MODULE 3

Application of 1st Order

Differential Equations
3.1. Decomposition/ Growth

Laws of Growth and Decay

“The rate at which the substances change is proportional to the quantity of substance present at
any time.”

EXPONENTIAL GROWTH
𝑑𝑃
= 𝑘𝑃
𝑑𝑡

𝑑𝑃
( = 𝑘𝑃) 𝑑𝑡
𝑑𝑡

𝑑𝑃 = 𝑘𝑃𝑑𝑡

𝑑𝑃
= 𝑘𝑑𝑡
𝑃

𝑑𝑃
∫ = ∫ 𝑘𝑑𝑡
𝑃

ln 𝑃 = 𝑘𝑡 + 𝑐

𝑒 ln 𝑃 = 𝑒 𝑘𝑡+𝑐
𝑒 ln 𝑃 = 𝑒 𝑘𝑡 • 𝑒 𝑐
𝑃 = 𝑐𝑒 𝑘𝑡
So, when 𝑡 = 0
𝑃(𝑡) = 𝑐𝑒 𝑘𝑡
𝑃(0) = 𝑐𝑒 𝑘(0)
𝑃(0) = 𝑐𝑒 0
𝑃(0) = 𝑐
𝑐 = 𝑃0

Therefore,

𝑷 = 𝑷𝟎 𝒆𝒌𝒕

Systems that exhibit exponential growth increase according to the mathematical model,

𝑷 = 𝑷𝟎 𝒆𝒌𝒕

where 𝑷𝟎 represents the initial state of the system and 𝒌 is a positive constant (𝒌 > 𝟎), called
the growth constant.

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 1

 Population growth is a common example of exponential growth. Consider a population of
bacteria, for instance. It seems plausible that the rate of population growth would be proportional
to the size of the population. After all, the more bacteria there are to reproduce, the faster the
population grows.
The exponential 𝑒 is used when modeling continuous growth that occurs naturally such as
populations, bacteria, radioactive decay, etc. You can think of 𝑒 like a universal constant
representing how fast you could possibly grow using a continuous process. And, the beauty
of e is that not only is it used to represent continuous growth, but it can also represent growth
measured periodically across time.

Examples:
1. The table below shows the rabbit population on a certain island where 𝑡 is the number of
years beginning with year 2000.
(a) Determine the relative growth rate.
(b) Write a general equation for the population 𝑃(𝑡).
(c) Estimate the population in 2010.
(d) How many years will it take for the population to double?

YEAR POPULATION

2000 1500

2001 1577

2002 1658

2003 1743

2004 1832

2005 1926

Solution:

𝑷(𝒕) = 𝑷𝟎 𝒆𝒌𝒕
To solve for the value of the growth rate
We have to solve for the value of 𝑘.
𝑘, first is to determine 𝑃0 .
𝑃(𝑡) = 𝑃0 𝑒 𝑘𝑡
Base on the table given, 𝑃(0) = 1500 at
First is to identify 𝑃0 , year 2000.
𝑃(0) = 1500
Using the model, we can get the value of
When 𝑡 = 0, 𝑃(0) 𝑃0 .
𝑃(0) = 𝑃0 𝑒 𝑘(0)

1500 = 𝑃0 𝑒 𝑘(0)
1500 = 𝑃0 (1)

𝑃0 = 1500

𝑷(𝒕) = 𝟏𝟓𝟎𝟎𝒆𝒌𝒕

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 2

 (a) Use another point to calculate for 𝒌:

𝑷(𝒕) = 𝟏𝟓𝟎𝟎𝒆𝒌𝒕
Use 𝑡 = 1; 𝑃(1) = 1577 at year 2001

1577 = 1500𝑒 𝑘(1)

1577
= 𝑒𝑘
1500
1.0513 = 𝑒 𝑘

ln 1.0513 = ln 𝑒 𝑘
0. 05 = 𝑘
Answer:
𝒌 = 𝟎.05, which means that the population increases at the rate of 5% every year.

