LEARNING-MODULE-2-ENDIFF Study Materials

COLLEGE OF ENGINEERING
LEARNING MODULE 2
IN
DIFFERENTIAL EQUATIONS
(ENDIFF)
Prepared by:
Engr. Albert Anthony M. Delgado, MECE
APRIL 2022
Carlos Hilado Memorial State College
Alijis Campus | Binalbagan Campus | Fortune Towne Campus | Talisay Campus
To be a leading GREEN institution of higher learning in the global community by 2030
(Good governance, Research-oriented, Extension-driven, Education for Sustainable Development & Nation-building)

LEARNING MODULE – DIFFERENTIAL EQUATIONS (ENDIFF)
MODULE II – APPLICATIONS OF FIRST ORDER DIFFERENTIAL
EQUATIONS
2
TABLE OF CONTENTS
MODULE II. Application of 1st Order Differential Equations
1. Decomposition/ Growth
2. Newton’s Law of Cooling
3. Mixing
4. Electric Circuits

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3
II. Application of 1st Order Differential Equations
In the previous module, we have discussed the different methods on how to find the
solution to a given differential equation. It is also important to note that the functions
must exist within the interval of the differential equations. The rationale behind this
constant reiteration is that, when we are to apply them to real-world scenarios as
mathematical models, we can manipulate our parameters to achieve a certain objective.
Applications of DE are very prominent in field of motion, research, chemistry, logistics,
economics, growth and decay, telecommunications, electrical and electronic circuits,
physics, thermodynamics, strength of materials, civil and materials engineering, medical,
and many more. As long as there is a rate of change involved, we can immediately include
differential equations to find solutions to certain problems. Regardless if you are studying
science and engineering or not, you already have a grasp of differential equations. A very
known example of a DE is Newton’s Second Law of Motion.
𝐹 = 𝑚𝑎
Force is mass multiplied by the acceleration and acceleration is defined as the rate of
change of the velocity with respect to time. Since velocity is the rate of change of the
distance covered by an object with respect to time towards a certain direction. We can
say that:
𝑎 =
𝑑𝑣
𝑑𝑡
𝑜𝑟 𝑎 =
𝑑2
𝑠
𝑑𝑡2
Hence, we can write 𝐹𝑜𝑟𝑐𝑒 as:
𝐹 = 𝑚
𝑑𝑣
𝑑𝑡
or
𝐹 = 𝑚
𝑑2
𝑠
𝑑𝑡2
In this module, we will be discussing some of the basic applications of first order
differential equations namely:
• Decomposition/ Growth
• Newton’s Law of Cooling
• Mixing
• Electric Circuits.
Other applications may require differential equations of an order 𝑛 which will be on for
the next topic. Let’s start with the first application – Decomposition and Growth.

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1. Decomposition and Growth
Decomposition refers to the gradual or drastic process disintegration or breaking down of
complex substances into to simpler parts while growth is the process of increasing,
developing, and/or populating whether in size, value, or quantity. In differential
equations, we can accurately predict the growth or decay of a certain matter. It is
governed by the principle where the rate at which a substance or event changes is
proportional to the quantity of the substance present at any time. Hence, the relationship
below.
𝑑𝑥
𝑑𝑡
𝛼 𝑥
𝑤ℎ𝑒𝑟𝑒:
𝑑𝑥
𝑑𝑡
𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑔𝑟𝑜𝑤𝑡ℎ 𝑜𝑟 𝑑𝑒𝑐𝑎𝑦
𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑟 𝑒𝑣𝑒𝑛𝑡 𝑖𝑛 𝑠𝑡𝑢𝑑𝑦
If we are to form an equation about this relationship, then we must include a parameter
or a factor (𝑘) contributing to its growth or decay.
𝑑𝑥
𝑑𝑡
= 𝑘𝑥
Applying variable separable method,
𝑑𝑥
𝑥
= 𝑘𝑑𝑡
Integrating both sides,
∫
𝑑𝑥
𝑥
= ∫ 𝑘𝑑𝑡
𝑙𝑛𝑥 = 𝑘𝑡 + 𝐶
Raising both sides to 𝑒,
𝑒𝑙𝑛𝑥
= 𝑒𝑘𝑡+𝐶
𝑥 = 𝑒𝑘𝑡
. 𝑒𝐶
𝑥 = 𝐶𝑒𝑘𝑡
We have generated a general solution or a mathematical model for the growth or decay
of a certain substance. With certain initial conditions, we can solve for the particular
solution of this problem. For growth and decay, the main element affecting the substance

