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WEEK 4-5 ULOa
Week 4-5:
Unit Learning Outcomes (ULO): At the end of the unit, you are expected
to
a. solve problems involving population growth, decay, cooling and
heating and flow.
b. solve problems involving mechanics and electric circuits.
Big Picture in Focus: ULOa. solve problems involving population
growth, decay, cooling and heating and flow.
Metalanguage
In this section will serve as your word bank where the most essential terms
relevant to the introduction of the application of first-order differential equations and
ULO-a will be operationally defined to establish a common frame of reference. You
will encounter these terms as we delve deeper to the study of differential equation.
Please refer to these definitions in case you will encounter difficulty to understand
mathematical concepts in relation with applications of first–order differential
equations.
1. Formulating proportionality problems:
a. The decomposition rate is dR/dt, which is proportional to R. Thus, dR/dt =
kR, where k is a constant of proportionality and R(t) denote the amount
present at time t.
b. The growth rate is dN/dt, which is proportional to N. Thus, dN/dt = kN,
where k is a constant of proportionality and N(t) denote the amount present
at time t.
2. Newton’s law of cooling. It is the rate at which a hot body cools is proportional
to the difference in temperature between the body and the (cooler) surrounding
medium.
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3. Flow problems: It is the time rate of change of dQ/dt, equals the rate at which
salt enters the tank minus the rate at which salt leaves the tank. Salt enters the
tank at the rate of be lb/min.
Please proceed immediately to the “Essential Knowledge” part since
the first lesson is also definition of essential terms.
Essential Knowledge
To perform the aforesaid big picture (unit learning outcomes) for the first
three (3) weeks of the course, you need to fully understand the following essential
knowledge that will be laid down in the succeeding pages. Please note that
you are not limited to exclusively refer to these resources. Thus, you are
expected to utilize other books, research articles and other resources that are
available in the university’s library (refer to the library contact on page 3).
Applications of first- order differential equations
a. Population Growth Problems
Example 38:
A certain population of bacteria is known to grow at a rate
proportional to the amount present in a culture that provides plentiful food
and space. Initially there are 250 bacteria, and after seven hours 800
bacteria are observed in the culture. Find an expression for the
approximate number of bacteria present in the culture at any time t and
determine the approximate number of bacteria that will be present in the
culture after 24 hours.
Solution:
Let N = the number of bacteria present
dN/dt = the growth rate at any time t
dN/dt = kN where, k is a constant of proportionality
Using separable
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∫dN/N = ∫kdt → ln N = kt + c →N = ekt + c = ekt ec = c1ekt
where, c1 = ec
Therefore, N = c1ekt
Applying initial condition
At t = 0, we are given N = 250
250 = c1ek(0) → c1= 250
So the solution becomes N = 250ekt
At t = 7 hours, we are given N = 800
800 = 250ek(7)
k = [ln(800/250)]/7 = 0.166
The particular solution is
N = 250e0.166t
To determine the approximate number of bacteria present at any time t
At t = 24 hours, N = ?
N = 250e0.166x24
N = 13,433
Example 39:
The population of a certain country is known to increase at a rate
proportional to the number of people presently living in the country. If after
2 years the population has doubled, and after 3 years the population is
20,000, find the number of people initially living in the country.
Solution:
Let N = the number of people present
N0 = the number of people initially living in the country
dN/dt = the growth rate at any time t
dN/dt = kN where, k is a constant of proportionality
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Using separable
∫dN/N = ∫kdt → ln N = kt + c →N = ekt + c = ekt ec = c1ekt
where, c1 = ec
Therefore, N = c1ekt
Applying initial condition
At t = 0, we are given N = N0
N0 = c1ek(0) → c1= N0
So the solution becomes N = N0ekt
At t = 2, N = 2N0
We get
2N0 = N0e2k →k = 0.347
The solution finally becomes N = N0e0.347t
At t = 3, N = 20,000
Therefore,
N0 = 7062 the number of people initially living in the country
b. Decay Problems
Example 40:
A certain radioactive material is known to decay at a rate
proportional to the amount present. If initially there is 100 mg of the
material present and if after 2 years it is observed that 5 percent of the
original mass has decayed, find an expression for the mass at any time t
and determine the time necessary for 10 per cent of the original mass to
decay.
