UNIVERSITY OF MINDANAO
College of Engineering Education
Physically Distanced but Academically Engaged
Self-Instructional Manual (SIM) for Self-Directed Learning (SDL)
Course/Subject: CEE 104 – Differential Equations
Name of Authors:
CAMACHO, BERRNARD V.
JOHN C. BACUS
LUIS CARLO GAYAK
THIS SIM/SDL MANUAL IS A DRAFT VERSION ONLY; NOT FOR
REPRODUCTION AND DISTRIBUTION OUTSIDE OF ITS INTENDED USE.
THIS IS INTENDED ONLY FOR THE USE OF THE STUDENTS WHO ARE
OFFICIALLY ENROLLED IN THE COURSE/SUBJECT.
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Tel no. (082)300-5456 local 133
TABLE OF CONTENTS
PAGE
Cover Page ………………………………………………………………………………………………..
1
Table of Contents……………………………………………………………………………………….
2
Course Outline…………………………………………………………………………………………...
3
Course Outline Policy…………………………………………………………………………………
3
Course Information……………………………………………………………………………………
7
Topic/ Activity
Unit Learning Outcomes- Unit 1………………………………………………………………….
11
Big Picture in Focus: ULO-a…………………………………………………………………..…….
11
Metalanguage…………………………………………………………………………………...
11
Essential Knowledge…………………………………………………………………………
11
Self-Help…………………………………………………………………………………………..
15
In a Nutshell……………………………………………………………………………………..
16
Big Picture in Focus: ULO-b…………………………………………………………………..…….
17
Metalanguage…………………………………………………………………………………...
17
Essential Knowledge…………………………………………………………………………
28
Self-Help…………………………………………………………………………………………..
33
College Of Engineering Education
2nd Floor Business and Engineering Building
Matina Campus, Davao City
Tel no. (082)300-5456 local 133
Course Outline: CEE 104 – Differential Equations
Course Coordinator:
Email:
Student Consultation:
Mobile:
Phone:
Effectivity Date:
Mode of Delivery:
sessions)
Time Frame:
Student Workload:
Requisites:
Credit:
Attendance Requirements:
Elena G. Matillan0
ematillano@umindanao.edu.ph
By online (LMS) or thru text, emails or calls
09956105089
(082) 296-1084 local 133
May 25, 2020
Blended (On-Line with face to face or virtual
54 Hours
Expected Self-Directed Learning
None
3
.
Course Outline Policy
Areas of Concern
Contact and Non-contact Hours
Details
This 3-unit course self-instructional manual is designed
for blended learning mode of instructional delivery with
scheduled face to face or virtual sessions. The
expected number of hours will be 54 including the face
to face or virtual sessions. The face to face sessions
shall include the summative assessment tasks (exams)
since this course is crucial in the licensure examination
for teachers.
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Assessment Task Submission
Submission of assessment tasks shall be on 3rd, 5th, 7th
and 9th week of the term. The assessment paper shall
be attached with a cover page indicating the title of the
assessment task (if the task is performance), the
name of the course coordinator, date of submission and
name of the student. The document should be emailed
to the course coordinator. It is also expected that you
already paid your tuition and other fees before the
submission of the assessment task.
If the assessment task is done in real time through the
features in the Blackboard Learning Management
System, the schedule shall be arranged ahead of time
by the course coordinator.
Policies and Guidelines
Turnitin Submission
(if necessary)
•
Virtual Sessions shall be done once a week with
an announcement 1 before the schedule.
• These meetings are for lectures, discussions,
reportings, etc.
• The official online platform is the Blackboard LMS
and BB Colab.
• Attendance is counted from the first day of class.
• Cheating is strictly prohibited. Any form of
dishonesty shall be dealt with accordingly. Honesty
is called for at all times.
• Valid examination permit will be checked in taking
the examinations as scheduled.
Since this course is included in the licensure examination
for engineers, you will be required to take the MultipleChoice Question exam inside the University. This should
be scheduled ahead of time by your course coordinator.
This is non-negotiable for all licensure-based programs.
To ensure honesty and authenticity, all assessment
tasks are required to be submitted through Turnitin
with a maximum similarity index of 30% allowed. This
means that if your paper goes beyond 30%, the
students will either opt to redo her/his paper or explain
in writing addressed to the course coordinator the
reasons for the similarity. In addition, if the paper has
reached more than 30% similarity index, the student
may be called for a disciplinary action in accordance
with the University’s OPM on Intellectual and Academic
Honesty.
Please note that academic dishonesty such as cheating
and commissioning other students or people to
complete the task for you have severe punishments
(reprimand, warning, expulsion).
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Penalties for Late
Assignments/Assessments
Return of Assignments/
Assessments
Assignment Resubmission
Re-marking of Assessment Papers
and Appeal
Grading System
Referencing Style
The score for an assessment item submitted after the
designated time on the due date, without an approved
extension of time, will be reduced by 5% of the possible
maximum score for that assessment item for each day
or part day that the assessment item is late.
However, if the late submission of assessment paper
has a valid reason, a letter of explanation should be
submitted and approved by the course coordinator. If
necessary, you will also be required to present/attach
evidences.
Assessment tasks will be returned to you two (2) weeks
after the submission. This will be returned by email or
via Blackboard portal.
For group assessment tasks, the course coordinator will
require some or few of the students for online or virtual
sessions to ask clarificatory questions to validate the
originality of the assessment task submitted and to
ensure that all the group members are involved.
You should request in writing addressed to the course
coordinator his/her intention to resubmit an assessment
task. The resubmission is premised on the student’s
failure to comply with the similarity index and other
reasonable grounds such as academic literacy
standards or other reasonable circumstances e.g.
illness, accidents financial constraints.
You should request in writing addressed to the program
coordinator your intention to appeal or contest the score
given to an assessment task. The letter should explicitly
explain the reasons/points to contest the grade. The
program coordinator shall communicate with the students
on
the approval
and disapproval
of the
request.
All culled
from BlackBoard
sessions
and
traditional
contact
If disapproved
by the course coordinator,
you can elevate
Course
discussions/exercises
– 30% 1st formative
your case to –the
program
or the
dean with the
assessment
10%
2nd head
formative
assessment
– 10%
original
letter of
request. The
final decision will come
3rd
formative
assessment
– 10%
from
dean of the college.
IEEEthe
Referencing.
All culled from on-campus/onsite sessions (TBA):
Final exam – 40%
Student Communication
Submission
of thetofinal
grades
shall follow
the usual
You are required
create
a umindanao
email
account
University
system
and
procedures.
which is a requirement to access the BlackBoard portal.
Then, the course coordinator shall enroll the
students to have access to the materials and resources
of the course. All communication formats: chat,
submission of assessment tasks, requests etc. shall be
through the portal and other university recognized
platforms.
You can also meet the course coordinator in person
through the scheduled face to face sessions to raise your
issues and concerns.