(b) General equation for 𝑷(𝒕)

Answer:

𝑷(𝒕) = 𝟏𝟓𝟎𝟎𝒆𝟎.𝟎𝟓𝒕
(c) Population in 2010

Simply get 𝑃(10), from 2000-2010 is 10 years

𝑃(10) = 1500𝑒 0.05(10)

𝑃(10) = 1500𝑒 0.5

𝑃(10) = 2473
Answer:
𝟐𝟒𝟕𝟑 rabbits in the year 2010.

(d) How many years will it take for the population to double?
This means that we will have to get the number of years it will take to have a total
population of 3000 which is twice of 1500.

3000 = 1500𝑒 0.05𝑡

2 = 𝑒 0.05𝑡

ln 2 = ln 𝑒 0.05𝑡

ln 2 = 0.05𝑡 ln 𝑒
ln 2 = 0.05𝑡
ln 2
𝑡=
0.05
𝑡 = 13.86 𝑦𝑒𝑎𝑟𝑠

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 3

 Answer:
It will take almost 14 years or it will be at the year 2014 to double the population.

2. A biologist is researching a newly-discovered species of bacteria. At time t = 0 hours, he puts

one hundred bacteria into what he has determined to be a favorable growth medium. Six
hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth
constant "k" for the bacteria? (Round k to two decimal places.)
Solution:

For this example, the units on time 𝑡 will be hours, because the growth is being measured in
terms of hours. The beginning amount at 𝑡 = 0 so, for this problem, 𝑃0 = 100. The ending
amount is 𝑃 = 450 at 𝑡 = 6.

𝑃 = 𝑃0 𝑒 𝑘𝑡
At time=0, 𝑃(0) = 𝑃0 = 100
After 6 hours, 𝑃(6) = 450
𝑃(6) = 100𝑒 𝑘(6)

450 = 100𝑒 𝑘(6)

450
= 𝑒 6𝑘
100

4.5 = 𝑒 6𝑘

ln 4.5 = ln 𝑒 6𝑘

ln 4.5 = 6𝑘

𝑘 = 0.25067

Answer:

The growth constant is 0.25/hour

3. A 25-year old student is offered an opportunity to invest some money in a retirement account
that pays 5 annual interest compounded continuously. How much does the student need to
invest today to have 1,000,000 when she retires at age of 65? What if she could earn 6
annual interest compounded continuously instead?
Solution:
We have,
𝑃 = 1,000,000
𝑡 = 65 − 25 = 40

𝑘 = 0.05 (5 annual interest)

𝑃 = 𝑃0 𝑒 𝑘𝑡

1,000,000 = 𝑃0 𝑒 0.05(40)

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 4

 1,000,000
𝑃0 =
𝑒 0.05(40)
Answer;
𝑷𝟎 = 𝟗𝟎, 𝟕𝟏𝟕. 𝟗𝟓
Doubling Time

If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to
double. It is given by

𝐥𝐧 𝟐
Doubling Time =
𝒌
Example:

Assume a population of fish grows exponentially. A pond is stocked initially with 500 fish. After 6
months, there are 1000 fish in the pond. The owner will allow his friend and neighbors to fish on
his pond after the fish population reaches 10,000. When will the owner’s friends be allowed to
fish?

Solution:

It took the population of fish 6 months to double in size. So, if 𝑡 represents time in months, by the
doubling-time formula,
ln 2
6=
𝑘

ln 2
𝑘=
6
Thus, the population is given by,
𝑃 = 𝑃0 𝑒 𝑘𝑡

ln 2
( 6 )(𝑡)
𝑃 = 𝑃0 𝑒

ln 2
( 6 )(𝑡)
1,000,000 = 500𝑒

ln 2
( 6 )(𝑡)
20 = 𝑒

ln 2
( 6 )(𝑡)
ln 20 = ln 𝑒

ln 2
ln 20 = ( )𝑡
6

6 ln 20
=𝑡
ln 2

𝑡 = 25.93

Answer:

The owner’s friends have to wait 𝟐𝟓. 𝟗𝟑 months to fish in the pond

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 5

 EXPONENTIAL DECAY

Exponential functions can also be used to model populations that shrink (from disease, for
example), or chemical compounds that break down over time. We say that such systems exhibit
exponential decay, rather than exponential growth. The model is nearly the same, except there
is negative sign in the exponent.

