Barrons AP Calculus - David Bock

where (in both) f (t) is the amount at time t and k and A are both positive. We may conclude that

(a) f (t) is increasing (Fig. N9–8a): (b) f (t) is decreasing (Fig. N9–8b):

f (t) = A − ce−kt f (t) = A+ ce−kt

for some positive constant c.

FIGURE N9–8a

FIGURE N9–8b
Here is how we solve the d.e. for Case II(a), where A − y > 0. If the quantity at time t is denoted
by y and k is the positive constant of proportionality, then

Case II (b) can be solved similarly.