College Of Engineering Education
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Tel no. (082)300-5456 local 133
Contact Details of the Dean
Dr. Charlito L. Canesares
Email: clcanesares@umindanao.edu.ph
Phone:
Contact Details of the Program
Head
Engr. Rolieven P. Canizares
Email: rolieven_canizares@umindanao.edu.ph
Phone: 09286383164
Students with Special Needs
Students with special needs shall communicate with the
course coordinator about the nature of his or her special
needs. Depending on the nature of the need, the course
coordinator with the approval of the program coordinator
may provide alternative assessment tasks or extension
of theare
deadline
of to
submission
assessment
You
required
enroll in a of
specific
tutorialtasks.
time for
However,
assessment
tasksPlease
shouldnote
still be
this
coursethe
viaalternative
the www.cte.edu.ph
portal.
in the
service
achieving
desired to
course
learning
that
there
is a of
deadline
for the
enrollment
the tutorial.
outcomes.
Online Tutorial Registration
Help Desk Contact
CEE BLACKBOARD ADMINISTRATOR
Engr. Jethron J. Adtoon
Email: jadtoon@umindanao.edu.ph
Phone: 09055267834
CEE
Frida Santa O. Dagatan
Email: cee@umindanao.edu.ph
Mobile: 09562082442
Phone: 082-2272902
LIC
Brigada E. Bacani
Email: library@umindanao.edu.ph
Mobile: 0951-376-6681
GSTC
Ronadora E. Deala, RPsy, RPm, RGC, LPT
Email: ronadora_deala@umindanao.edu.ph
09212122846
Silvino P. Josol
Email: gstcmain@umindanao.edu.ph
09060757721
Course Information – see/download course syllabus in the Black Board LMS
CC’s Voice: Hello future engineer! Welcome to this course CEE 104: Differential
Equations. By now, I am confident that you really wanted to become
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an engineer and that you have acquired enough knowledge in
Engineering Calculus to solve Differential Equations.
CO
CO 1. Solve correctly different types of differential equations.
CO 2. Apply differential equations to selected engineering problems.
Let us begin!
Big Picture
Week 1-3: Unit Learning Outcomes (ULO): At the end of the unit, you are
expected to
a. Define and classify the different types of Differential Equation.
b. Determine the different types of first order Differential Equation and solve
for the general and particular solution.
Big Picture in Focus
ULOa.
Define and classify the different types of Differential Equation.
Metalanguage
In this section, the essential terms and symbols relevant in the study of
Differential Equations will be discussed to demonstrate ULOa. Please refer to
these definitions in case you will encounter difficulty in understanding educational
concepts.
1. Differential Equations. Equations containing derivatives or differentials.
2. Ordinary Differential Equations (ODEs). Differential equations that depend on a single
variable.
3. Partial Differential Equations (PDEs). Differential equations that depend on a single
variable.
4. Order. Refers to the highest - ordered derivative in the given differential equation.
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5.
6.
7.
8.
9.
Degree. The exponent of the highest - ordered derivative.
Dependent Variable.
Independent Variable.
Parameters.
First Order ODE. An ODE that contains the first derivative only of the unknown function.
Essential Knowledge
To perform the aforesaid big picture (unit learning outcomes) for the first three
(3) weeks of the course, you need to review the fundamental concepts in Calculus
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 e.g. ebrary, search.proquest.com etc.
Differential Equation
A differential equation is any equation which contains derivatives, either ordinary
derivatives or partial derivatives.
There is one differential equation that everybody probably knows, that is Newton’s
Second Law of Motion. If an object of mass m is moving with acceleration a and being
acted on with force F then Newton’s Second Law tells us.
Example 1
𝐅 = 𝐦𝐚
Rewriting this equation to prove that it a differential equation,
𝐚=
𝐝𝐯
𝐝𝐭
𝐝𝟐 𝐬
= (𝐝𝐭 𝟐 )
At this point v is the velocity and s is the position of the particle at any time t. In addition,
force F may also be a function of time, velocity, and/or position. So, having this the
Newtons Second Law of Motion can now be written as a differential equation in terms
of 𝐝𝐯/𝐝𝐭 and 𝐝𝐬/𝐝𝐭
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𝐅(𝐭, 𝐯) = 𝐦
𝐅(𝐭, 𝐬
𝐝𝐯
𝐝𝐭
𝐝𝐬
𝐝𝟐 𝐬
)=𝐦 𝟐
𝐝𝐭
𝐝𝐭
This type of equation will be used in the succeeding topics.
Here are some other differential equation.
𝐚𝐲" + 𝐛𝐲′ + 𝐜𝐲 = 𝐠(𝐭)
𝐬𝐢𝐧 𝐲
𝐝𝟐 𝐲
𝐝𝐲
= (𝟏 − 𝐲)
+ 𝐲 𝟐 𝐞−𝟓𝐲
𝟐
𝐝𝐱
𝐝𝐱
𝐲 (𝟒) + 𝟏𝟎𝐲 ′′′ − 𝟒𝐲 ′ + 𝟐𝐲 = 𝐜𝐨𝐬 𝐭
∝
𝛛𝟐 𝐮 𝛛𝐮
=
𝛛𝟐 𝐱 𝛛𝐱
𝐚𝟐 𝐮𝐱𝐱 = 𝐮𝐭𝐭
𝟐
𝛛𝟑 𝐮
𝛛𝐮
( 𝟐
) =𝟏+
𝛛 𝐱𝛛𝐭
𝛛𝐭
Definition of Integration.
Integration is the inverse operation to differentiation. In differentiation, we
solve for the differential of a given function whereas in integration, we solve
for the function corresponding to a given differential. The resulting function is
called the integral of the differential.
Indefinite Integral
The collection of all the possible antiderivatives of a given function is called
the indefinite integral. The indefinite integral comprises of the antiderivative
and the constant of integration (𝑪). The presence of this constant of
integration (in indefinite integration) introduces a family of functions which
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have the same derivative in all points in their domain.
If 𝑭(𝒙) is a function whose derivative 𝑭′ (𝒙) = 𝒇(𝒙) on a certain interval of the
x axis, then 𝑭(𝒙) is called an antiderivative or indefinite integral of 𝒇(𝒙).
The indefinite integral of a given function is not unique; for example, 𝒙𝟐 , 𝒙𝟐 −
𝟔, 𝒙𝟐 + 𝟑 are all indefinite integrals of 𝒇(𝒙) = 𝟐𝒙. All indefinite integrals of
𝒇(𝒙) = 𝟐𝒙 are then included in 𝑭(𝒙) = 𝒙𝟐 + 𝑪.
The Notation
The symbol ∫ 𝒇(𝒙)𝒅𝒙 is used to indicate the indefinite integral of 𝒇(𝒙). Thus,
we write ∫ 𝟐𝒙 𝒅𝒙 = 𝒙𝟐 + 𝑪. In the expression ∫ 𝒇(𝒙)𝒅𝒙, the function 𝒇(𝒙) is
called the integrand and is mathematically expressed as:
The Concept of Integration
For example, find the indefinite integral for 𝒇(𝒙) = 𝒙𝟓 .
