4.4.1 applications of first oder ODE (population growth & bacterial colony)

Applications of first order ODE
 Population Growth & Bacterial Colony
 Radioactive Decay
 Newton’s Law of Cooling
 Mixing Problems
 Kirchhoff Law

MOOC MAT438/ UiTM

At the end of this session, the students should be able to
❶ Understand the terms ‘exponential growth/decay’, ‘growth/decay constant’,

‘doubling time’ and ‘half-life’
❷ Solve problems involving applications of Separable Differential Equation

MOOC MAT438/ UiTM

Using Separable equation

Where,

∶ Population @ radioactive mass Doubling time : =? when = 2 0

Triple : =? when = 3 0

∶ time Quadruple : =? when = 4 0

Population Growth &

> 0 constant growth rate Bacterial Colony

< 0 constant decay rate Half-life : =? when = 1 0
2
MOOC MAT438/ UiTM

/ MAT235/ Sep 2014/ Q3c/ 7 marks

200 crabs are placed in a pond that can support a population of 2000 crabs. After one month, the
population of the crabs had increased to 500. How long will it take for the pond to reach full capacity?

= 0, = 200 Integrating both sides, 
= 1 , = 500
=? = 2000 1
=
= +
From =
= +
Separating variables t and P, = ∙
=
1
=

MOOC MAT438/ UiTM

/ MAT235/ Sep 2014/ Q3c/ 7 marks

200 crabs are placed in a pond that can support a population of 2000 crabs. After one month, the
population of the crabs had increased to 500. How long will it take for the pond to reach full capacity?

= 0, = 200 to find C To find k, substitute = 1 , = 500 into 
= 1 , = 500 to find k
=? = 2000 500 = 200 1

=  = 500
200

To find C, substitute = 0, = 200 into  = 2.5
from =
200 = 0 1 = 2.5 
= 200 (into ) = 0.91629 (into )
→ = 200 0.91629

→ = 200 

MOOC MAT438/ UiTM

/ MAT235/ Sep 2014/ Q3c/ 7 marks

200 crabs are placed in a pond that can support a population of 2000 crabs. After one month, the
population of the crabs had increased to 500. How long will it take for the pond to reach full capacity?

= 0, = 200 to find C 0.91629 = 10

= 1 , = 500 to find k

=? = 2000 Into  0.91629 = 10

An expressions for the population at any time t, 10 = 2.513 months
= 0.91629

= 200 0.91629 

it took 2.513 months for the pond to reach full capacity.

substitute =? = 2000 into 

2000 = 200 0.91629

0.91629 = 2000
200

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

Extract the information
(t and P)

A population of butterfly in a garden grow at a rate that is proportional to their current population. If there
are initially 300 butterflies in the garden and in two weeks time , the population become 500, estimate
i) the population after one month.
ii) the doubling time for the butterfly’s population.

= 0, = 300 Integrating both sides,

= 14 , = 500 1
=
i) = 30 , = ? = +

ii) =? , = 600 Doubling = +
time = ∙
=

From =

Separating variables t and P, 

1
=

MOOC MAT438/ UiTM

A population of butterfly in a garden grow at a rate that is proportional to their current population. If there
are initially 300 butterflies in the garden and in two weeks time , the population become 500, estimate
i) the population after one month.
ii) the doubling time for the butterfly’s population.

= 0, = 300 to find C To find k, substitute = 14 , = 500 into 
to find k
= 14 , = 500 500 = 300 14

i) = 30 , = ? 500
300
ii) =? , = 600 14 =

=  14 = 5
3

To find C, substitute = 0, = 300 into  5
3
from = 14 =
300 = 0 1
= 300 into  = 0.03648 into

→ = 300 → = 300 0.03648 



MOOC MAT438/ UiTM

A population of butterfly in a garden grow at a rate that is proportional to their current population. If there
are initially 300 butterflies in the garden and in two weeks time , the population become 500, estimate
i) the population after one month.
ii) the doubling time for the butterfly’s population.

= 0, = 300 to find C ii) substitute =? , = 600 into 

= 14 , = 500 to find k = 300 0.03648

i) = 30 , = ? Into  600 = 300 0.03648

ii) =? , = 600 0.03648 = 600
300
An expressions for the population at any time t,
0.03648 = 2
= 300 0.03648 

i) substitute = 30 , =? into  0.03648 = 2
= 300 0.03648(30) = 896.21691
= 19 days
There are 896 butterflies after 1 month.
Thus, it took 19 days for the population to become
double from the original population.

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

The initial number of a bacteria colony is 100. After 5 hours, the numbers of bacteria become quadruple.
Assuming that the number of bacteria increases at a rate proportional to the number of the bacteria at time
t, find the number of bacteria after 1 day.

= 0, = 100 quadruple Integrating both sides,
= 5 ℎ, = 400
= 24 ℎ, =? 1
=
= +
From =
= +
Separating variables t and P, = ∙ 
=
1
=

MOOC MAT438/ UiTM

The initial number of a bacteria colony is 100. After 5 hours, the numbers of bacteria become quadruple.
Assuming that the number of bacteria increases at a rate proportional to the number of the bacteria at time
t, find the number of bacteria after 1 day.

= 0, = 100 to find C To find k, substitute = 5 ℎ , = 400 into 
= 5 ℎ, = 400 to find k
= 24 ℎ, =? 400 = 100 5

=  5 = 400
100

To find C, substitute = 0, = 100 into  5 = 4
from =
100 = 0 1 5 = 4 into
= 100 into  = 0.27725

→ = 100 0.27725 

→ = 100 

MOOC MAT438/ UiTM

The initial number of a bacteria colony is 100. After 5 hours, the numbers of bacteria become quadruple.
Assuming that the number of bacteria increases at a rate proportional to the number of the bacteria at time
t, find the number of bacteria after 1 day.

= 0, = 100 to find C
= 5 ℎ, = 400 to find k
= 24 ℎ, =?
Into 

An expressions for the population at any time t,

= 100 0.27725 

substitute = 24 ℎ, =? into 

= 100 0.27725(24)
= 77,588.165
There are 77,588 bacteria after 1 day.

MOOC MAT438/ UiTM

The initial number of a bacteria colony is 100. After 5 hours, the numbers of bacteria become quadruple.
Assuming that the number of bacteria increases at a rate proportional to the number of the bacteria at time
t, find the number of bacteria after 1 day.

= 100 0.27725(24)
= 77,588.165
There are 77,588 bacteria after 1 day.

0.27725
100 24

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM