The graph of the solution is shown above.  Click on the graph to see an animation of the direction field vectors moving across the screen for increasing values of x along with an animated solution point.

y_o = 1   y_o = 2   y_o = 3   y_o = 4

QT  animation as y(0) varies from 0 to 40.

dy/dx = .01y(100-y),   y(0) = 10.

The solution is y = 100e^x / (9 + e^x).

Here is an Animation (Quicktime version) of the changing graph of the solution as y(0) varies from -50 to 150 with x between 0 and 10 and here is an Animation (Quicktime version) with x between -10 and 10.  Notice the significance of the blue horizontal lines and their relationship to the zeroes of .01y(100-y).  Look at examples 3 and 4 in Section 2.1 discussing autonomous first order differential equations (DE's of the form F(y,y') = 0 or in normal form dy/dx = f(y)).  If the DE was modeling a population then y(0) would have to be positive.  If y(0) = y_o then

y = 100 is a singular solution (see pages 7-8 in your text).

Note:  y = 0.9 is a singular solution that could not be obtained from the general solution shown above.  Click here to see animated solution graphs as y_o varies from -2.1 to 3.9 demonstrating the "missing" singular solution.  Quicktime version

Solution graphs for y_o = 1

Click here to see animated solution graphs as y_o varies from -3 to 3.  Quicktime version

The picture on the right shows the graphs of particular solutions with y₀ = 1/4 (blue) and y₀ = -1/4 (red).  Click here or on the picture to see an animation of solutions as y₀ varies from -3 to 3.  In the case of the blue solution on the right, does y continue to increase as x continues to increase beyond 8?  Click here for a pictorial answer.

Quicktime Version