•
Base on the definition if 𝑭(𝒙) is such a function we should have:
′
(𝑭(𝒙)) = 𝒙𝟓
Obviously, the function
𝒙𝟔
𝒙𝟔
𝒙𝟔
𝟔
satisfies the above equation but other functions
such as 𝟔 + 𝟏, 𝟔 − 𝟑, and etc. can be considered as an answer so we can
write the answer generally as:
∫ 𝒙𝟒 𝒅𝒙 =
•
𝒙𝟓
+𝑪
𝟓
Using the definition of the indefinite integral, we can find the integral of a
simple functions directly:
o
∫ 𝟎 𝒅𝒙 = 𝑪
o ∫ 𝟏 𝒅𝒙 = 𝒙 + 𝑪
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o ∫ 𝒙𝒏 𝒅𝒙 =
𝒙𝒏+𝟏
𝒏+𝟏
+ 𝑪
(𝒏 ≠ −𝟏)
𝟏
o ∫ 𝒙−𝟏 𝒅𝒙 = ∫ 𝒙 𝒅𝒙 = 𝒍𝒏 |𝒙| + 𝑪
𝒂𝒙
o ∫ 𝒂𝒙 𝒅𝒙 = ∫ 𝒍𝒏 𝒂 + 𝑪
(𝒙 ≠ 𝟎)
(𝒂 ≠ 𝟏, 𝒂 > 𝟎)
o ∫ 𝒄𝒐𝒔 𝒙 𝒅𝒙 = 𝒔𝒊𝒏 𝒙 + 𝑪
o ∫ 𝒔𝒊𝒏 𝒙 𝒅𝒙 = −𝒄𝒐𝒔 𝒙 + 𝑪
Rules of Integration
1. The derivative of the indefinite integral is the integrand:
′
(∫ 𝒇(𝒙)𝒅𝒙) = 𝒇(𝒙)
2. The differential of the indefinite integral is equal to the element of the
integration:
𝒅 (∫ 𝒇(𝒙)𝒅𝒙) = 𝒇(𝒙)𝒅𝒙
3. The indefinite integral of a differential of a function is equal to that function plus
a constant:
∫ 𝒅(𝑭(𝒙)) = 𝑭(𝒙) + 𝑪
4. If 𝒂 ≠ 𝟎 and is a constant, then:
∫ 𝒂𝒇(𝒙)𝒅𝒙 = 𝒂 ∫ 𝒇(𝒙)𝒅𝒙
➢ A constant coefficient goes in and comes out of the integral sign
5. The indefinite integral of the sum or difference of two integrable functions is
equal to the sum or difference of their individual indefinite integral:
∫[𝒇(𝒙) ± 𝒈(𝒙)]𝒅𝒙 = ∫ 𝒇(𝒙)𝒅𝒙 ± ∫ 𝒈(𝒙)𝒅𝒙
6. If ∫ 𝒇(𝒙)𝒅𝒙 = 𝑭(𝒙) + 𝑪, then:
𝟏
∫ 𝒇(𝒂𝒙 + 𝒃)𝒅𝒙 = 𝒂 𝑭(𝒂𝒙 + 𝒃 ) + 𝑪
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And if 𝒃 = 𝟎, then
𝟏
∫ 𝒇(𝒂𝒙 )𝒅𝒙 = 𝒂 𝑭(𝒂𝒙 ) + 𝑪
o Using the last rule, we can easily calculate some integrals without
applying a specific method:
Example:
a. ∫ 𝒆𝒂𝒙 𝒅𝒙 =
𝟏
𝒂
𝒆𝒂𝒙 + 𝑪
𝒅𝒙
b. ∫ 𝒙−𝒂 = 𝒍𝒏 |𝒙 − 𝒂| + 𝑪
c. ∫ 𝒔𝒊𝒏(𝒂𝒙 )𝒅𝒙 =
−𝒄𝒐𝒔 (𝒂𝒙 )
𝒂
+𝑪
The Substitution Method of Integration
o If the integrand is in the form of 𝒇(𝒈(𝒙))𝒈′(𝒙) 𝒅𝒙, and substituting 𝒖 =
𝒈(𝒙), then we will have
∫ 𝒇(𝒈(𝒙))𝒈′(𝒙) 𝒅𝒙 = ∫ 𝒇(𝒖)𝒖′ 𝒅𝒙 = ∫ 𝒇(𝒖) 𝒅𝒖
o And if ∫ 𝒇(𝒖) 𝒅𝒖 = 𝑭(𝒖) + 𝑪, then:
∫ 𝒇(𝒈(𝒙))𝒈′(𝒙) 𝒅𝒙 = 𝑭(𝒈(𝒙)) + 𝑪
Example:
1. Find the indefinite integral ∫
𝒙
𝟏+𝒙𝟐
𝒅𝒙.
Solution:
Let 𝒖 = 𝟏 + 𝒙𝟐 , then 𝒅𝒖 = 𝟐𝒙𝒅𝒙, and we will have:
𝟏
𝒙
𝟐 𝒅𝒖 = 𝟏 ∫ 𝒅𝒖 = 𝟏 𝒍𝒏 |𝒖| + 𝑪
∫
𝒅𝒙
=
∫
𝟏 + 𝒙𝟐
𝒖
𝟐 𝒖
𝟐
since 𝒖 = 𝟏 + 𝒙𝟐 , thus,
𝟏
𝟏
𝒍𝒏 |𝒖| + 𝑪 = 𝒍𝒏 |𝟏 + 𝒙𝟐 | + 𝑪
𝟐
𝟐
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𝟐
2. Find ∫ 𝒙𝒆𝒙 𝒅𝒙.
Solution:
Let 𝒖 = 𝒙𝟐 , then 𝒅𝒖 = 𝟐𝒙𝒅𝒙, and:
𝟏
𝟐
∫ 𝒙𝒆𝒙 𝒅𝒙 = ∫ 𝒆𝒖 (𝟐 𝒅𝒖) =
3. Find ∫
𝒅𝒙
𝒙𝒍𝒏 𝒙
,
𝟏
𝟐
𝒆𝒖 + 𝑪 =
𝟏
𝟐
𝟐
𝒆𝒙 + 𝑪
(𝒙 > 𝟎)
Solution:
Let 𝒖 = 𝒍𝒏 𝒙, then 𝒅𝒖 =
𝒅𝒙
∫ 𝒙𝒍𝒏 𝒙 = ∫
𝒙𝒅𝒖
𝒙𝒖
𝒅𝒙
= ∫
, and:
𝒙
𝒅𝒖
𝒖
= 𝒍𝒏 |𝒖| + 𝑪 = 𝒍𝒏 |𝒍𝒏 (𝒙)| + 𝑪
Note that, having success with this method requires finding a relevant substitution,
which comes after lots of practice.
ORDER
It is the largest derivative present in the differential equation. As per listed above a is
a first order DE, while b, c, f, and g are second order DE, h is a third order DE, and e
is a fourth order DE.
DEGREE
It is the represented by the power of the highest order derivative in the given differential
equation and which the differential coefficients are free from radicals and fractions.
From the examples from c to g it is considered as first degree while h is considered as
a second degree.
TYPE OF DIFFERENTIAL EQUATION
As stated earlier at the equations that exists is either an ordinary differential equation
(ODE) or a partial differential equation (PDE). ODE are those equations that contains
ordinary differential signs e.g. 𝐝𝐲/𝐝𝐱, 𝐲 ′ , 𝐮𝐱𝐲 , while in the other hand PDE has partial
differential signs like e.g.
𝛛𝟐 𝐮
𝛛𝟐 𝐱
. From the examples c to e are ODE and f to h are PDE.
