3.

MODELING WITH
FIRST ORDER DIFFERENTIAL EQUATIONS

Chapter 2- have learnt several method to solve first order ODE.

Chapter 3- will use those methods to solve linear and nonlinear
ODE that commonly arise in application.

3.1 LINEAR EQUATIONS

Recall: The standard form of Linear ODE

dy
 P( x) y  f ( x)
dx  P ( x ) dx
To solve, need to find integrating factor: v( x)  e
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Important applications are:

a) Exponential Growth and Decay Problem

Assumption:
The rate of growth/decay of certain populations is
proportional to the population present at any time.
Notation:
P(t)= the population present at time t
dP
 the rate of growth/decay of certain population
dt
dP
Model: P dP
 k P where k is a constant
dt dt of proportionality.
Note:
k  0  growth constant , k  0  decay constant 2
 Examples
1. The population of bacteria in a culture grows at a rate
proportional to the number of bacteria present any time.
After 3 hours it is observed that there are 400 bacteria
present and there are 2000 bacteria after 10 hours.
What was the initial number of bacteria?
Solution
Let P(t) be the number of bacteria present at time t.
dP
k P
dt dP
Put it in standard form:  kP  0
dt d
v(t )  e  kdt
e  kt
 kt
[e P ]  0
dt 3
 e PC  kt P  Ce kt
P(t )  Ce kt

Given P(3)  400 Ce3k  400 ……..(i)

P(10)  2000 Ce 10k

 2000 ……..(ii)

Want the initial number of bacteria.

i.e What is P when t=0@ what is P(0)=P0
1
(ii)/(i): e7 k  5 7k  ln 5 k  ln 5
7
From (i): Ce 3 / 7 ln 5
 400 C  400 / e 3 / 7 ln 5
 201
ln 5
t
P(t )  201e 7 Hence P(0)  201

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 Application in physics - Half life problem
-- is a measure of stability of a radioactive substance
Half life- the time that the substance takes for 1/2 of
the initial amount of the atom to
disintegrate.

2. The radioactive isotope of lead, Pb-209, decays at a rate

propotional to the amount present at any time and has a
half-life of 3.3 hours.
If 1 gram of lead is present initially, how long will it take for
90% of the lead to decay?
Solution
Let A(t) be the amount of lead present at time t.
dA
k A k  0 - - decaying problem
dt 5
 Have A(t )  Ce kt

Given A(0)  A0  1 ……..(i)

A(3.3)  1 / 2 A0  1 / 2 ……..(ii)
Need to find t when A(t )  0.1A0  0.1
Since A(t )  Ce kt
So A(0)  C  1 From (i)

A(t )  e kt
A(3.3)  e 3.3k
 1/ 2 From (ii)

3.3k  ln 1 / 2 k  0.21
0.21t
So A(t )  e
0.21t
Want t when A(t )  0.1 e  0.1
t  ln 0.1 /  0.21  10.96 hrs.
(In about 11 hours, 90% of lead decayed) 6
 b) Newton’s Law of Cooling

Assumption:
According to Newton’s law of cooling, the rate at which a
body cools is proportional to the difference between the
temperature of the body and the temperature of the
surrounding medium (ambient temperature).
Notation:
T(t)= the temperature of a body at any time t
Tm= the constant temperature of the surrounding medium
dT
 the rate of which the body cools
dt
dT dT
Model:  (T  Tm )  k (T  Tm ) k0
dt dt 7
 Examples
1. When a cake is removed from an oven, its temperature is
measured at 3000F. Three minutes later, at room
temperature of 700F, its temperature is 2000F.
How long will it take for the cake to cool off to 730F?
Solution
Let T(t) be the temperature of the cake at time t
dT
Have  k (T-Tm )
dt
dT
Given Tm  70  k (T-70) ……..(i)
dt
T (0)  300 ……..(ii)
T (3)  200 ……..(iii)
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 dT
From (i)  k dt
T  70
ln T  70  kt  c T  70  Ce kt

T (t )  Ce kt  70
T (0)  C  70  300 From (ii)
C  230
T (3)  Ce  70  230e  70  200
3k 3k
From (iii)

e  130/230
3k
k  -0.19
0.19t
T (t )  230e  70
Need to find t when T (t )  73
0.19t
230e  70  73 t  22.8 min.
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2. It was noon on a cold December day in Tampa:160C.
Detective Taylor arrived at the crime scene to find the
sergeant leaning over the body. The sergent said there
were several suspects. If they know the exact time of
death, they could narrow the list.
Detective Taylor took out the thermometer and measured
the temperature of the body: 34.50C. He then left for lunch.
Upon returning at 1.00pm, he found the body temperature
to be 33.70C.
When did the murder occur?
(Hint: Use Newton’s Law Of Cooling & normal body
temperature is 370C)

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 Given Tm  16
T (0)  34.5 Consider at 12 noon t=0
T (1)  33.7
Need to find t when T (t )  37
dT
Have  k (T-Tm )  k (T  16)
dt
ln T  16  kt  c T  16  Ce kt

T (t )  Ce  16
kt

T (0)  C  16  34.5 C  18.5

T (t )  18.5ekt  16
T (1)  18.5e  16  33.7
k e  17.7/18.5
k

k  -0.044 11
 So have T (t )  18.5e0.044t  16
Since we need to find t when T (t )  37
0.044t
Hence 18.5e  16  37
e0.044t  21/18.5
ln (21 / 18.5)
t
0.044
 2.9 hrs.
So the murder occurred at 2.9 hours before 12 i.e 9.06 am

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c) SERIES CIRCUITS
For circuit containing only a resistor and an inductor

L R LR Series Circuit
E

Kirchhoff’s second law states that the sum of the voltage

drop across the inductor (L(di/dt)) and the voltage drop
across the resistor (iR) is the same as the impressed
voltage (E(t)) on the circuit.
Thus the linear DE for the current i(t),
di
L  Ri  E (t ) First order DE
dt
where L and R are constants known as inductance and resistance.
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The current i (t) is also called the response of the system.
For circuit containing only a resistor and a capasitor

E R RC Series Circuit

C
The voltage drop across a capacitor with capacitance C
=q (t)/C, where q is the charge on the capacitor.
Kirchhoff’s second law gives
1
Ri  q  E (t )
C
But i =dq/dt So have:
dq 1
R  q  E (t ) First order DE
dt C
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 Units in series circuit

quantity unit
Capasitance- C farad (F)
Inductance -L henry (H)
Resistance - R ohms (  )
Current- i Ampere (A)
Charge -q coulombs (C)

Voltage (E) volts

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 Examples
1. A 12-volt battery is connected to a series circuit in which
the inductance is ½ henry and the resistance is 10 ohms.
Determine the current i if the initial current is zero.
Solution Have LR series circuit.
di
We have L  Ri  E (t )
dt
di
So have: 1 / 2  10i  12
dt
di
 20i  24
dt

v(t )  e  20dt
e 20t
d 20t
[e i]  24e 20t

dt 16
 d 20t
[e i]  24e 20t

dt
Integrate both side: e i  24 / 20e  C
20t 20t

20t
i(t )  6 / 5  Ce
Given when t=0, i=0 C  6 / 5
20t
So i(t )  6 / 5  6 / 5e

Suggested Exe.:3.1:1-10;13-20, 29-34.

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