Solution:
Let N = the number of radioactive material present
dN/dt = the decomposition rate at any time t
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dN/dt = kN where, k is a constant of proportionality
Using separable
∫dN/N = ∫kdt → ln N = kt + c →N = ekt + c = ekt ec = c1ekt
where, c1 = ec
Therefore, N = c1ekt
Applying initial condition
At t = 0, we are given N = 100
100 = c1ek(0)→c1 = 100
So the solution becomes N = 100ekt
At t = 2, 5 percent of the original mass of 100 mg, or 5 mg, has decayed
N = 100 – 5 = 95 mg
We get,
95 = 100e2k →k = [ln(95/100)]/2 = - 0.0256
The amount of radioactive material present at any time t
N = 100e – 0.0256t
When 10 percent of the original mass to decay
N = 100 – 0.1(100) = 90 mg t = ?
90 = 100e – 0.0256t
t = 4.12 years
Example 41:
A certain radioactive material is known to decay at rate proportional
to the amount present. If after 1 h it is observed that 10 percent of the
material has decayed, find the half-life of the material.
Solution:
Let N = the amount of the material present at time t.
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N0 = the amount of the material initially present.
dN/dt = the decomposition rate at any time t
dN/dt = kN where, k is a constant of proportionality
Using separable
∫dN/N = ∫kdt → ln N = kt + c →N = ekt + c = ekt ec = c1ekt
where, c1 = ec
Therefore, N = c1ekt
Applying initial condition
At t = 0, we are given N = N0
N0 = c1ek(0) → c1= N0
So the solution becomes N = N0ekt
At t = 1, 10 percent of the original mass N0 has decayed, so 90 percent
remains. Hence N = 0.9N0
0.9N0 = N0ek(1) →k = - 0.105
The solution finally becomes N = N0e – 0.105t
When N is half-life N = ½ N0 t = ?
Therefore,
½ N0 = N0e – 0.105t
t = 6.60 h
c. Cooling and Heating Problems
Newton’s law of cooling states that the time rate of change of the
temperature of a body is proportional to the temperature difference
between the body and its surrounding medium.
Newton’s law of cooling can be formulated dT/dt = - k(T – Tm)
Where T = temperature of the body
Tm= temperature of the surrounding medium
dT/dt = the time rate of change of the temperature of the body
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k = is a constant of proportionality
Example 42:
A metal bar at a temperature of 1000F is place in a room at a constant
temperature of 00F. If after 20 min the temperature of the bar is 50 0F, find
an expression for the temperature of the bar at any time and the time it will
take for the bar to reach a temperature of 250F.
Solution:
Let T = temperature of the body
Tm= temperature of the surrounding medium
dT/dt = the time rate of change of the temperature of the body
k = is a constant of proportionality
Using Newton’s law of cooling formula
dT/dt = - k(T – Tm) → note: Tm = 00F
dT/dt = - k(T – 0)
Apply separable
∫dT/T = - ∫kdt
Integrating
ln T = - kt + c → T = e – kt + c = e – kt ec
T = c1e – kt where c1 = ec
Apply initial condition at t = 0 T = 1000F
1000F = c1e – k(0) →c1 = 1000F
Therefore, T = 100e- kt
At t = 20 min T = 500F
50 = 100e- k(20) →k = [ln(50/100)]/-20 = 0.035
The temperature of the bar at any time t
T = 100e- 0.035t
When T = 25 t = ?
25 = 100e- 0.035t →t = [ln(25/100)]/-0.035
t = 39.6 min.