Systems that exhibit exponential decay behave according to the model,

𝑷 = 𝑷𝟎 𝒆−𝒌𝒕

where 𝑷𝟎 represents the initial state of the system and 𝒌 > 𝟎 is a constant, called the decay
constant.

Half-Life

If a quantity decays exponentially, the half-life is the amount of the time it takes the quantity to be
reduced by half. It is given by

𝐥𝐧 𝟐
Half-Life =
𝒌

Example:

(Radiocarbon Dating)

One of the most common applications of an exponential decay model is carbon dating. Carbon-
14 decays (emits a radioactive particle) at a regular and consistent exponential rate. Therefore,
if we know how much carbon was originally present in an object and how much carbon remains,
we can determine the age of the object. The half-life of carbon-14 is approximately 5730
years—meaning, after that many years, half the material has converted from the original
carbon-14 to the new nonradioactive nitrogen-14. If we have 100 g carbon-14 today, how much
is left in 50 years? If an artifact that originally contained 100 g of carbon now contains 10 g of
carbon, how old is it? Round the answer to the nearest hundred years.

Solution:

ln 2
5730 =
𝑘

ln 2
𝑘=
5730
So,
𝑃 = 𝑃0 𝑒 𝑘𝑡

ln 2
−(5730)𝑡
𝑃 = 100𝑒
In 50 years,
ln 2
−(
𝑃 = 100𝑒 5730)(50)

𝑃 = 99.40

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 6

 Answer;

In 50 years, 99.40 g of carbon-14 remains.

To determine the age of the artifact,

ln 2
−(5730)𝑡
10 = 100𝑒

1 ln 2
−( )𝑡
= 𝑒 5730
10

𝑡 = 19,035
Answer:

The artifact is 19,035 years old.

Exercise 1.

1. Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the
substance is left after 6 hours?

2. Nadia owns a chain of fast food restaurants that operated 200 stores in 2014. If the rate of
increase is 8 annually, how many stores does the restaurant operate in 2022?

3. A colony of bacteria increases from 300 to 250 in 14 days. Write an exponential growth
function for the number of bacteria in the colony as a function of time.

4. In 1990, a small town had a population of 15,000 people. Since 1990, the town has
increased 3% per year. What was the population in 2019?

3.2. Newton’s Law of Cooling

Newton’s law of cooling says that an object cools at a rate proportional to the difference between
the temperature of the object and the temperature of the surroundings.

Let 𝒖 = temperature of the body at time 𝑡

𝒖𝒂 = temperature of the surrounding/environment

𝑑𝑢
∝ ( 𝑢 − 𝑢𝑜 )
𝑑𝑡

𝑑𝑢
= − 𝑘 ( 𝑢 − 𝑢𝑜 )
𝑑𝑡

𝑑𝑢
= − 𝑘 𝑑𝑡
( 𝑢 − 𝑢𝑜 )

𝑑𝑢
∫ = −𝑘 ∫ 𝑑𝑡
( 𝑢 − 𝑢𝑜 )

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 7

 ln( 𝑢 − 𝑢𝑜 ) = −𝑘𝑡 + 𝑐
ln( 𝑢 − 𝑢𝑜 ) = −𝑘𝑡 + ln 𝑐

ln( 𝑢 − 𝑢𝑜 ) − ln 𝑐 = −𝑘𝑡

( 𝑢 − 𝑢𝑜 )
ln = −𝑘𝑡
𝑐

( 𝑢−𝑢𝑜 )
𝑒 ln 𝑐 = 𝑒 −𝑘𝑡

( 𝑢−𝑢𝑜 )
= 𝑒 −𝑘𝑡
𝑐

𝑢 − 𝑢𝑜 = 𝑐𝑒 −𝑘𝑡

𝒖 = 𝒖𝒂 + 𝒄𝒆−𝒌𝒕

Examples:

1. A thermometer reading 18ºF is brought into a room where the temperature is 70ºF; 1 minute
later the thermometer reading is 31ºF. Determine the temperature reading as a function of
time and in particular. Find the temperature reading 5 minutes after the thermometer is first
brought into the room.