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or event is time. Thus, with initial conditions we can when a certain material will disappear
or what will its size or value be in a given period of time. This application can touch a lot
of fields in science and research from bacteria, culture, population, lifespan, forensics and
so on. Let’s see some examples below.
EXAMPLE 1.
Radium decomposes in air at the rate proportional to the present amount. If initially,
there 20 grams and after 10 years, only 0.6% of the original amount decomposed,
determine half-life of this initial amount and what will be left after 900 years.
SOLUTION:
1. Solve for the arbitrary constant and the parameter k.
Given:
𝑡 = 0 𝑥(0) = 20𝑔
𝑥(0) = 𝐶𝑒𝑘
20 = 𝐶𝑒𝑘(0)
20 = 𝐶(1)
𝐶 = 20
𝑡 = 10 𝑦𝑒𝑎𝑟𝑠 𝑥(10) = 𝑥 − 0.006𝑥
𝑥(10) = 𝐶𝑒𝑘𝑡
𝑥 − 0.006𝑥 = 20𝑒𝑘(10)
20 − 0.006(20) = 20𝑒10𝑘
19.88
20
=
20𝑒10𝑘
20
𝑙𝑛0.994 = 𝑙𝑛𝑒10𝑘
−
0.006
10
=
10𝑘
10
𝑘 = −0.0006

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6
Now that you have the parameter and arbitrary constant, you can solve for
the amount of radium for any time t.
2. Solve for the half-life of this amount of radium.
𝑥
2
= 𝐶𝑒𝑘𝑡
20
2
= 20𝑒−0.0006𝑡
1
2
= 𝑒−0.0006𝑡
ln
1
2
= 𝑙𝑛𝑒−0.0006𝑡
−0.6931
−0.0006
= −
−0.0006𝑡
−0.0006
𝑡 = 1,155.17 𝑦𝑒𝑎𝑟𝑠
Thus, this 20 g of radium will become half of its quantity or 10 g only
approximately after 1,155.17 𝑦𝑒𝑎𝑟𝑠 provided there are no other factors that
will affect this rate.
3. Solve for the amount of radium after 900 years.
𝑥 = 𝐶𝑒𝑘𝑡
𝑥 = 20𝑒−0.0006(900)
𝑥 = 11.65 𝑔
After 900 years, there would still be 11.65 grams of radium in air. Thus, we
have to be careful and mindful of whatever substance we introduce into our
environment.
EXAMPLE 2.
A colony of bacteria is growing exponentially. At time t=0 it has 10 bacteria in it, and at
time t=4 hours it has 2000. At what time will it have 100,000 bacteria?
1. Solve for the parameter and the arbitrary constant.
Given:
𝑡 = 0 𝑥(0) = 10

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10 = 𝐶𝑒𝑘(0)
𝐶 = 10
𝑡 = 4 ℎ𝑜𝑢𝑟𝑠 𝑥(4) = 2000
2000 = 10𝑒𝑘(4)
2000
10
=
10𝑒𝑘(4)
10
𝑙𝑛200 = 𝑙𝑛𝑒4𝑘
5.298 = 4𝑘
𝑘 = 1.3246
2. Solve for when this colony of bacteria becomes 100,000 in number.
𝑥(𝑡) = 𝐶𝑒𝑘𝑡
100,000 = 10𝑒1.3246𝑡
100,000
10
=
10𝑒1.3246𝑡
10
𝑙𝑛10000 = 𝑙𝑛𝑒1.3246𝑡
9.21 = 1.3246𝑡
𝑡 ≈ 6.95 ℎ𝑜𝑢𝑟𝑠
In approximately, 6.95 hours the colony of bacteria has gone to 100,000 in number.
Although, these are random given problems, or arbitrary in nature, this model has been
very helpful in a lot of ways to researchers and scientists in various fields.
Thus, for reference, this mathematical model solved through differential equations has
been the basis of news reports and articles published by experts in their field which
directly impacts our decision making and plans to mitigate or improve the effects of decay
or growth.