LINEAR DIFFERENTIAL EQUATIONS
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A linear differential equation is any differential equation that can be written in the
following form.
𝐚𝐧 (𝐭)𝐲 𝐧 (𝐭) + 𝐚𝐧−𝟏 (𝐭)𝐲 𝐧−𝟏 (𝐭) + ⋯ + 𝐚𝟏 (𝐭)𝐲′ (𝐭) + 𝐚𝟎 (𝐭)𝐲(𝐭) = 𝐠(𝐭)
(1)
The important thing to note about linear differential equations is that there are no
products of the function, 𝐲(𝐭), and its derivatives and neither the function or its
derivatives occur to any power other than the first power. Also note that neither the
function or its derivatives are “inside” another function, for example, √𝐲′ or 𝐞𝐲
The coefficients 𝐚𝟎 (𝐭), 𝐚𝐧 (𝐭) and 𝐠(𝐭) can be zero or non-zero functions, constant or
non-constant functions, linear or non-linear functions. Only the function 𝐲(𝐭), and its
derivatives are used in determining if a differential equation is linear.
If a differential equation cannot be written in the form of equation 1 then its called a
non – linear differential equation.
In the ODE group from c to e, c and e are considered as linear while d is non – linear.
SOLUTION OF DIFFERENTIAL EQUATIONS
When we first performed integrations, we obtained a general solution (involving a
constant, C).
We obtained a particular solution by substituting known values for 𝐱 and 𝐲. These
known conditions are called boundary conditions (or initial conditions).
It is the same concept when solving differential equations - find general solution first,
then substitute given numbers to find particular solutions.
Example 2
Solve for the a.) general solution of 𝐝𝐲 + 𝟕𝐱𝐝𝐱 = 𝟎 and b.) its particular solution if
𝐲(𝟎) = 𝟑
For the general solution, primarily we can transfer either 𝐝𝐲 or 𝟕𝐱𝐝𝐱 at the left side,
either will have the same outcome.
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𝐝𝐲 = −𝟕𝐱𝐝𝐱
Then at this point we will integrate the two terms individually,
∫ 𝐝𝐲 = ∫ −𝟕𝐱𝐝𝐱
Which will give us an answer of,
𝟕
𝐲 = − 𝐱𝟐 + 𝐂
𝟐
At this point we have solved the general solution of the given differential equation, now
we can easily solve the particular solution by substituting the values of 𝐱 and 𝐲 from
the given.
The values of 𝐲 = 𝟑 , while the value of 𝐱 = 𝟎 that will give us a value for our arbitrary
constant,
𝟕
𝟑 = − (𝟎) + 𝐂
𝟐
𝐂=𝟑
The complete the solution simply plug in the value of 𝐂 to the general solution which
gives us an answer of,
𝟕
𝐲𝐩 = − 𝐱 𝟐 + 𝟑
𝟐
We denoted the particular solution as 𝐲𝐩 to clarity purposes.
𝟕
In summary, our general solution is 𝐲 = − 𝟐 𝐱 𝟐 + 𝐂 while our particular solution is 𝐲𝐩 =
𝟕
− 𝟐 𝐱 𝟐 + 𝟑.
Example 3
Find the particular solution of 𝐲 ′ = 𝟓 given that when 𝐱 = 𝟎, 𝐲 = 𝟐
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Here we expand 𝐲′ to 𝐝𝐲/𝐝𝐱 and multiplying both side by 𝐝𝐱 we will have,
𝐝𝐲 = 𝟓𝐝𝐱
We integrate both sides which gives us,
𝐲 = 𝟓𝐱 + 𝐂
Plugging in the values of 𝐱 and 𝐲 results
𝐲𝐩 = 𝟓𝐱 + 𝟐
In summary, general solution 𝐲 = 𝟓𝐱 + 𝐂 and 𝐲𝐩 = 𝟓𝐱 + 𝟐.
NOTE: You can refer to other resources below for further understanding
Cioranescu , Doina. (2012). Introduction to classical and variational partial differential
equations
Quezon City: The University of the Philippines Press
Zill, Dennis G.(2009). A first course in differential equations with modeling
applications. 9th Australia: Brooks/Cole Cengage Learning
LET’S CHECK! (PRACTICE PROBLEMS)
Problem 1
Solve the following
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Differential Equation
Order Degree
ODE/PDE
Linear/Non linear
𝐲 ′ − 𝐜𝐨𝐬 𝐱 = 𝟎
𝐝𝟐 𝐲
𝐝𝐲 𝟐
𝐝𝐲
𝐱𝐲 (𝐝𝐱𝟐 ) + 𝐱 (𝐝𝐱) − 𝐲 𝐝𝐱 = 𝟎
𝐱(𝐮𝐱𝐱𝐲 )𝟐 + 𝟓(𝐮𝐱𝐲 )𝟑 = 𝟎
𝐮𝐱𝐱 = (𝐱 𝟒 + 𝐲 𝟒 )𝐮𝐲𝐲 + 𝐬𝐢𝐧(𝐱𝐲)𝐮
𝐲 ′′ = 𝐲𝐲 ′ − 𝟗
Problem 2
𝐝𝐲
𝟏
Solve for the general solution of 𝐝𝐱 = 𝐱𝟐 and the particular solution if 𝐲(𝟏) = 𝟒
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Problem 3
𝐝𝐲
𝛑
Solve for the general solution of 𝐝𝐱 = −𝟐 𝐬𝐢𝐧 𝐱 and the particular solution if 𝐲 (𝟐) = 𝛑
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Problem 4
𝐝𝐱
Solve for the general solution of 𝐝𝐲 − 𝐲 𝟐 + 𝟏 = 𝟎 and the particular solution if 𝐲(𝟎) = 𝟏
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LET’S ANALYZE! (PRACTICE PROBLEMS)
Problem 1
Differential Equation
Order
Degree
ODE/P
DE
Linear/
Non linear
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𝐝𝟐 𝐲
𝟐
𝐝𝐲 𝟒
𝐱 𝟑 (𝐝𝐱𝟐 ) + 𝐱 (𝐝𝐱) = 𝟎
𝐝𝟐 𝐲
𝟑
𝐝𝐲 𝟒
𝐱 (𝐝𝐱𝟐 ) + 𝐱𝐞𝐱 (𝐝𝐱) = 𝟎
𝐝𝐲
𝐝𝟐 𝐲
𝐝𝟑 𝐲
𝐞𝐝𝐱 + 𝐬𝐢𝐧 (𝐱 𝐝𝐱𝟐 ) + 𝐜𝐨𝐬 (𝐱 𝐝𝐱𝟑 ) + 𝟕𝐱𝐲 = 𝟎
𝐝𝟐 𝐲
𝟑
𝐱 (𝐝𝐱𝟐 )
𝐝𝐲
𝐝𝟐 𝐲
− 𝟓𝐞𝐱 (𝐝𝐱𝟐 ) + 𝐲 𝐥𝐨𝐠(𝐱) = 𝟎
𝐝𝐱
𝐝𝐲 𝟐
𝐝𝟐 𝐲
𝟒
𝐝𝐲 𝟕
𝟐 𝐥𝐨𝐠 𝐱 (𝐝𝐱) + 𝟕 𝐜𝐨𝐬 𝐱 (𝐝𝐱𝟐 ) (𝐝𝐱) + 𝐱𝐲 = 𝟎
Problem 2
Solve for the general solution of 𝛉𝟐 𝐝𝛉 = 𝐬𝐢𝐧(𝐭 + 𝟎. 𝟐)𝐝𝐭 and the particular solution if
𝛉(𝛑) = 𝟐
Problem 3
𝐝𝐲
Solve for the general solution of 𝐝𝐱 = 𝟑𝐱 𝟐 − 𝟏 and the particular solution if 𝐲(𝟎) = 𝟒
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Problem 4
𝐝𝐲
𝟏
Solve for the general solution of 𝐝𝐱 − 𝐱𝟐 = 𝟎 and the particular solution if 𝐲(𝟏) = 𝟒
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IN A NUTSHELL!