Solution:
𝒕 𝒖 𝒖𝒂

0 18ºF 70ºF

1 31ºF 70ºF

From the general equation:

𝒖 = 𝒖𝒂 + 𝒄𝒆−𝒌𝒕

When 𝑡 = 0, 𝑢 = 18°F, 𝑢𝑎 = 70°F

18 = 70 + 𝒄𝒆−𝒌(𝟎)

18 = 70 + 𝒄

𝑐 = −52

Thus,
𝒖 = 𝒖𝒂 − 𝟓𝟐𝒆−𝒌𝒕

When 𝑡 = 1, 𝑢 = 31°F, 𝑢𝑎 = 70°F

31 = 70 − 52𝒆−𝒌(𝟏)

31 = 70 − 52𝒆−𝒌

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 8

 31 − 70
𝑒 −𝑘 =
−52

39
𝑒 −𝑘 =
52
Thus,
39
ln 𝑒 −𝑘 = ln
52

39
−𝑘 = ln
52

𝑘 = 0.29
Substitute,

Answer:
𝒖 = 𝒖𝒂 − 𝟓𝟐𝒆−𝟎.𝟐𝟗𝒕

When 𝑡 = 5 𝑚𝑖𝑛𝑠, 𝑢 =? , 𝑢𝑎 = 70°F

𝑢 = 70 − 52𝑒 −0.29(5)

Answer:
𝒖 = 𝟓𝟕. 𝟔𝟔°𝐅 ≈ 𝟓𝟖°𝐅

2. A thermometer reading 75°F is taken out where the temperature is 20°F. The reading is 30°F,
4 minutes later. Find (a) the thermometer reading 7 minutes after the thermometer was
brought outside, and (b) the time taken for the reading to drop from 75°F to within a half
degree of the air temperature.

Solution:

𝒕 𝒖 𝒖𝒂
0 𝟕𝟓°𝐅 𝟐𝟎°𝐅

4 𝟑𝟎°𝐅 𝟐𝟎°𝐅

7 ? 𝟐𝟎°𝐅

? 𝟐𝟎. 𝟓°𝐅 𝟐𝟎°𝐅

𝒖 = 𝒖𝒂 + 𝒄𝒆−𝒌𝒕

When 𝑡 = 0, 𝑢 = 75°F, 𝑢𝑎 = 20°F

75 = 20 + 𝑐𝑒 −𝑘(0)

𝑐 = 75 − 20

𝑐 = 55
Substitute,

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 9

 𝑢 = 𝑢𝑎 + 55𝑒 −𝑘𝑡

When 𝑡 = 4, 𝑢 = 30°F, 𝑢𝑎 = 20°F

30 = 20 + 55𝑒 −𝑘(4)
30 − 20
𝑒 −4𝑘 =
55
1
1 10 4
(𝑒 −4𝑘 )4 =( )
55
1
10 4
𝑒 −𝑘 =( )
55
𝑡
10 4
𝑢 = 𝑢𝑎 + 55 ( )
55

(a) When 𝑡 = 7 𝑚𝑖𝑛𝑠, 𝑢 =?, 𝑢𝑎 = 20°F

7
10 4
𝑢 = 20 + 55 ( )
55
Answer:
𝒖 = 𝟐𝟐. 𝟕𝟖°𝐅 ≈ 𝟐𝟑°𝐅

(b) When 𝑡 =?, 𝑢 = 20.5°F, 𝑢𝑎 = 20°F

𝑡
10 4
20.5 = 20 + 55 ( )
55
𝑡
10 4 20.5 − 20
( ) =
55 55
𝑡
10 4 0.5
ln ( ) = ln
55 55

𝑡 10 0.5
ln = ln
4 55 55

0.5
𝑡 ln 55
=
4 ln 10
55

0.5
4 ln
𝑡= 55
10
ln
55
Answer:
𝒕 = 𝟏𝟏. 𝟎𝟑 𝒎𝒊𝒏𝒔

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 10

 Exercise 2.
1. It takes 12 minutes for an object at 100°C to cool to 80°C in a room at 50°C. How much
longer will it take for its temperature to decrease to 70°C?

2. At 1:00 PM, a thermometer reading 70°F is taken outside where the air temperature is
−10°F (10 below zero). At 1:02 PM, the reading is 26°F. At 1:05 PM, the thermometer is
taken back indoors where the air temperature is 70°F. What is the thermometer reading
at 1:09 PM?