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2. Newton’s Law of Cooling
Newton’s Law of Cooling is primarily used in the field of chemistry and thermodynamics.
Though it only applies to convective heat transfer and not thermal radiation as there are
a lot of factors to consider when it comes to the latter and a different guide or model is
suited for those kinds of problems.
Newton’s Law of Cooling states that the rate at which a body/object cools is directly
proportional to the difference between the body and that of its medium or surroundings.
It can be presented as
𝑑𝑇
𝑑𝑡
𝛼 𝑇 − 𝑇𝑚
where:
𝑇 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦
𝑇𝑚 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑜𝑟 𝑚𝑒𝑑𝑖𝑢𝑚, 𝑎𝑚𝑏𝑖𝑒𝑛𝑡 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
𝑡 = 𝑒𝑙𝑎𝑝𝑠𝑒𝑑 𝑡𝑖𝑚𝑒.
Introducing a constant of proportionality (𝑘) which, in this case, may refer to the
characteristic of the body to lose or gain heat, the equation then becomes,
𝑑𝑇
𝑑𝑡
= 𝑘(𝑇 − 𝑇𝑚)
As it is, Newton’s Law of Cooling operates on tow major principles, the first law of
thermodynamics (law of conservation of energy) and the second law of thermodynamics.
Since energy is conserved, the total energy of this system remains the same regardless if
the object losses heat or the medium gains heat energy from the object. A thing to note
also as the concept of entropy comes to the picture, the object here does not gain
coldness for heat transfer is always going to be from a hotter body or medium to a colder
body or medium.
Hence, we can form to general solutions if either the temperature of the body or the
ambient temperature is hotter than the other. Let us some examples below.
EXAMPLE 1.
A metal is heated up to a temperature of 500°C. It is then exposed to an ambient
temperature of 38°C. After 2 minutes, the temperature of the metal becomes 190°C.
When will it become 100°C? What is the temperature after 4 mins?

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9
SOLUTION:
𝑑𝑇
𝑑𝑡
= 𝑘(𝑇 − 𝑇𝑚)
1. Find the mathematical model of the solution to the DE.
𝑑𝑇
𝑑𝑡
= 𝑘(𝑇 − 𝑇𝑚)
𝑑𝑇
𝑇 − 𝑇𝑚
= 𝑘𝑑𝑡
This is turned out to be a variable separable differential equation. Integrate both
sides to find the general solution.
∫
𝑑𝑇
𝑇 − 𝑇𝑚
= ∫ 𝑘𝑑𝑡
ln(𝑇 − 𝑇𝑚) = 𝑘𝑡 + 𝐶
Raise both sides to e and simplify,
𝑒ln(𝑇−𝑇𝑚)
= 𝑒𝑘𝑡+𝐶
𝑇 − 𝑇𝑚 = 𝑒𝑘𝑡
. 𝑒𝐶
𝑇 − 𝑇𝑚 = 𝐶𝑒𝑘𝑡
2. Substitute initial conditions to solve for the arbitrary constant and the
parameter 𝑘.
Given:
𝑡 = 0 𝑚𝑖𝑛, 𝑇 = 500°𝐶, 𝑇𝑚 = 38°𝐶
500 − 38 = 𝐶𝑒𝑘(0)
462 = 𝐶(1)
𝐶 = 462°𝐶
𝑡 = 2 𝑚𝑖𝑛𝑠, 𝑇 = 190°𝐶, 𝑇𝑚 = 38°𝐶
190 − 38 = 462𝑒𝑘(2)
152 = 462𝑒2𝑘
152
462
=
462𝑒2𝑘
462

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0.329 = 𝑒2𝑘
𝑙𝑛0.329 = 𝑙𝑛𝑒2𝑘
−1.1117 = 2𝑘
𝑘 = −
1.1117
2
𝑘 = −0.5558
Thus, the particular solution to the DE is 𝑇 − 𝑇𝑚 = 462𝑒−0.5558𝑡
. Now, you can solve
when this metal will reach a certain temperature or what will its temperature be at any
given time for all values of 𝑡 > 0.
3. Solve for the time when the temperature of the object is 100°C.
𝑇 − 𝑇𝑚 = 462𝑒−0.5558𝑡
100 − 38 = 462𝑒−0.5558𝑡
62 = 462𝑒−0.5558𝑡
62
462
=
462𝑒−0.5558𝑡
462
0.1342 = 𝑒−0.5558𝑡
𝑙𝑛0.1342 = 𝑙𝑛𝑒−0.5558𝑡
−2.008 = −0.5558𝑡
𝑡 =
−2.008
−0.5558
𝑡 ≈ 3.61 𝑚𝑖𝑛𝑠.
At approximately 3.61 minutes, the metal will be cooled to 100°𝐶.
4. Solve for the temperature after 4 mins.
𝑇 − 𝑇𝑚 = 462𝑒−0.5558𝑡
𝑇 − 38 = 462𝑒−0.5558(4)
𝑇 = 38 + 462𝑒−0.5558(4)
𝑇 = 88.017°𝐶

