Brief discussion of Problem 18, page 628

We start first with the graph of the right hand side of the equation:

> plot((y+2)*(y+1)*(y-1)*(y-2),y=-3..3,title=`Graph of f(y)`);

Keep in mind that the important information here is the sign of f(y) (and consequently that of dy/dt).

We notice that if y(0) is in one of the intervals (minus_infinity, -2), (-1,1), or (2, plus_infinity), the

derivative is positive so the solution is increasing (towards -2, 1, and plus_infinity, respectively).

If y(0) is in one of the intervals (-2, -1) or (1, 2), the derivative is negative so the solution is

decresing (towards -2, and 1, respectively). Because of this behavior, -2 and 1 are called

stable critical points (`the system tends to approach these values'), where -1 and 2 are

called unstable critical points (`the system will not remain near these values'). This is best

viewed by drawing the direction field of the DE:

> with(DEtools):

> DE := diff(y(t),t)=(y+2)*(y+1)*(y-1)*(y-2):

> DEplot( DE ,y(t), t=-5..5, y=-3..3,
[[y(0)=-2],[y(0)=-1],[y(0)=2],[y(0)=1]], arrows=SMALL,
dirgrid=[30, 30], stepsize=.1,
linecolor=black, color=BLUE);

Warning, y is present as both a dependent variable and a name. Inconsistent specification of the dependent variable is deprecated, and will be removed in the next release.

Note . The black curves on the picture represent the "critical points"=constant solutions.