3.3. Mixing (non-reacting fluids)

Rate of Change of
Substance in a Volume = Rate of Entrance – Rate of Exit

Mathematically,
𝒅𝒙
= 𝒓 𝒊 𝒄𝒊 − 𝒓 𝒐 𝒄𝒐
𝒅𝒕
Where:
𝑟𝑖 − volumetric flow rate at the entrance

𝑐𝑖 − concentration of substance rate at the entrance

𝑟𝑜 − volumetric flow rate at the exit

𝑐𝑜 − concentration of substance rate at the exit

𝒙
𝒄𝒐 =
𝑽
𝑽 = 𝑽𝒐 + (𝒓𝒊 − 𝒓𝒐 )𝒕
Where:
𝑉 − final volume of the solution at any time 𝑡

𝑉𝑜 − initial volume
𝑥 − the amount of substance at any time 𝑡

Example.
1. A tank contains 80 gallons (gal) of pure water. A brine solution with 2 lb/gal of salt enters at 2
gal/min, and the well stirred mixtures leaves at the same rate. Find (a) the amount of salt in the tank
at any time, and (b) the time at which the brine leasing will contain 1 lb/gal or salt.
Solution:
Brine Solution:
Stirrer
𝑐1 = 2 𝑙𝑏/𝑔𝑎𝑙

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 11

 𝑣𝑜 = 2 𝑔𝑎𝑙/𝑚𝑖𝑛

IN

OUT

Mixture
𝑐1 = 2 𝑙𝑏/𝑔𝑎𝑙
𝑣𝑜 = 2 𝑔𝑎𝑙/𝑚𝑖𝑛

Exercise 3:
1. A tank contains 200 liters of fluid in which 30 grams of salt are dissolved. Brine containing 1 gram of
salt per liter is then pumped into the tank at the rate of 4 liters per minute and the solution mixed well
is pumped out of the same rate.
(a) Find the number of grams of salt in the tank at any time t in minutes.
(b) Find the amount of salt in the tank after 5 minutes.

3.4. Other applications

Escape Velocity from the Earth
By Newton’s Universal Law of Gravitation:

𝑮𝒎𝟏 𝒎𝟐
𝑭=
𝒓𝟐
Where:
𝐹 − force of attraction between two heavenly bodies
𝐺 − the universal gravitational constant
𝑚1 , 𝑚2 − respective masses
𝑟 − distance between centers of mass

Since 𝐺, 𝑚1 , 𝑚2 are constants

Let 𝐾 ′ = 𝐺𝑚1 𝑚2
𝑲′
𝑭= − − − − − (𝟏)
𝒓𝟐
By Newton’s Second Law:
𝐹 = 𝑚𝑎
Where:
𝐹 − net force acting on the body
𝑚 − mass of the body
𝑎 − resulting acceleration

From (1)
𝐾′
= 𝑚𝑎
𝑟2

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 12

 𝐾′
𝑎=
𝑟2 𝑚
𝐾′
Since is constant,
𝑚
𝐾′
Let 𝑘 =
𝑚
𝑘
𝑎=
𝑟2
𝑑𝑣
But 𝑎 = ,
𝑑𝑡
Then,
𝑑𝑣 𝑘
= 2
𝑑𝑡 𝑟

Consider a particle projected radically outward from the earth:

𝑎 = −𝑔

Since the direction of force of the weight of the body is opposite the direction of motion.
𝑟 = 𝑅 (at the surface)
Thus,
𝑘
−𝑔 = 2
𝑅

𝑘 = −𝑔𝑅 2
From
𝑘
𝑎=
𝑟2

−𝑔𝑅 2
𝑎=
𝑟2
By the Chain Rule:
𝑑𝑣 𝑑𝑟 𝑑𝑣 𝑣𝑑𝑣
𝑎= = =
𝑑𝑡 𝑑𝑡 𝑑𝑟 𝑑𝑟
And so,
𝑑𝑣 −𝑔𝑅 2
𝑣 =
𝑑𝑟 𝑟2