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Thus, the temperature of the metal after 4 minutes is 88.017°C. Variations to the
problem or required is also possible.
EXAMPLE 2.
A thermometer reading 10℃ is brought in a room whose temperature is 18°C. One
minute later, the thermometer reading is 14°C. How long does it take until the reading
becomes 16°C?
SOLUTION:
Note: We will use the mathematical model we previously derived and proceed to step 2.
2. Substitute initial conditions to solve for the arbitrary constant and the parameter 𝑘.
𝑡 = 0 𝑚𝑖𝑛, 𝑇 = 10°𝐶, 𝑇𝑚 = 18°𝐶
10 − 18 = 𝐶𝑒𝑘(0)
−8 = 𝐶(1)
𝐶 = −8
𝑡 = 1 𝑚𝑖𝑛, 𝑇 = 14°𝐶, 𝑇𝑚 = 18°𝐶
14 − 18 = −8𝑒𝑘(1)
−4 = −8𝑒𝑘
1
2
= 𝑒𝑘
ln
1
2
= 𝑙𝑛𝑒𝑘
𝑘 = −0.6931
Thus, the particular solution to this problem is given by the equation:
𝑇 − 𝑇𝑚 = −8𝑒−0.6931𝑡
3. How long does it take until the reading becomes 16°C?
𝑇 − 𝑇𝑚 = −8𝑒−0.6931𝑡
16 − 18 = −8𝑒−0.6931𝑡

EQUATIONS
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−2 = −8𝑒−0.6931𝑡
1
4
= 𝑒−0.6931𝑡
ln
1
4
= 𝑙𝑛𝑒−0.6931𝑡
−1.3863 = −0.6931𝑡
𝑡 =
−1.3863
−0.6931
𝑡 = 2 𝑚𝑖𝑛𝑠
Thus, in two minutes the thermometer reading will be 16°C.
Note that in these examples, the ambient temperature is held constant or it means that
the heat lost by the body and gained by the system is not enough to change the
temperature of the surrounding medium. Size or area of the medium might be a good
explanation or the size and area of the object in study.
3. Mixing
The applications of mixing in the food industry and processing, chemistry, fabrication,
materials engineering and many more has been revolutionary. The goal of mixing is to
slowly change the concentration of a solution until the desired percentage of the medium
is achieved.
In order to administer this gradual balancing of substances, there has to be a container
with a solvent, typically water, which allows for entry and exit of solute and the solution.
The main factors in play will be the volumetric flow rate of the solute that enters the
medium and its concentration and the volumetric flow rate of the solution that exits the
medium and its concentration. Normally, a measuring device, ppm, will determine if the
desired percentage of solution is achieved and with the aid of control system, closed the
valve for entry and exit, or signals the process to resume again once the medium is
emptied. The mathematical model for mixing is given by:
𝑑𝑥
𝑑𝑡
= 𝑟𝑖𝑐𝑖 − 𝑟𝑜𝑐𝑜

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13
where:
𝑟𝑖 = 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑡𝑟𝑎𝑛𝑐𝑒
𝑐𝑖 = 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑡𝑟𝑎𝑛𝑐𝑒
𝑟𝑜 = 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑥𝑖𝑡
𝑐𝑜 = 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑥𝑖𝑡
It is also important to note that the concentration of the substance at the exit is equal to
the amount of substance (𝑥) per unit volume (𝑉) which we can also say as the solution.
𝑐𝑜 =
𝑥
𝑉
The volume of the solution inside the container at any time 𝑡 is given by
𝑉 = 𝑉𝑜 + (𝑟𝑖 − 𝑟𝑜)𝑡
Unlike, growth, decay, and newton’s law of cooling, mixing problems and its solution may
differ from one problem to another. Let us look at some examples below.
EXAMPLE 1
A tank contains 100L of fluid in which 20 grams of salt are dissolved. Brine consisting of 1
gram of salt per liter is then pumped into the tank at a rate of 4L per minute. The solution
mixed well is then pumped out at the same rate.
a. Find the amount of salt in grams at any time t.
b. Find the amount of salt in the tank after 5 minutes
SOLUTION:
1. Draw a simple illustration of the problem.
2. Solve for the volume of the fluid in the container.
𝑐𝑖 =
1𝑔
𝐿
𝑟𝑖 = 4𝐿/𝑚𝑖𝑛
𝑟𝑜 = 𝑟𝑖
𝑐𝑜 =
𝑥
𝑉
𝑉𝑜 = 100𝐿
𝑥𝑜 = 20𝑔

