Suppose we have a first order autonomous equation  y' = f(y).
In many cases f(y) is quite complicated, and it is very reasonable
to make an approximation assuming that y is near  a fixed value p.
We will carry out this discussion for an initial value of t = 0.

The first approximation would be  y' = f(p) assuming h = y-p is
small.  Then the solutions are solutions to h' = f(p) or
y(t) - p = h(t) = y(0)- p  + f(p)t, or y(t) = y(0) +tf(p). This
is the equation of the line segment used for the direction fields,
and is valid only near y = p.

When p is a critical point,  f(p) = 0 and this is not very much
information.  So we use the next approximation to f, namely near
y = p we have

       f(y)= f(p) + f'(p)(y - p)  plus error of size (y-p)^2.


Since at a critical  point f(p) = 0, the approximate differential
equation would be

       y' ~    f'(p)(y - p).

This is ONLY valid for  h = y-p small.  In terms of h, the equation
is    

       h' = f'(p)h.

Then   h = kexp(f'(p)t    and y ~ h + p = p + k exp (f'(p)t).

When f'(p) < 0, solutions decay exponentially towards the critical point
p near p (a stable equilibrium).  When f'(p) > 0, solutions decay
exponentially away from the critical point p (unstable equilibrium).



Example:   Find the linear approximation to the equation

  y' = 2 y^2 - 5 y   at the critical point y = 0.

This is easy.  We assume y is small, so y^2 is very small compared
to y, so the approximation  at y = 0 is

        y' = - 5y.

Example:  y ' =- y (3 -4y + y^2).  The critical points are p = 0, 1, 3.

We compute  f'(y) = -3 + 8y -3y^2.  the the approximations are:

    h' =-3h   for h = y near y = 0  (stable)
    h' = 2h  for h = y- 1 near y = 1 (unstable)
    h' = -6h  for h = y-3  near y = 3 (stable).

If you are asked to approximated compute the solution y(t) 
for y(0) = 3.2, a reasonable guess would be to use the
approximation   y(t) = h(t) + 3, h small and y(t) near 3 so

       y(t) = h(t) + 3 = h(0)exp(-6t)+ 3 = .2exp(-6t) + 3.

We not only see that solutions  are converging to the critical
value of 3 as t goes to infinity, but we can read off the rate of
convergence.(Note...sorry students...the -6 was a -15 the first time
around).