𝑣𝑑𝑣 = −𝑔𝑅 2 𝑟 −2 𝑑𝑟

∫ 𝑣𝑑𝑣 = − 𝑔𝑅 2 ∫ 𝑟 −2 𝑑𝑟

𝑣2
= 𝑔𝑅 2 𝑟 −1 + 𝑐
2

𝑣 2 = 2𝑔𝑅 2 𝑟 −1 + 2𝑐
At the earth surface,
𝑣 = 𝑣0 , 𝑟 = 𝑅

𝑣0 2 = 2𝑔𝑅 2 𝑅 −1 + 2𝑐

2𝑐 = 𝑣0 2 − 2𝑔𝑅
Thus,

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 13

 𝑅2
𝑣 2 = 2𝑔 + (𝑣0 2 − 2𝑔𝑅)
𝑟2

From the equation, it can be seen that V will remain positive (in outward direction) if and only if
𝑣0 2 − 2𝑔𝑅 ≥ 0. Otherwise the particle will reach a distance r where V will be zero or the moon
will stop after which the body then proceeds back to the earth.

∴ 𝑣0 2 − 2𝑔𝑅 = 0

The minimum velocity of projection or escape velocity should be:

𝒗𝒆 = √𝟐𝒈𝑹

Example:
1. The radius of the moon is roughly 1080 miles. The acceleration of gravity at the surface of
the moon is about 0.165 g, where g is the acceleration of gravity at the surface of the earth.
Determine the velocity of escape for the moon.

Solution:
𝑔𝑚 = 0.165𝑔

𝑚𝑖
𝑔𝑚 = 0.165 (6.09 x 10− 3 )
𝑠𝑒𝑐 2

𝑚𝑖
𝑔𝑚 = 0.165 (6.09 x 10− 3 )
𝑠𝑒𝑐 2

𝑚𝑖
𝑔𝑚 = 1.00485 x 10−3
𝑠𝑒𝑐 2

𝑅𝑚 = 1080 𝑚𝑖𝑙𝑒𝑠

𝑉𝑒 = √2(1.00485 x 10−3 )(1080)

Answer:
𝒎𝒊
𝑽𝒆 = 𝟏. 𝟒𝟕𝟑
𝒔𝒆𝒄𝟐

Exercise 4:
1. Determine, to two significant figures, the velocity of escape of the earth of the celestial
𝑚𝑖
bodies listed below. The data are rough and g may be taken to be 6.1 x 10− 3 .
𝑠𝑒𝑐 2

Acceleration of gravity at Radius (miles)

Celestial Body
the surface
Venus 0.85 g 3800

Mars 0.38 g 2100

Jupiter 2.60 g 43000

Sun 28 g 432000

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 14

 Ganymede 0.12 g 1780

Simple Chemical Conversion

The time rate of change of the amount x of unconverted substance is proportional n to x.
𝑑𝑥
∝𝑥
𝑑𝑡

𝑑𝑥
= −𝑘𝑥
𝑑𝑡

𝑑𝑥
= −𝑘𝑑𝑡
𝑥
𝑑𝑥
∫ = −𝑘 ∫ 𝑑𝑡
𝑥
ln 𝑥 = −𝑘𝑡 + 𝑐1
Let 𝑐1 = ln 𝑐,

ln 𝑥 = −𝑘𝑡 + ln 𝑐
𝑥
ln = −𝑘𝑡
𝑐
𝑥
𝑒 ln 𝑐 = 𝑒 −𝑘𝑡

𝒙 = 𝒄𝒆−𝒌𝒕
Where:
𝑥 − amount of unconverted substance

Example:
1. Suppose that a chemical reaction proceeds according to the law given. If half of the substance A
has been converted at the end of 10 seconds, find when the nine-tenths of the substance will have
seen converted.