Activity 1
The ORDER of a differential equation depends on the derivative of the highest order
in the equation. The DEGREE of a differential equation, similarly, is determined by the
highest exponent on any variables involved. Can you explain the process of
determining the ORDER and DEGREE of a differential equation? In addition, what is
the relationship of ORDER and DEGREE?
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Activity 2
Arbitrary constants are introduced in integral calculus, it represents all the constant
values. Why is it important to solve the particular solution of a differential equation?
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Big Picture Focus in ULO-b. demonstrate the solution of first order differential
equation.
Metalanguage
In the section, majority of your knowledge in integral calculus will be utilized,
understanding the different ways to solve a differential equation. From ULOa it is
discussed the on how to solve the general solution as well as the particular solution,
this topic is a key in solving the problems.
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Essential Knowledge
Process of integration
Step 1: Simplify the integrand, if possible.
Step 2: See if a “simple” substitution will work
Step 3: Identify the type of integral
Step 4: Can we relate the integral to an integral we already know how to do?
Step 5: Do we need to use multiple techniques?
FIRST ORDER DIFFERENTIAL EQUATION
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A first order differential equation is an equation
𝐝𝐲
𝐝𝐱
= 𝐟(𝐱, 𝐲)
(1)
in which ƒ(x, y) is a function of two variables defined on a region in the 𝐱𝐲-plane. The
equation is of first order because it involves only the first derivative
𝐝𝐲
𝐝𝐱
(and not
higher-order derivatives). We point out that the equations
𝐝
𝐲 ′ = 𝐟(𝐱, 𝐲) and 𝐝𝐱 𝐲 = 𝐟(𝐱, 𝐲)
are equivalent to Equation (1) and all three forms will be used interchangeably in the
text. A solution of Equation (1) is a differentiable function defined on an interval I of x
- values (perhaps infinite) such that
𝐝
𝐲(𝐱) = 𝐟(𝐱, 𝐲(𝐱))
𝐝𝐱
on that interval. That is, when y(x) and its derivative are substituted into Equation (1),
the resulting equation is true for all x over the interval I. The general solution to a first
order differential equation is a solution that contains all possible solutions.
The general solution always contains an arbitrary constant, but having this property
doesn’t mean a solution is the general solution. That is, a solution may contain an
arbitrary constant without being the general solution. Establishing that a solution is the
general solution may require deeper results from the theory of differential equations
and is best studied in a more advanced course.
Variable Separable Differential Equation
if the function 𝐟(𝐱, 𝐲) can be factored into the product of two functions of 𝐱 and 𝐲
𝐟(𝐱, 𝐲) = 𝐩(𝐱)𝐡(𝐲)
Where 𝐩(𝐱) and 𝐡(𝐲) are continuous functions.
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𝐝𝐲
Considering the derivative 𝐲′ as the ratio of two differentials 𝐝𝐱 , we move dx to the
right side and divide the equation by 𝐡(𝐲):
𝐝𝐲
𝐝𝐲
= 𝐩(𝐱)𝐡(𝐲) →
= 𝐩(𝐱)𝐝𝐱
𝐝𝐱
𝐡(𝐲)
Of course, we need to make sure that 𝐡(𝐲) ≠ 𝟎. If there’s a number 𝐲𝐨 such
that 𝐡(𝐲𝐨 ) = 𝟎, then this number will also be a solution of the differential equation.
Division by 𝐡(𝐲) causes loss of this solution.
∫
𝐝𝐲
= ∫ 𝐩(𝐱)𝐝𝐱 + 𝐂
𝐡(𝐲)
Calculating the integrals, we get the expression
𝐇(𝐲) = 𝐏(𝐱) + 𝐂,
Represents the general solution of the variable separable DE.
Example 1
Solve the differential equation of 𝐲 ′ = −𝐱𝐞𝐲
We transform and rearrange the equation into
𝐝𝐲
= −𝐱𝐞𝐲
𝐝𝐱
Integrating them separately
→
𝐞−𝐲 𝐝𝐲 = −𝐱𝐝𝐱
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∫ 𝐞−𝐲 𝐝𝐲 = ∫ −𝐱𝐝𝐱 + 𝐂
−𝐞
−𝐲
𝐱𝟐
=− +𝐂
𝟐
→
𝐞
−𝐲
𝐱𝟐
=
+𝐂
𝟐
𝐱𝟐
𝐱𝟐
𝐥𝐧 𝐞−𝐲 = 𝐥𝐧 ( + 𝐂) → −𝐲 = 𝐥𝐧 ( + 𝐂)
𝟐
𝟐
After simplification using rules in logarithm, we have the general solution
𝐱𝟐
𝐲 = −𝐥𝐧 ( + 𝐂)
𝟐
Example 2
Solve the particular solution of the equation (𝟏 + 𝐞𝐱 )𝐲 ′ = 𝐞𝐱 satisfying the initial
condition 𝐲(𝟎) = 𝟎
We write this equation in the following way
(𝟏 + 𝐞𝐱 )
𝐝𝐲
= 𝐞𝐱 → (𝟏 + 𝐞𝐱 )𝐝𝐲 = 𝐞𝐱 𝐝𝐱
𝐝𝐱
Dividing both sides by 𝟏 + 𝐞𝐱 .
𝐞𝐱 𝐝𝐱
𝐝𝐲 =
(𝟏 + 𝐞𝐱 )
Since 𝟏 + 𝐞𝐱 > 𝟎, then we did not miss solutions of the original equation. Integrating
this equation yields:
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∫ 𝐝𝐲 = ∫
𝐞𝐱 𝐝𝐱
𝐞𝐱 𝐝𝐱
+
𝐂
→
𝐲
=
∫
+𝐂
(𝟏 + 𝐞𝐱 )
(𝟏 + 𝐞𝐱 )
Obtaining the general solution of the equation,
𝐲 = 𝐥𝐧(𝟏 + 𝐞𝐱 ) + 𝐂
Now we will substitute the initial values 𝐲 = 𝟎 and 𝐱 = 𝟎.