Example 43:
A body at an unknown temperature is placed in a room which is
held at a constant temperature of 300F. If after 10 min the temperature of
the body is 00F and after 20 min the temperature of the body is 150F, find
an expression for the temperature of the body at time t and the initial
temperature of the body, just as it is placed into the room.
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Solution:
Let T = temperature of the body
Tm= temperature of the surrounding medium
dT/dt = the time rate of change of the temperature of the body
k = is a constant of proportionality
Using Newton’s law of cooling formula
dT/dt = - k(T – Tm) → note: Tm = 300F
dT/dt = - k(T – 30) →dT/dt + kT = 30k
Apply linear first order D.E. p(t) = k q(t) = 30k
Formula: Te∫p(t)dt = ∫e∫p(t)dt q(t)dt + c
Te∫(k)dt = ∫e∫(k)dt (30k)dt + c
Tekt = ∫ekt(30k)dt + c
Tekt = ekt(30k)/k + c
T = 30 + ce – kt
Apply the condition
At t = 10 T = 0
0 = 30 + ce-k(10) →ce-10k = - 30 ……. eqn. 1
At t = 20
T = 15
15 = 30 + ce- 20k →ce-20k = - 15 ……. eqn. 2
Divide eqn. 1 and eqn. 2
ce-10k = - 30
ce-20k - 15
e10k = 2 → k = 0.069
c = - 30e10(0.069) = - 30(2) = -60
Therefore,
T = 30 – 60e – 0.069t for the temperature of the body at any time t
AT t = 0 T = ?
T = 30 – 60e- 0.069(0)
T = - 300F
d. Flow Problems
A tank initially holds V0 gal of brine that contains a lb of salt. Another
brine solution, containing b lb of salt per gallon, is poured into the tank at the
rate of e gal/min while, simultaneously, the well-stirred solution leaves the
tank at the rate of f gal/min.
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Fig. 1
Let Q denote the amount (in pounds) of salt in the tank at any time.
The rate of change of Q, dQ/dt, equals the rate at which salt enters the
tank minus the rate at which salt leaves the tank. Salt enters the tank at
the rate of be lb/min. To determine the rate at which salt leaves the tank,
we first calculate the volume of brine in the tank at any time t, which the
initial volume V0 plus the volume of brine added et minus the volume of
brine removed ft. Thus the volume of brine in the tank at any time is V0 +
et – ft. The concentration of salt in the tank at any time is then Q/(V0 + et –
ft), from which it follows that salt leaves the tank at the rate of f[Q/(V0 + et
– ft)] lb/min. Thus, dQ/dt = be - f[Q/(V0 + et – ft)], so that
dQ + f
Q = be
dt V0 + (e – f)t
At t = 0, Q = a, so we also have the initial condition Q(0) = a
Example 44:
A tank initially holds 100 gal of a brine solution containing 20 lb of
salt. At t = 0, fresh water poured into the tank at the rate of 5 gal/min, while
the well-stirred mixture leaves the tank at the same rate. Find the amount
of salt in the tank at any time t.
Solution:
Given: V0 = 100, a = 20, b = 0, e = f = 5
Formula: dQ + f
Q = be
dt V0 + (e – f)t
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dQ/dt + 5Q/[ 100 + (5 – 5)0] = (0)(5)
dQ/dt + Q/20 = 0
By separable
∫dQ/Q = - ∫dt/20
lnQ = - t/20 + c → Q = e- t/20 + c = e- t/20 ec
Q = c1e- t/20
Apply initial condition
At t = 0 Q = a = 20
20 = c1e- (0)/20 →c1 = 20
Therefore,
Q = 20e – t/20
Example 45:
A 50-gal tank initially contains 10 gal of fresh water. At t = 0, a brine
solution containing 1 lb of salt per gallon is poured into the tank at the rate
of 4 gal/min, while the well-stirred mixture leaves the tank at the rate of 2
gal/min. Find the amount of time required for overflow to occur and the
amount of salt in the tank at the moment of overflow.
