6.2 Differential Equations: Growth and Decay

```6.2
Growth and Decay
Law of Exponential Growth and Decay
y  Ce
kt
C = initial value
k = constant of proportionality
if k &gt; 0, exponential growth occurs
if k &lt; 0, exponential decay occurs
Solving a Differential Equation
y'
2x
Multiply both sides by y.
y
yy '  2 x
Integrate both sides with respect to x.

yy 'dx 

y dy 
y
 2 x dx
dy = y’dx because dy/dx = y’
 2 x dx
2
 x C
2
Multiply both sides by 2.
2
y  2x  C
2
2
C = 2C
Exponential Growth and Decay Model
If y is a differential function of t such that y &gt;0
and y’ = ky, for some constant k, then
Y = Cekt
C is the initial value of y, and k is the
proportionality constant. Exponential growth
occurs when k &gt; 0, and exponential decay
occurs when k &lt; 0.
Proof:
y’ = ky
y'
k
y

y'

1
dy
dt 
 k dt
dy 
 k dt
y
y
ln y  kt  C
ye e
kt
C1
 y'
dt
 dy  y ' dt

y  Ce
kt
So, all solutions of y’ = ky are in the form of y = Cekt
Ex.
The rate of change of y is proportional to y.
When t = 0, y = 2. When t = 2, y = 4. What
is the value of y when t = 3?
Because y’ = ky we know that y = Cekt.
We can find the values of C and k by applying the
initial conditions.
2 = Ce0
4=
C=2
k 
2e2k
1
2
ln 2  .3466
y  2e
When t = 3, the value of y is 2e0.3466(3) = 5.657
So, the model is

0.3466 t
Newton’s Law of Cooling
Let y represent the temperature (in oF) of an object
in a room whose temperature is kept at a constant 60o.
If the object cools from 100o to 90o in 10 minutes, how
much longer will it take for its temperature to decrease
to 80o?
From Newton’s Law of Cooling, the rate of change in
y is proportional to the difference between y and 60.
y '  k  y  60 
dy
dt
 k  y  60 
dy
y  60
 kdt
Separate variables first.

dy
y  60

 kdt
ln y  60  kt  C 1
ln y  60  kt  C 1
y  60  e
The model is: y  60  40 e
kt  C 1
y  60  Ce
Take e to both sides.
C e
kt
.02877 t
C1
Using y = 100 when t = 0,solve for C.
100  60  Ce
40 = C
k( 0)
Since y = 90 when t = 10
90  60  40 e
30  40 e
 60  C
k (10 )
k (10 )
Finally, when y = 80, you obtain
80  60  40 e
 .02877 t
t  24 .09 min
So, it will require 14.09 more minutes
k = -0.02877 for the object to cool to 80o.
Ex. 2 Money is deposited in an account for which
interest is compounded continuously. If the balance
doubles in 6 years, what is the annual percentage rate?
A = Pert
2P = Pert
2 = e6r
ln 2 = 6r
r  . 1155  11 . 55 %
Ex. 1 A sample contains 1 gram of radium. How
much radium will remain after 1000 years? (Use
a half-life of 1620 years.)
1
 1e
k 1 , 620

2
ln
ln
First we need to find k,
the constant of proportionality.
Take the ln of both sides.
1
2
1
2
 ln e
1620 k
k   . 0004279
y  1e
 1620 k
 . 0004279 (1000 )
y  . 652 g
Ex.
Suppose that 10 grams of the plutonium isotope
Pu-239 was released in the Chernobyl nuclear
accident. How long will it take for the 10 grams
to decay to 1 gram? Pu-239 has a half life of 24,100 years
y = Cekt
We know that C = 10 grams at
time t = 0. First, find k.
5 = 10ek(24,100)
1
2
e
24 ,100 k
k
1
24 ,100
ln
1
2
k  .000028761
So, the model is
To find the time it would take for
y = 10e-.000028761t  10 grams to decay to 1 gram, solve
for t in 1 = 10e-.000028761


t = 80,059 years
```