𝟎 = 𝐥𝐧(𝟏 + 𝐞𝟎 ) + 𝐂
𝐂 = − 𝐥𝐧 𝟐
Homogeneous Differential Equation
A function 𝐏(𝐱, 𝐲) and 𝐐(𝐱, 𝐲) is called a homogeneous function of the degree n and
m are equal
𝐧 𝐦
𝐏(𝐭 𝐧𝐱 , 𝐭 𝐦
𝐲 ) = 𝐐(𝐭 𝐱 , 𝐭 𝐲 )
*Solving process of Homogeneous Equation
In solving a homogeneous equation, its either let 𝐱 = 𝐲𝐯 and 𝐝𝐱 = 𝐯𝐝𝐲 + 𝐲𝐝𝐯 or 𝐲 =
𝐱𝐯 and 𝐝𝐲 = 𝐯𝐝𝐱 + 𝐱𝐝𝐯. After plugging in the values, expand the equation and
eliminate/combine terms to reduce the equation, then as a result you can integrate
them separately.
Example 3:
Solve the differential equation (𝟐𝐱 + 𝐲)𝐝𝐱 − 𝐱𝐝𝐲 = 𝟎
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Let 𝐲 = 𝐱𝐯 and 𝐝𝐲 = 𝐱𝐝𝐯 + 𝐯𝐝𝐱 and substitute to the given equation, we will get
(𝟐𝐱 + 𝐱𝐯)𝐝𝐱 − 𝐱(𝐱𝐝𝐯 + 𝐯𝐝𝐱) = 𝟎 → 𝟐𝐱𝐝𝐱 + 𝐱𝐯𝐝𝐱 − 𝐱 𝟐 𝐝𝐯 − 𝐱𝐯𝐝𝐱 = 𝟎
Now we will eliminate/combine terms to reduce the equation
𝟐𝐱𝐝𝐱 − 𝐱 𝟐 𝐝𝐯 = 𝟎 → 𝟐𝐱𝐝𝐱 = 𝐱 𝟐 𝐝𝐯
If we evaluate the resulting equation it is noticed that it can be solved by variable
separable by grouping 𝐱 and 𝐯
𝟐∫
𝐝𝐱
= ∫ 𝐝𝐯 + 𝐂
𝐱
Which gives us an answer of
𝟐 𝐥𝐧 𝐱 = 𝐯 + 𝐂 → 𝐥𝐧 𝐱 𝟐 = 𝐯 + 𝐂
At this point we observed that the equation contains 𝐱 and 𝐯 variable, were it supposed
to be 𝐱 and 𝐲. But we know that earlier we let 𝐲 = 𝐱𝐯, rearranging this we will have 𝐯 =
𝐲
, so we simply substitute it to our current equation to generate the general solution
𝐱
which is
𝐥𝐧 𝐱 𝟐 =
𝐲
+𝐂 →
𝐱
𝐲 = 𝐱 𝐥𝐧 𝐱 𝟐 + 𝐂
Example 4
Solve the differential equation (𝐱𝐲 + 𝐲 𝟐 )𝐲 ′ = 𝐲 𝟐
Let 𝐲 = 𝐱𝐯 and 𝐝𝐲 = 𝐱𝐝𝐯 + 𝐯𝐝𝐱 and expand
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(𝐱(𝐱𝐯) + (𝐱𝐯)𝟐 )
(𝐱𝐝𝐯 + 𝐯𝐝𝐱)
= (𝐱𝐯)𝟐
𝐝𝐱
(𝐱 𝟐 𝐯 + 𝐱 𝟐 𝐯 𝟐 )(𝐱𝐝𝐯 + 𝐯𝐝𝐱) = 𝐱 𝟐 𝐯 𝟐 𝐝𝐱
𝐱 𝟑 𝐯𝐝𝐯 + 𝐱 𝟐 𝐯 𝟐 𝐝𝐱 + 𝐱 𝟑 𝐯 𝟐 𝐝𝐱 + 𝐱 𝟐 𝐯 𝟐 𝐝𝐱 = 𝐱 𝟐 𝐯 𝟐 𝐝𝐱
Now we can observe that each term 𝐱 𝟐 is common, we can divide all the equation by
𝐱 𝟐 , resulting.
𝐱𝐯𝐝𝐯 + 𝐯 𝟐 𝐝𝐱 + 𝐱𝐯 𝟐 𝐝𝐱 + 𝐯 𝟐 𝐝𝐱 = 𝐯 𝟐 𝐝𝐱
At this point we combine/eliminate like terms and group 𝐱 and 𝐯.
𝐱𝐯𝐝𝐯 + 𝐯 𝟐 𝐝𝐱 + 𝐱𝐯 𝟐 𝐝𝐱+= 𝟎 →
𝐱𝐯𝐝𝐯 + 𝐯 𝟐 (𝟏 + 𝐱)𝐝𝐱
𝐝𝐯 𝟏 + 𝐱
𝐝𝐯 𝟏
=
𝐝𝐱 →
= 𝐝𝐱 + 𝐝𝐱
𝐯
𝐱
𝐯
𝐱
𝐲
𝐥𝐧 𝐯 = 𝐥𝐧 𝐱 + 𝐱 + 𝐂 → 𝐥𝐧 = 𝐥𝐧 𝐱 + 𝐱 + 𝐂
𝐱
To simplify it we can multiply 𝐞 to the whole equation, having the final equation
𝐲
= 𝐱 + 𝐞𝐱 + 𝐂 → 𝐲 = 𝐱 𝟐 + 𝐞𝐱 + 𝐂
𝐱
Exact Differential Equation
Is a type of a differential equation
𝐏(𝐱, 𝐲)𝐝𝐱 + 𝐐(𝐱, 𝐲)𝐝𝐲 = 𝟎
when two variables (𝐱, 𝐲) with continuous partial derivative.
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*Solving process of Exact Differential Equation
Primarily, we will test the exactness of the equation by partially differentiating the terms
with, the key factor is the differential signs 𝐝𝐲 and 𝐝𝐱.
𝐏(𝐱, 𝐲)𝐝𝐱 + 𝐐(𝐱, 𝐲)𝐝𝐲 = 𝟎
Here, we will partially differentiate 𝐏(𝐱, 𝐲)𝐝𝐱 to 𝛛𝐲 while 𝐐(𝐱, 𝐲)𝐝𝐲 to 𝛛𝐱, in other words
we will differentiate the two terms with their opposite differential signs. Then the
outcome should be equal.
𝛛𝐏 𝛛𝐐
=
𝛛𝐲 𝛛𝐱
The exactness test is not limited into two terms, it depends on how many are there.
As long as you proved correctly that they are exact it can be solved by this type of
differential equation.
As soon as we proved their exactness, we will make two equations 𝐅𝐱 and 𝐅𝐲
separating the two terms which it will be like this.
𝐅𝐱 = ∫ 𝐏(𝐱, 𝐲)𝛛𝐱 + 𝐠 𝐲 and 𝐅𝐲 = ∫ 𝐐(𝐱, 𝐲)𝛛𝐲 + 𝐡𝐱
From here, we will operate partial integration for each equation.
𝐅𝐱 = 𝐏(𝐱, 𝐲) + 𝐠 𝐲 and 𝐅𝐲 = 𝐐(𝐱, 𝐲) + 𝐡𝐱
Our goal here is to identify the values of 𝐠 𝐲 and 𝐡𝐱 by comparing the unlike terms of
the two partially integrated equations. For 𝐠 𝐲 , it is the unlike 𝐲 variable term at 𝐅𝐲 , while
𝐡𝐱 , is the unlike 𝐱 variable at 𝐅𝐱 .