EQUATIONS
14
𝑉 = 100 + (4 − 4)𝑡
𝑉 = 100𝐿
Since the rate of entry and exit is the same, then it is as if there is no reduction of
volume in the container.
3. Substitute the values to the mathematical model for mixing and simplify.
𝑑𝑥
𝑑𝑡
𝑑𝑥
𝑑𝑡
= 4(1) − 4 (
𝑥
𝑉
)
𝑑𝑥
𝑑𝑡
= 4 − 4 (
𝑥
100
)
𝑑𝑥
𝑑𝑡
= 4 − (
𝑥
25
)
𝑑𝑥
𝑑𝑡
=
(100 − 𝑥)
25
𝑑𝑥
100 − 𝑥
=
𝑑𝑡
25
4. Use the appropriate method to evaluate the resulting expression and simplify.
∫
𝑑𝑥
100 − 𝑥
= ∫
𝑑𝑡
25
− ln(100 − 𝑥) =
1
25
𝑡 + 𝐶
ln(100 − 𝑥)−1
=
𝑡
25
+ 𝐶
𝑒ln(100−𝑥)−1
= 𝑒
𝑡
25
+𝐶
(100 − 𝑥)−1
= 𝑒
𝑡
25. 𝑒𝐶
1
100 − 𝑥
= 𝐶𝑒
𝑡
25
[
1
100 − 𝑥
= 𝐶𝑒
𝑡
25] (100 − 𝑥)
1 = 100𝐶𝑒
𝑡
25 − 𝑥𝐶𝑒𝑡/25

EQUATIONS
15
𝑥𝐶𝑒
𝑡
25 = 100𝐶𝑒
𝑡
25 − 1
[𝑥𝐶𝑒
𝑡
25 = 100𝐶𝑒
𝑡
25 − 1]
𝐶𝑒
𝑡
25
𝑥 = 100 − 𝐶𝑒−
𝑡
25
Thus, the general solution for this mixing problem is 𝑥 = 100 − 𝐶𝑒−
𝑡
25.
a. Find the amount of salt in grams at any time t.
𝑡 = 0 𝑚𝑖𝑛, 𝑥 = 20𝑔
𝑥 = 100 − 𝐶𝑒−
𝑡
25
20 = 100 − 𝐶𝑒−
0
25
20 − 100 = −𝐶(1)
𝐶 = 80
Substitute this value to the general solution to get the particular solution of the DE.
Thus, at any given time t, the amount of salt in the container is:
𝑥 = 100 − 80𝑒−
𝑡
25
b. Find the amount of salt in the tank after 5 minutes.
𝑥 = 100 − 80𝑒−
𝑡
25
𝑥 = 100 − 80𝑒−
5
25
𝑥 = 34.4 𝑔
Thus, after 5 minutes there is 34.4 𝑔𝑟𝑎𝑚𝑠 of salt in the container.
EXAMPLE 2
Brine containing 2lbs per gal of salt enters a large tank at the rate of 2 gal/min. The
mixture well stirred leaves at 1 gal/min. If the tank contains initially 100 gallons of water
with 4lbs of dissolved salt,
a. Find the amount of salt at any time t
b. Find the amount of salt in the tank after 4 minutes
SOLUTION:

EQUATIONS
16
1. Draw a simple illustration of the problem.
2. Solve for the volume of the fluid in the container.
𝑉 = 100 + (2 − 1)𝑡
𝑉 = 100 + 𝑡
3. Substitute the values to the mathematical model for mixing and simplify.
𝑑𝑥
𝑑𝑡
𝑑𝑥
𝑑𝑡
= 2(2) − 1(
𝑥
100 + 𝑡
)
𝑑𝑥
𝑑𝑡
= 4 − (
𝑥
100 + 𝑡
)
𝑑𝑥
𝑑𝑡
+ (
𝑥
100 + 𝑡
) = 4
𝑑𝑥
𝑑𝑡
+ (
1
100 + 𝑡
) 𝑥 = 4
The resulting expression is linear in x with 𝑃(𝑡) =
1
100+𝑡
and 𝑄(𝑡) = 4.
As previously mentioned, depending on the initial conditions and parameters set by the
problem, methods in finding the solution the DE may involve approaches that are not
limited to variable separable method.
Hence, the particular solution of a certain differential equation may be used to solve or
either the amount of substance or at what time will the amount of substance be a certain
value in the container.
4. Use the appropriate method to evaluate the resulting expression and simplify.
𝑐𝑖 = 2𝑙𝑏𝑠/𝑔𝑎𝑙
𝑟𝑖 = 2 𝑔𝑎𝑙/𝑚𝑖𝑛
𝑟𝑜 = 1 𝑔𝑎𝑙/𝑚𝑖𝑛
𝑐𝑜 =
𝑥
𝑉
𝑉𝑜 = 100 𝑔𝑎𝑙
𝑥𝑜 = 4𝑙𝑏𝑠