Solution:
Given: V0 = 10, a = 0, b = 1, e = 4, f = 2
From the volume of brine in the tank at any time is
V0 + et – ft = 10 + 4t – 2t = 10 + 2t
IF we required t when overflow to occur.
Therefore, 10 + 2t = 50 → t = 20 min.
If we asked the amount of salt in the tank at the moment of overflow
Formula: dQ + f
Q = be
dt V0 + (e – f)t
dQ/dt + 2Q/[ 10 + (4 –2)t] = (1)(4)
dQ/dt + 2Q/(10 + 2t) = 4
By linear differential equation
p(t) = 2/(10 + 2t)
q(t) = 4
∫p(t)dt
Formula: Qe
= ∫e∫p(t)dt q(t)dt + c
∫2dt/(10
+
2t)
∫2dt/(10
+ 2t) (4)dt + c
Qe
= ∫e
Integrating,
Qeln(10 + 2t) = ∫eln(10 + 2t) (4)dt + c note: eln(10 + 2t) = Ι10 + 2tǀ
Q(10 + 2t) = ∫(10 + 2t)(4)dt + c = 40t + 4t2 + c
Simplifying,
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Q= (40t + 4t2 + c)/(10 + 2t)
Apply the initial condition
At t = 0, Q = a = 0
0 = [40(0) + 4(0)2 + c]/[10 + 2(0)]
c=0
Q = (40t + 4t2)/(10 + 2t)
If we required Q at the moment of overflow, which t = 20
Q = [40(20) + 4(20)2]/[10 + 2(20)]
Q = 48 lb.
Self-Help: You can refer to the sources below to help you further understand the
lesson
Ferland, K. (2009). Differential Equation. Hougton Mifflin.
DuChateau, P. & Zachmann D. (2011), Schaum’s of Partrial Differential
Equations (3rd Edition). New York: McGraw-Hill.
Dobrushkin, V. A. (2015). Applied Differential Equations: the primary course,
CRC Press.
Strauss, W. (2008). Partial Differential Equation: an Introduction, John Willey &
Sons.
Let's Check
Activity4: Since you are now armed with basic knowledge to solve
problems involving population growth, decay, cooling and heating and flow, it is
now your turn to prove what you have learned in the previous discussion. Solve
the following problems.
a). Population growth problems
1. A bacteria culture is known to grow at a rate proportional to the
amount present. After one hour, 1000 bacteria are observed in the
culture; and after four hours, 3000. Determine the number of bacteria
originally in the culture. Ans.: 694
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2. A bacteria culture is known to grow at a rate proportional to the
amount present. Find an expression for the approximate number of
bacteria in such a culture if the initial number is 300 and if it is
observed that the population has increased by 20 percent after 2 h.
Ans.: 300e0.09116t; 2675
b). Decay problems
1. A certain radioactive material is known to decay at a rate proportional
to the amount present. If initially there is 50 mg of the material present
and after 2 h it is observed that the material has lost 10 percent of its
original mass, determine the mass of the material after 4 h. Ans.: 40.5
mg
2. Radium decomposes at a rate proportional to the amount present. If
half the original amount disappears in 1600 years, find the percentage
lost in 100 years. Ans.: 4.2 percent
c). Cooling and heating problems
1. A body at a temperature of 500F is placed outdoors where the
temperature is 1000F. If after 5 min the temperature of the body is 60 0F,
determine how long it will take the body to reach a temperature of 75 0F.
Ans.: 15.4 min
2. A body at a temperature of 00F is placed in a room whose
temperature is kept at 1000F. If after 10 min the temperature of the body
is 250F, find the time needed for the body to reach a temperature of
500F. Ans.: 24.1 min.
d). Flow problems
1. A tank initially holds 100 gal of a brine solution containing 1 lb of salt.
At t = 0 another brine solution containing 1 lb of salt per gallon is poured
into the tank at the rate of 3 gal/min, while the well-stirred mixture leaves
the tank at the same rate. Find the time at which the mixture contains 2
lb of salt. Ans.:0.338 min.