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For the general solution, simply plug in either the values of 𝐠 𝐲 and 𝐡𝐱 to 𝐅𝐲 and 𝐅𝐱
respectively. You can also observe if you plug in the other value the answer will be the
same.
Example 5
Solve the differential equation 𝟐𝐱𝐲𝐝𝐱 + (𝐱 𝟐 + 𝟑𝐲 𝟐 ) 𝐝𝐲 = 𝟎.
We perform the exactness test each term.
𝐏 = 𝟐𝐱𝐲 (partially differentiating by 𝐲)
𝛛𝐏
= 𝟐𝐱
𝛛𝐲
𝐐 = (𝐱 𝟐 + 𝟑𝐲 𝟐 ) (partially differentiating by 𝐱)
𝛛𝐐
= 𝟐𝐱
𝛛𝐱
Now that we proved their exactness, we can proceed. We will let 𝐅𝐱 and 𝐅𝐲 be equal
to the respective terms of 𝐏 and 𝐐.
𝐅𝐱 = ∫ 𝟐𝐱𝐲 𝛛𝐱 + 𝐠 𝐲 and 𝐅𝐲 = ∫(𝐱 𝟐 + 𝟑𝐲 𝟐 )𝛛𝐲 + 𝐡𝐱
At this point we will operate partial integration for 𝐅𝐱 and 𝐅𝐲 .
𝐅𝐱 = ∫ 𝟐𝐱𝐲 𝛛𝐱 + 𝐠 𝐲 → 𝐅𝐱 = 𝐱 𝟐 𝐲 + 𝐠 𝐲
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𝐅𝐲 = ∫ 𝐱 𝟐 𝛛𝐲 + ∫ 𝟑𝐲 𝟐 𝛛𝐲 + 𝐡𝐱 → 𝐅𝐲 = 𝐱 𝟐 𝐲 + 𝐲 𝟑
So, as we now evaluate the current equation, comparing 𝐅𝐱 and 𝐅𝐱 they have a
common term which is 𝐱 𝟐 𝐲 while the difference is 𝐲 𝟑 from 𝐅𝐲 which will be the value
for 𝐠 𝐲 . For 𝐡𝐱 , since there are no other terms found at 𝐅𝐱 it will be equal to 𝟎.
𝐠 𝐲 = 𝐲 𝟑 and 𝐡𝐱 = 𝟎
Finally, the general solution will be either from 𝐅𝐱 or 𝐅𝐱
𝐅𝐱 = 𝐱 𝟐 𝐲 + 𝐠 𝐲 → 𝐅𝐱 = 𝐱 𝟐 𝐲 + 𝐲 𝟑 + 𝐂
𝐅𝐲 = 𝐱 𝟐 𝐲 + 𝐲 𝟑 + 𝐂
𝐱𝟐𝐲 + 𝐲𝟑 + 𝐂 = 𝟎
Example 6
Solve the equation 𝐞𝐲 𝐝𝐱 + (𝟐𝐲 + 𝐱𝐞𝐲 )𝐝𝐲 = 𝟎.
Partially differentiate the first terms, and prove the exactness
𝛛𝐏
𝛛𝐲
=𝐞𝐲 and
𝛛𝐐
𝛛𝐱
=𝐞𝐲
Now, we partially integrate the terms,
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𝐅𝐱 = ∫ 𝐞𝐲 𝛛𝐱 + 𝐠 𝐲 and 𝐅𝐲 = ∫(𝟐𝐱 + 𝐱𝐞𝐲 )𝛛𝐲
The results will be
𝐅𝐱 = 𝐱𝐞𝐲 + 𝐠 𝐲 and 𝐅𝐲 = 𝟐𝐱𝐲 + 𝐱𝐞𝐲 + 𝐡𝐱
𝐠 𝐲 = 𝟐𝐱𝐲 and 𝐡𝐱 = 𝟎
The general solution will be
𝟐𝐱𝐲 + 𝐱𝐞𝐲 + 𝐂 = 𝟎
Linear Differential Equation
A differential equation of type
𝐲 ′ + 𝐲𝐠(𝐱) = 𝐟(𝐱)
(1)
where 𝐠(𝐱) and 𝐟(𝐱) are continuous functions of 𝐱, is called a linear differential
equation of first order. An indication that it can be solved by this method is the
arrangement of the equations. We have the 𝐲 ′ , the term 𝐠(𝐱, 𝐲) where found at the left
𝐝𝐱
side and 𝐟(𝐱) at the right side. This can be altered if your arrangement is 𝐝𝐲 which will
𝐝𝐲
be a lot different to 𝐝𝐱, you will have another arrangement of
𝐱 ′ + 𝐱𝐠(𝐲) = 𝐟(𝐲)
*Solving Linear Differential Equation
After arranging the desired equation, we will introduce two variable which are 𝐏(𝐱) and
𝐐(𝐱). Using equation (1), for 𝐏(𝐱) = 𝐠(𝐱) only, variable y is not included, while 𝐐(𝐱) =
𝐟(𝐱).
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In solving the general solution of a linear differential equation, it requires 𝐏(𝐱), 𝐐(𝐱)
and ∅.
𝐲∅ = ∫ ∅𝐐(𝐱)𝐝𝐱 + 𝐂
Where:
∅ = 𝐈𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐧𝐠 𝐅𝐚𝐜𝐭𝐨𝐫 = 𝐞∫ 𝐏(𝐱)𝐝𝐱
Example 7
Solve the equation 𝐲′ − 𝐲 − 𝐱𝐞𝐱 = 𝟎
Primarily we need to rearrange the equation
𝐲 ′ − 𝐲(𝟏) = 𝐱𝐞𝐱
Now, we let 𝐏(𝐱) = −𝟏 and 𝐐(𝐱) = 𝐱𝐞𝐱 and solve for the integrating factor which will
be 𝐞−𝐱.
∅ = 𝐞∫(−𝟏)𝐝𝐱 = 𝐞−𝐱
At this point we will just need to substitute it to our general solution formula and
integrate, then we will have an answer.
𝐲(𝐞−𝐱 ) = ∫ 𝐱𝐞𝐱 (𝐞−𝐱 )𝐝𝐱 + 𝐂
𝐲 = 𝐞𝐱 (
→
𝐲(𝐞−𝐱 ) = ∫ 𝐱𝐝𝐱 + 𝐂
𝐱𝟐
+ 𝐂)
𝟐
Example 8
Solve the differential equation (IVP) 𝐲 ′ +
𝟑𝐲
𝐱
=
𝟐
𝐱𝟐
with the initial condition 𝐲(𝟏) = 𝟐.