EQUATIONS
17
𝑥𝑒∫ 𝑃(𝑡)𝑑𝑡
= ∫ 𝑄(𝑡) 𝑒∫ 𝑃(𝑡)𝑑𝑡
𝑑𝑡
𝑥𝑒∫
1
100+𝑡𝑑𝑡
= ∫ 4 𝑒∫
1
100+𝑡𝑑𝑡
𝑑𝑡
𝑥𝑒ln (100+𝑡)
= ∫ 4 𝑒ln (100+t)
𝑑𝑡
𝑥(100 + 𝑡) = 4 ∫(100 + 𝑡) 𝑑𝑡
𝑥(100 + 𝑡) = 400𝑡 + 2𝑡2
+ 𝐶
𝑥(100 + 𝑡)
100 + 𝑡
=
400𝑡 + 2𝑡2
+ 𝐶
100 + 𝑡
𝑥 =
2𝑡2
+ 400𝑡 + 𝐶
100 + 𝑡
This is now the general solution of the DE.
a. Find the amount of salt at any time t
𝑡 = 0 𝑚𝑖𝑛, 𝑥 = 4𝑙𝑏𝑠
𝑥 =
2𝑡2
+ 400𝑡 + 𝐶
100 + 𝑡
4 =
2(0)2
+ 400(0) + 𝐶
100 + (0)
4 =
𝐶
100
𝐶 = 400
Substitute this value to the general solution to get the particular solution of the DE.
Thus, at any given time t, the amount of salt in the container is:
𝑥 =
2𝑡2
+ 400𝑡 + 400
100 + 𝑡
b. Find the amount of salt in the tank after 4 minutes.
𝑥 =
2𝑡2
+ 400𝑡 + 400
100 + 𝑡
𝑥 =
2(4)2 + 400(4) + 400
100 + (4)
𝑥 = 19.54𝑙𝑏𝑠

EQUATIONS
18
4. Electric Circuits
In line with this program, the most related application of differential equation is in finding
solutions for electric circuits. The application of DE in electrical circuits revolves around
the concept of Kirchhoff’s Law and the behavior of the parameters, in this case, circuit
components with respect to current or voltage. Although since the rate of change for first
order differential equation is inclusive in a current-inductor and voltage-capacitor
relationship, the differential equation we will be setting up later will only contain RL or
RC network. The general solution for a series RLC circuit with respect to time falls under
the second order DE, which will be covered in the next module, and is given by:
𝑑2
𝑖
𝑑𝑡2
+ 𝑅
𝑑𝑖
𝑑𝑡
+
1
𝐶
𝑖(𝑡) = 𝐸𝑐𝑜𝑠𝜔𝑡
In a series RL circuit, as the current enters the circuit, the inductor will try to resist any
change in current. This property is called inductance. Thus, the voltage drop across the
inductive component is dependent on that change of current value with respect to time.
At this rate, the inductor is able to store energy due to its increasing interaction with
current. Conversely, the voltage decreases on the inductor when current is removed from
the circuit, thereby causing the inductor to discharge.
Figure 1. Sample RL Circuit.
Employing Kirchhoff’s Voltage Law, the sum of all voltages in the circuit is equal to 0. Thus,
∑ 𝐸 = 0
𝐸 = 𝐸𝑅 + 𝐸𝐿
𝐸 = 𝑖𝑅 + 𝐿
𝑑𝑖
𝑑𝑡
where:
𝐿 = 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝐻𝑒𝑛𝑟𝑦, 𝐻
𝑅 = 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑜ℎ𝑚, 𝛺
𝑖 = 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛 𝑎𝑚𝑝𝑒𝑟𝑒𝑠, 𝐴
𝐸 = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑠𝑜𝑢𝑟𝑐𝑒 𝑖𝑛 𝑣𝑜𝑙𝑡𝑠, 𝑉

EQUATIONS
19
Rewriting the above equation will give us a mathematical model that resembles a DE
linear in 𝑖.
𝐿
𝑑𝑖
𝑑𝑡
= 𝐸 − 𝑖𝑅
Consequently, in a series RC circuit, there will be a voltage rise time which we can view as
a delay as the potential difference across the capacitor plates starts to increase. This state
the capacitor is said to be charging and will be discharging when the supply is removed –
a very helpful characteristic in electronics. Thus, the current will be presented as the rate
of charge per unit time and the voltage across the capacitor is given the ratio between
the charge and its capacitance value.
Figure 2. Sample RC Circuit.
Employing Kirchhoff’s Voltage Law, the sum of all voltages in the circuit is equal to 0. Thus,
∑ 𝐸 = 0
𝐸 = 𝐸𝑅 + 𝐸𝐶
𝐸 = 𝑅
𝑑𝑄
𝑑𝑡
+
𝑄
𝐶
where:
𝐶 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝐹𝑎𝑟𝑎𝑑, 𝐹
𝑅 = 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑜ℎ𝑚, 𝛺
𝑄 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑖𝑛 𝐶𝑜𝑢𝑙𝑜𝑚𝑏, 𝐶
𝐸 = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑠𝑜𝑢𝑟𝑐𝑒 𝑖𝑛 𝑣𝑜𝑙𝑡𝑠, 𝑉
Rewriting the above equation will give us a mathematical model that resembles a DE
linear in 𝑄. The unknown quantities may vary depending on the problem and the resulting
DE may also have other approaches or methods of finding the solution other variable
separable. Let us take a look at the examples below.

