6.3 Differential Equations of Exponential Growth and Decay
Let y  f x  . Then, the relative rate of growth is
rate of growth
f ' x  d ln  f x 


.
size of f x 
f x 
dx
Suppose the relative rate of growth is a constant. Then,
f ' x 
   f ' x   f x   y '  y .
f x 
Next question is how to find y  f x  ? There are two approaches to
find y  f x  :
(approach 1):
d ln  f x  f ' x 

   ln  f x   x  c1  f x   exp x  c1 
dx
f x 
 f x   c exp x , c  exp c1 
(approach 2):
f ' x   f x   f ' x exp  x   f x exp  x 
 f ' x exp  x   f x exp  x  
 exp  x  f x   c
d exp  x  f x 
0
dx
 f x   c exp x 
Example 1:
Suppose 10 grams Pu-239 were released in the Chernobyl nuclear accident. How long
will it take for the 10 grams to decay to 1 gram? Note that the half-life of Pu-239 is
24360 years.
[solutions:]
Let
yt 
be the mass of Pu-239.
1




y t   c exp t   y 0  10  c exp   0  c  10
1
1
10  exp   24360  10    exp 24360 
2
2
 ln 1 t 
ln 1
2
2   10exp  0.000028454t 

 y t   10 exp 
24360
 24360 


1  10exp  0.000028454t 
ln 10
t
 80922
0.000028454
 
 
Example 2:
Suppose an experimental population of fruit flies increases according to the law of
exponential growth. There were 100 flies after the second day of the experiment and
300 flies after the fourth day. Approximately how many flies were in the original
population?
[solutions:]
Let
yt 




be the number of flies at time t.
 y 2  100  c exp   2
y t   c exp t   
 y 4  300  c exp   4
y 2 1
ln 3
  exp  2    
 0.5493
y 4 3
2
100  c exp 0.5493  2  c  100 exp  1.0986  33
y t   33 exp 0.5493t 
y 0  33 exp 0.5493  0  33
Example 3:
Four months after it stopped advertising, a manufacturing company noticed that its
sales had dropped from 100000 units per month to 80000 units per month. If the sales
follow an exponential pattern of decline, what will they be after another 2 months?
[solutions:]
Let
yt 
be the sales at time t.
2
y t   c exp t   100000  c exp   0  c  100000
 80000  100000exp   4   
ln 0.8
 0.0558
4
 y t   100000 exp  0.0558t 
 y 6  100000 exp  0.0558  6  71500
3