Lecture – 06
Applications of First Order Des(Growth and Decay)
Applications of First Order Differential Equations
In order to translate a physical phenomenon in terms of mathematics, we strive for a set
of equations that describe the system adequately. This set of equations is called a Model
for the phenomenon. The basic steps in building such a model consist of the following
steps:
Step 1: We clearly state the assumptions on which the model will be based. These
assumptions should describe the relationships among the quantities to be studied.
Step 2: Completely describe the parameters and variables to be used in the model.
Step 3: Use the assumptions (from Step 1) to derive mathematical equations relating the
parameters and variables (from Step 2).
The mathematical models for physical phenomenon often lead to a differential equation
or a set of differential equations. The applications of the differential equations we will
discuss in next two lectures include:





Population dynamics.
Radioactive decay.
Carbon dating.
Newton’s Law of cooling.
Orthogonal Trajectories.
etc.
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Lecture – 06
Applications of First Order Des(Growth and Decay)
Population Dynamics
Some natural questions related to population problems are the following:


What will the population of a certain country after e.g. ten years?
How are we protecting the resources from extinction?
The easiest population dynamics model is the exponential model. This model is based
on the assumption:
The rate of change of the population is proportional to the existing population.
If P (t ) measures the population of a species at any time
mentioned assumption we can write
t
then because of the above
dP
 kP
dt
where the rate k is constant of proportionality. Clearly the above equation is linear as
well as separable. To solve this equation we multiply the equation with the integrating
factor e  kt to obtain
d   kt 
Pe
0

dt 
Integrating both sides we obtain
P e  kt  C
or
P  C e kt
If P0 is the initial population then P(0)  P0 . So that C  P0 and obtain
P(t )  P0 e kt
Clearly, we must have
k 0
for growth and
k 0
for the decay.
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Lecture – 06
Applications of First Order Des(Growth and Decay)
Illustration
Example:
The population of a certain community is known to increase at a rate proportional to the
number of people present at any time. The population has doubled in 5 years, how long
would it take to triple?. If it is known that the population of the community is 10,000
after 3 years. What was the initial population? What will be the population in 30 years?
Solution:
Suppose that
time
P0 is initial population of the community and P (t ) the population at any
t then the population growth is governed by the differential equation
dP
 kP
dt
As we know solution of the differential equation is given by
P(t )  P0 e kt
Since P(5)  2 P0 . Therefore, from the last equation we have
2P0  P0 e5k  e5k  2
This means that
5k  ln 2  0.69315 or
k
0.69315
 0.13863
5
Therefore, the solution of the equation becomes
P(t )  P0 e0.13863 t
If t1 is the time taken for the population to triple then
3P0  P0 e
t1 
0.1386t1
e
0.1386t1
3
ln 3
 7.9265  8 years
0.1386
Now using the information P(3)  10,000 , we obtain from the solution that
10,000  P0 e
(0.13863 )(3)
 P0 
10,000
e 0.41589
Therefore, the initial population of the community was
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Lecture – 06
Applications of First Order Des(Growth and Decay)
P0  6598
Hence solution of the model is
P(t )  6598 e 0.13863 t
So that the population in 30 years is given by
P(30)  6598e
(30)(0.13863 )
 6598e 4.1589
P30  659864.0011
or
P30  422279
or
Radioactive Decay
In physics a radioactive substance disintegrates or transmutes into the atoms of another
element. Many radioactive materials disintegrate at a rate proportional to the amount
present. Therefore, if A(t ) is the amount of a radioactive substance present at time t ,
then the rate of change of A(t ) with respect to time t is given by
dA
 kA
dt
where
k is a constant of proportionality.
Let the initial amount of the material be A0
then A(0)  A0 . As discussed in the population growth model the solution of the
differential equation is
A(t )  A0 e kt
The constant
k can be determined using half-life of the radioactive material.
The half-life of a radioactive substance is the time it takes for one-half of the atoms in an
initial amount A0 to disintegrate or transmute into atoms of another element. The halflife measures stability of a radioactive substance. The longer the half-life of a substance,
the more stable it is. If
T
denotes the half-life then
A(T ) 
A0
2
Therefore, using this condition and the solution of the model we obtain
A0
 A0 e kt
2
So that
kT   ln 2
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Lecture – 06
Applications of First Order Des(Growth and Decay)
Therefore, if we know T , we can get k and vice-versa. The half-life of some important
radioactive materials is given in many textbooks of Physics and Chemistry. For example
the half-life of C  14 is 5568  30 years.
Example:
A radioactive isotope has a half-life of 16 days. We have 30 g at the end of 30 days.
How much radioisotope was initially present?
Solution: Let A(t ) be the amount present at time t and A0 the initial amount of the
isotope. Then we have to solve the initial value problem.
dA
 kA,
dt
A(30)  30
We know that the solution of the IVP is given by
A(t )  A0 e kt
If
T
the half-life then the constant is given
kT   ln 2
Now using the condition
k
or k  
by
ln 2
ln 2

T
16
A(30)  30 , we have
30  A0e30k
So that the initial amount is given by
30 ln 2
A0  30e  30k  30e 16  110.04 g
Example:
A breeder reactor converts the relatively stable uranium 238 into the isotope plutonium
239. After 15 years it is determined that 0.043% of the initial amount
A0
of the
plutonium has disintegrated. Find the half-life of this isotope if the rate of disintegration
is proportional to the amount remaining.
Solution:
Let A(t ) denotes the amount remaining at any time t , then we need to find solution to
the initial value problem
dA
 kA,
dt
A(0)  A0
which we know is given by
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Lecture – 06
Applications of First Order Des(Growth and Decay)
A(t )  A0 e kt
If 0.043% disintegration of the atoms of A0 means that 99.957% of the substance
remains. Further 99.957% of A0 equals (0.99957) A0 . So that
A(15)  0.99957 A0
So that
A0 e15k  0.99957 A0
15k  ln( 0.99957)
ln( 0.99957)
 0.00002867
15
A(t )  A0 e  0.00002867 t
k
Or
Hence
If T denotes the half-life then A(T ) 
A0
. Thus
2
A0
 A0e 0.00002867 T
2
1
 e 0.00002867 T
2
1
 0.00002867 T  ln    ln 2
 2
T
or
ln 2
 24,180 years
0.00002867
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