EQUATIONS
20
𝑖
EXAMPLE 1.
An inductor and a resistor with a value of 10 ohms are connected in series with a supply
of 100V. If the current is initially zero amperes and is equal to 9A after 1 second, find the
current after 0.5 second.
SOLUTION:
1. Find the mathematical model of the given circuit.
𝐿
𝑑𝑖
𝑑𝑡
= 𝐸 − 𝑖𝑅
𝐿
𝑑𝑖
𝑑𝑡
= 100 − 10𝑖
[𝐿
𝑑𝑖
𝑑𝑡
= 100 − 10𝑖]
1
𝐿
𝑑𝑖
𝑑𝑡
=
100
𝐿
−
10
𝐿
𝑖
𝑑𝑖
𝑑𝑡
+
10
𝐿
𝑖 =
100
𝐿
The resulting expression is linear with respect to 𝑖.
2. Solve for the general solution of the given circuit.
𝑖𝑒∫ 𝑃(𝑡)𝑑𝑡
= ∫ 𝑄(𝑡)𝑒∫ 𝑃(𝑡)𝑑𝑡
𝑑𝑡
𝑖𝑒∫
10
𝐿
𝑑𝑡
= ∫
100
𝐿
𝑒∫
10
𝐿
𝑑𝑡
𝑑𝑡
𝑖𝑒
10
𝐿
𝑡
= ∫
100
𝐿
𝑒
10
𝐿
𝑡
𝑑𝑡

EQUATIONS
21
𝑖𝑒
10
𝐿
𝑡
=
100
𝐿
.
𝐿
10
∫
10
𝐿
𝑒
10
𝐿
𝑡
𝑑𝑡
𝑖𝑒
10
𝐿
𝑡
= 10𝑒
10
𝐿
𝑡
+ 𝐶
[𝑖𝑒
10
𝐿
𝑡
= 10𝑒
10
𝐿
𝑡
+ 𝐶]
𝑒
10
𝐿
𝑡
𝑖=10+𝐶𝑒−10𝐿𝑡𝑖 = 10 + 𝐶𝑒−
10
𝐿
(𝑡)
Thus, the general solution for this circuit problem is 𝑖 = 10 + 𝐶𝑒−
10
𝐿
𝑡
.
3. Apply initial conditions to find the particular solution.
𝑡 = 0 𝑠𝑒𝑐𝑜𝑛𝑑, 𝑖 = 0𝐴
𝑒−10𝐿𝑡𝑖 = 10 + 𝐶𝑒−
10
𝐿
(𝑡)
0 = 10 + 𝐶𝑒−
10
𝐿
(0)
0 = 10 + 𝐶(1)
𝐶 = −10
𝑖 = 10 − 10𝑒−
10
𝐿
𝑡
𝑖 = 10 − 10𝑒−
10
𝐿
𝑡
9 = 10 − 10𝑒−
10
𝐿
(1)
9 − 10 = −10𝑒−
10
𝐿
−1
−10
= 𝑒−
10
𝐿
ln
1
10
= 𝑙𝑛𝑒−
10
𝐿
−2.3 = −
10
𝐿
𝐿 =
−10
−2.3
𝐿 ≈ 4.35 𝐻

EQUATIONS
22
𝑖
𝑖 = 10 − 10𝑒−
10
4.35
𝑡
𝑖 = 10 − 10𝑒−2.3𝑡
Thus, the particular solution to this DE is 𝑖 = 10 − 10𝑒−2.3𝑡
with current as a function of time.
4. Find the current after 0.5 second.
𝑖 = 10 − 10𝑒−2.3𝑡
𝑖 = 10 − 10𝑒−2.3(0.5)
𝑖 = 6.834 𝐴
EXAMPLE 2.
In a series RC circuit below, Find the charge at any time t. What is the charge after 10 seconds?
SOLUTION:
1. Find the mathematical model of the given circuit.
𝐸 = 𝑅
𝑑𝑄
𝑑𝑡
+
𝑄
𝐶
𝑅
𝑑𝑄
𝑑𝑡
+
𝑄
𝐶
= 𝐸
[𝑅
𝑑𝑄
𝑑𝑡
+
𝑄
𝐶
= 𝐸]
1
𝑅
𝑑𝑄
𝑑𝑡
+
𝑄
𝑅𝐶
=
𝐸
𝑅
𝑑𝑄
𝑑𝑡
+
𝑄
5(0.01)
=
100
5
𝑑𝑄
𝑑𝑡
+ 20𝑄 = 20