Solution:

𝑥 = 𝑐𝑒 −𝑘𝑡

When 𝑡 = 0, 𝑥 = 𝐴,

𝐴 = 𝑐𝑒 −𝑘(0)
𝑐=𝐴
Substitute,

𝑥 = 𝐴𝑒 −𝑘𝑡

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 15

 1
When 𝑡 = 10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠, 𝑥 = 𝐴,
2

1
𝐴 = 𝐴𝑒 −𝑘(10)
2
1
1 10
𝑒 −𝑘 =( )
2
Substitute,
𝑡
1 10
𝑥 = 𝐴( )
2
1
When 𝑥 = 𝐴 , 𝑡 =?,
10
𝑡
1 1 10
𝐴 = 𝐴( )
10 2
1 𝑡 1
ln = ln ( )
10 10 2
1
10 ln ( )
𝑡= 10
1
ln ( )
2
Answer:
𝒕 = 𝟑𝟑. 𝟐𝟐𝒔𝒆𝒄𝒐𝒏𝒅𝒔 ≈ 𝟑𝟑 𝒔𝒆𝒄𝒐𝒏𝒅𝒔
Exercise 5:
1. The conversion of substance B follows the law given above. If only a fourth of the substance has
been converted at the end of ten second, find when the nine-tenths of the substance will have
been converted.

Orthogonal Trajectories
The equation,

𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 0
Differentiate,
𝑑𝑦 𝑁
=
𝑑𝑥 𝑀
𝑁𝑑𝑥 − 𝑀𝑑𝑦 = 0
Example:
Find the orthogonal trajectories of the given families of curves.
1. 𝑥 − 4𝑦 = 𝑐

Solution:
𝑥 − 4𝑦 = 𝑐
Differentiate,

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 16

 𝑑𝑥 − 4𝑑𝑦 = 0

𝑑𝑦 1
=
𝑑𝑥 4
The orthogonal trajectories,
𝑑𝑦
= −4
𝑑𝑥

𝑑𝑦 = −4𝑑𝑥

4𝑑𝑥 + 𝑑𝑦 = 0

∫ 4𝑑𝑥 + ∫ 𝑑𝑦 = ∫ 0
Answer:
𝟒𝒙 + 𝒚 = 𝒌

2. 𝑥 2 − 𝑦 2 = 𝑐1

Solution:
𝑥 2 − 𝑦 2 = 𝑐1
Differentiate,
2𝑥𝑑𝑥 − 2𝑦𝑑𝑦 = 0

𝑑𝑦 𝑥
=
𝑑𝑥 𝑦
Orthogonal trajectories,
𝑑𝑦 −𝑦
=
𝑑𝑥 𝑥

𝑑𝑦 −𝑑𝑥
=
𝑦 𝑥

𝑑𝑦 𝑑𝑥
+ =0
𝑦 𝑥

𝑑𝑦 𝑑𝑥
∫ +∫ =∫ 0
𝑦 𝑥

ln 𝑦 + ln 𝑥 = ln 𝑐2

ln 𝑥𝑦 = ln 𝑐2

𝑒 ln 𝑥𝑦 = 𝑒 ln 𝑐2
Answer:
𝒙𝒚 = 𝒄𝟐

Exercise 6:

Find the orthogonal trajectories of the given families of curves.

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 17

 1. 𝑥 2 + 𝑦 2 = 𝑐 2
2. 𝑦 2 = 𝑐𝑥 3
Falling Object
An object is dropped from a height at time 𝑡 = 0. If ℎ(𝑡) is the height of the object at time 𝑡, 𝑎(𝑡) the
acceleration and 𝑣(𝑡) the velocity. The relationships between a, v and h are as follows:

𝒅𝒗 𝒅𝒉
𝒂(𝒕) = ; 𝒗(𝒕) =
𝒅𝒕 𝒅𝒕

For a falling object, 𝑎(𝑡) is constant and is equal to 𝑔 = −9.8 𝑚/𝑠.

Combining the above differential equations, we can easily deduce the following equation:

𝒅𝟐 𝒉
𝒈=
𝒅𝒕𝟐

Integrate both sides of the above equation to obtain,

𝑑ℎ
= 𝑔𝑡 + 𝑣0
𝑑𝑡
Integrate one more time to obtain,

𝟏 𝟐
𝒉(𝒕) = 𝒈𝒕 + 𝒗𝟎 𝒕 + 𝒉𝟎
𝟐

The above equation describes the height of a falling object, from an initial height ℎ0 at an initial velocity
𝑣0 , as a function of time.

Math 121 – Differential Equations | Instructor: Engr. Jennifer C. Paglingayen 18