2. A tank initially holds 10 gal of fresh water. At t = 0, a brine solution
containing ½ lb of salt per gallon is poured into the tank at a rate of 2
gal/min, while the well-stirred mixture leaves the tank at the same rate.
Find the amount of salt in the tank at any time t. Ans.: - 5e-t/5 + 5
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Let's Analyze
Activity 4. Getting acquainted with the problems involving population
growth, decay, cooling and heating and flow is not enough. This time shows the
concepts of the applications of first-order differential equations again and
explains its step-by-step procedure. Solve the following problems.
a). Population growth problems
1. A certain culture of bacteria grows at a rate that is proportional to the
number present. If it is found that the number doubles in 4 h, how many
may be expected at the end of 12 h? Ans.: 8x0
2. In a culture of yeast the amount of active ferment grows at a rate
proportional to the amount present. If the amount doubles in 1 h, how
many times the original amount may be anticipated at the end of 2.75 h?
Ans.: 6.72N0
b). Decay problems
1. A certain radioactive material is known to decay at a rate proportional to
the amount present. If initially ½ g of the material is present and 0.1 percent
of the original mass has decayed after 1 week, determine the half-life of the
material. Ans.: 693 weeks
2. A certain radioactive material is known to decay at a rate proportional to
the amount present. If initially 500 mg of the material is present and after 3
years 20 percent of the original mass has decayed, determine the amount
remaining after 25 years. Ans.: 77.9 mg
c). Cooling and heating problems
1. A body at a temperature of 500F is placed in an oven whose temperature
is kept at 1500F. if after 10 min the temperature of the body is 750F, find the
time required to reach a temperature 1000F. Ans.: 24.1 min
2. A body whose temperature is initially 1000C is allowed to cool in air whose
temperature remains at a constant 200C. Find a formula which gives the
temperature of the body as a function of time t if it is observed that after 10
min the body has cooled to 400C. Ans. 20 + 80e- 0.1386t
d). Flow problems
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1. A tank initially holds 80 gal of a brine solution containing 1/8 lb of salt per
gal. At t = 0, another brine solution containing 1 lb of salt per gallon is poured
into the tank at the rate of 4 gal/min, while the well-stirred mixture leaves
the tank at the rate of 8 gal/min. Find the amount the amount of salt in the
tank when the tank contains exactly 40 gal of brine. Ans.: 22.5 lb
2. A tank contains 100 gal of brine made by dissolving 80 lb of salt in water.
Pure water runs into the tank at the rate of 4 gal/min, and the mixture, kept
uniform by stirring, runs out at the same rate. Find the amount of salt in the
tank at any time t. Ans.: 80e- 0.04t
In a Nutshell
We are now done with the generalization and concept of first-order D.E.
which exploring and understanding the application of first-order differential
equation. Before proceeding to the next unit learning outcomes, be reminded of
some important points when dealing with application of first-order differential
equations.





A certain population of bacteria is known to grow at a rate proportional to the
amount present.
A certain radioactive material is known to decay at a rate proportional to the
amount present.
Newton’s law of cooling states that the time rate of change of the temperature of a body
is proportional to the temperature difference between the body and its surrounding
medium.
The time rate of change of Q, dQ/dt, equals the rate at which salt enters the tank minus
the rate at which salt leaves the tank.
The volume of brine in the tank at any time is V0 + et – ft.
Q & A-List
If you have any questions regarding application first-order differential
equation, kindly write down on the table provided.
QUESTIONS
ANSWERS
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1.
2.
3.
4.
5.
Key Words Index:
Population growth, proportional, decompose, Newton’s law of cooling and
heating, medium, time rate of change, brine solution and concentration of salt.