EQUATIONS
23
The resulting model is linear in Q.
2. Solve for the general solution of the given circuit. Note, as Q is the symbol representing
charge, do not be confused with Q(t) as the component of the linear DE.
𝑄𝑒∫ 𝑃(𝑡)𝑑𝑡
= ∫ 𝑄(𝑡)𝑒∫ 𝑃(𝑡)𝑑𝑡
𝑑𝑡
𝑄𝑒∫ 20𝑑𝑡
= ∫ 20𝑒∫ 20𝑑𝑡
𝑑𝑡
𝑄𝑒20𝑡
= ∫ 20𝑒20𝑡
𝑑𝑡
𝑄𝑒20𝑡
= ∫ 20𝑒20𝑡
𝑑𝑡
𝑄𝑒20𝑡
= 𝑒20𝑡
+ 𝐶
[𝑄𝑒20𝑡
= 𝑒20𝑡
+ 𝐶]
1
𝑒20𝑡
𝑄 = 1 + 𝐶𝑒−20𝑡
Thus, the general solution to the given problem is 𝑄 = 1 + 𝐶𝑒−20𝑡
.
3. Apply initial conditions to find the particular solution.
𝑄 = 1 + 𝐶𝑒−20𝑡
0 = 1 + 𝐶𝑒−20(0)
0 = 1 + 𝐶(1)
𝐶 = −1
𝑄 = 1 − 𝑒−20𝑡
Thus, the particular solution to the given problem is 𝑄 = 1 − 𝑒−20𝑡
.
4. What is the charge after 10 seconds?
𝑄 = 1 − 𝑒−20𝑡
𝑄 = 1 − 𝑒−20(10)
𝑄 = 1 𝐶𝑜𝑢𝑙𝑜𝑚𝑏
Thus, after 10 seconds the charge in the capacitor is 1 Coulomb.
Like mixing problems, the resulting DE may vary, hence, the approach is not limited to
variable separable. Also, the DE has to be derived depending on the given conditions or
parameters because not all problems have the same general or particular solution.

EQUATIONS
24
PROBLEM SET 1
GROWTH AND DECAY
1. In 2003, the population of a certain city is 1743. After a year, it became 1832. In 2005,
it is now 1926. Estimate the population in 2010. In what year will it be doubled? How
many years will it take before it reaches 10,000?
2. A slow economy caused a company’s annual revenues
to drop from Php530, 000 in 2018 to Php386,000 in 2020. If the revenue is following
an exponential pattern of decline, what is the expected revenue in 2022?
3. The number of bacteria in a culture is growing at a rate of per unit of time t. At t = 0,
the number of bacteria present was 7,500. Find the number present at t = 5.
NEWTON’S LAW OF COOLING
1. A cheesecake is taken out of the oven with an ideal temperature of 175 degrees
Fahrenheit and is placed into a refrigerator with a temperature of 30 degrees
Fahrenheit. After 10 minutes, the cheesecake has cooled to 150 degrees Fahrenheit.
If we must wait until the cheesecake has cooled to 70 degrees Fahrenheit before we
eat it, how long will we have to wait?
2. A pitcher of water at 45 degrees Fahrenheit is placed into a 75-degree Fahrenheit
room. One hour later, the temperature has risen to 50 degrees. How long will it take
for the temperature to rise to 60 degrees?
3. A body at temperature 40ºC is kept in a surrounding of constant temperature 20ºC. It
is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time
will it take for the body to attain a temperature of 30ºC.
MIXING PROBLEMS
1. A tank contains 1,000 L of water and 20 kg of dissolved salt. Fresh water is entering
the tank at 10 L/min (the solution stays perfectly mixed), and the solution drains at a
rate of 5 L/min. How much salt is in the tank at t minutes and at 10 minutes?
2. A tank contains 1,000 L of water and 15 kg of dissolved salt. Fresh water is entering
the tank at 5 L/min (the solution stays perfectly mixed), and the solution drains at the
same rate as it enters. How much salt is in the tank at t minutes and at 20 minutes?

EQUATIONS
25
CIRCUIT PROBLEMS
1. Consider the circuit below. At t = 0, 𝑖 = 1𝐴 and rises up to 5A at time t = 5 seconds.
Disregard the value of L in the diagram below. Find L and the current at after 10
seconds.
2. A series RC circuit with R = 5 W and C = 0.02 F is connected with a battery of E = 100
V. At t = 0, the I = 0. Find charge at any time t.

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