question isn't the one in the first equation. But even if we correct this, the StreamPlot looks bad because it cuts off the automatically generated streamlines before they are long enough to outline the shape of the slope field.

To remedy this, you can try specifying a minimum length for the streamlines, and also choose them to go through the interesting points in the plot. I've taken your (corrected) StreamPlot and added the necessary StreamPoints option:

StreamPlot[
 Normalize[{1, (-0.02 y + 0.00002 x y)/(0.08 x - 
      0.001 x y)}], {x, 0, 3000}, {y, 0, 150}, Axes -> True, 
 StreamPoints -> {Table[{1040, i}, {i, 13, 150, 5}], Automatic, 3000},
  PlotRange -> All]