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Rearrange the equation and identify 𝐏(𝐱), 𝐐(𝐱), and ∅
𝐝𝐲
𝟑
𝟐
+ 𝐲( ) = 𝟐
𝐝𝐱
𝐱
𝐱
𝐏(𝐱) =
𝟑
𝐱
𝐐(𝐱) =
𝟐
𝐱𝟐
For the integrating factor,
𝟑
∅ = 𝐞∫𝐱𝐝𝐱
→
𝐝𝐱
∅ = 𝐞𝟑 ∫ 𝐱
∅ = 𝐱𝟑
Substitute to our general solution formula,
𝟐
𝐲𝐱 𝟑 = ∫ 𝐱 𝟑 ( 𝟐 ) 𝐝𝐱 + 𝐂
𝐱
→
𝐲𝐱 𝟑 = ∫ 𝟐𝐱𝐝𝐱 + 𝐂
𝐲𝐱 𝟑 = 𝐱 𝟐 + 𝐂
Bernoulli’s Equation
Bernoulli equation is one of the well-known nonlinear differential equations of the first
order. It is written as
𝐲 ′ + 𝐲𝐠(𝐱) = 𝐡(𝐱)𝐲 𝐧
where 𝐠(𝐱) and 𝐡(𝐱) are continuous functions.
If 𝐧 = 𝟎. It is considered as linear differential equation like what we learned last time.
In case of 𝐧 = 𝟏, it will be solved by variable separable.
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Generally, Bernoulli’s Equation is like Linear Differential Equation in majority of ways
in solving, the different part is just from Bernoulli we will convert it into Linear utilizing
change of variable
𝐳 = 𝐲 𝟏−𝐧
𝐏(𝐱), 𝐐(𝐱), ∅ will be transformed by using the formula
𝐝𝐳
= 𝐳 ′ + 𝐳𝐏(𝐱)𝐑 = 𝐐(𝐱)𝐑
𝐝𝐱
Where:
𝐏(𝐱)𝐑 = (𝟏 − 𝐧)𝐏(𝐱)
𝐐(𝐱)𝐑 = (𝟏 − 𝐧)𝐐(𝐱)
∅ = 𝐞∫ 𝐏(𝐱)𝐑𝐝𝐱
As we can observe, the 𝐲 variable from the linear DE is just replaced by 𝐳. Which gives
us, a new general solution formula,
𝐳∅ = ∫ ∅𝐐(𝐱)𝐑 𝐝𝐱 + 𝐂
Example 9
Solve the general solution of the equation 𝐲′ − 𝐲 = 𝐲 𝟐 𝐞𝐱 .
Same as the process of linear DE, first we identify (𝐱) ,𝐐(𝐱), and 𝐧.
𝐏(𝐱) = −𝟏
𝐐(𝐱) = 𝐞𝐱
𝐧=𝟐
Take note that 𝐧 is just the power of our variable y at the left-hand side, where 𝐳 =
𝟏
𝐲 𝟏−𝐧 = 𝐲 −𝟏 . Where its derivative is 𝐳’ = (− 𝐲𝟐 ) 𝐲’
Divide the whole equation by −𝐲 𝟐
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𝐲 ′ − 𝐲 = 𝐲 𝟐 𝐞𝐱
𝟏
Now, we substitute 𝐳 = 𝐲 and 𝐳’ =
→ −
𝐲′ 𝟏
+ = −𝐞𝐱
𝐲𝟐 𝐲
−𝐲’
𝐲𝟐
.
𝐳 ′ + 𝐳(𝟏) = −𝐞𝐱
Next will be solving for the integrating factor,
∅ = 𝐞∫ 𝐝𝐱 = 𝐞𝐱
Then, substitute to our general solution formula 𝐐(𝐱)𝐑 = −𝐞𝐱 and ∅ = 𝐞𝐱
𝐳𝐞𝐱 = ∫ −𝐞𝐱 (𝐞𝐱 )𝐝𝐱 + 𝐂
→ 𝐳𝐞𝐱 = −
𝐞𝟐𝐱
+𝐂
𝟐
At this point what we need is to turn it back to its original variables 𝐱 and 𝐲. A while
𝟏
ago we learned that 𝐳 = 𝐲 𝟏−𝐧 or 𝐳 = 𝐲, plugging in it to our current equation, we will
have a final answer of.
𝟏 −𝐞𝐱 𝐂
=
+ 𝐱 → 𝟐𝐲 = 𝐂𝐞−𝐱 − 𝐞𝐱
𝐲
𝟐
𝐞
Example 10
Find the solution of the DE 𝟒𝐱𝐲𝐲′ = 𝐲 𝟐 + 𝐱 𝟐 , satisfying the initial condition 𝐲(𝟏) = 𝟏.
Here in first glance, its not obvious that it is under Bernoulli, but with a bit of equation
rearranging it can be Bernoulli.
Dividing the while term by 𝟒𝐱𝐲, we will have
𝐲
𝐱
𝐲 ′ − 𝟒𝐱 = 𝟒𝐲
𝟏
(1)
𝐱
𝟏
Next will be, we identified that 𝐏(𝐱) = − 𝟒𝐱 , 𝐐(𝐱) = 𝟒, and 𝐧 = −𝟏. So 𝐏(𝐱)𝐑 = 𝟒𝐱 and
𝐱
𝐐(𝐱)𝐑 = − 𝟒
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But as we can observe the value of 𝐧 = −𝟏, which results 𝐳 = 𝐲 𝟏−𝐧 = 𝐲 𝟐 , and the
resulting derivative of this is 𝐳’ = 𝟐𝐲𝐲’. In this case we will multiply the equation (1) by
𝟐𝐲
𝟐𝐲𝐲 ′ −
𝟐𝐲 𝟐 𝟐𝐲𝐱
=
𝟒𝐱
𝟒𝐲
→ 𝟐𝐲𝐲 ′ −
𝐲𝟐 𝐱
=
𝟐𝐱 𝟐
Then, replace 𝐳 = 𝐲 𝟐 and 𝐳’ = 𝟐𝐲𝐲’
𝐳 + 𝐳(
𝟏
𝐱
)=
𝟐𝐱
𝟐
𝐝𝐱
𝟏
At this point the integrating factor will be ∅ = 𝐞− ∫𝟐𝐱 = 𝐞−𝟐 𝐥𝐧 𝐱 = 𝐞
𝐥𝐧
𝟏
√𝐱
𝐱
=
𝟏
, then
√𝐱
substitute it to our general solution formula with 𝐐(𝐱)𝐑 = 𝟐.
𝟏
𝟏 𝐱
𝐳( ) = ∫
( ) 𝐝𝐱 + 𝐂
√𝐱
√𝐱 𝟐
We integrate the equation and we will have,
𝟏
𝟏
𝐳 ( ) = ∫ √𝐱 𝐝𝐱 + 𝐂
𝟐
√𝐱
𝟏
𝟑
𝟏 𝐱𝟐
→ 𝐳( ) = ( ) + 𝐂
𝟐 𝟑
√𝐱
𝟐
𝟏 𝟑
𝐱𝟐
𝟐
𝐳 ( ) = (𝐱 ) + 𝐂 → 𝐳 =
+ 𝐂√ 𝐱
𝟑
𝟑
√𝐱
𝟏
Substitute back the value of 𝐳 = 𝐲 𝟐 , and we will have the general solution
𝐲= √
𝐱𝟐
+ 𝐂√ 𝐱
𝟑
Finally, for the particular solution. Simply solve for the value of 𝐂 if 𝐲 = 𝟏 and 𝐱 = 𝟏
𝟐
where 𝐂 = 𝟑 that results into our final answer
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𝐲= √
𝐱 𝟐 𝟐√ 𝐱
+
𝟐
𝟑
NOTE: You can refer to other resources below for further understanding
Cioranescu , Doina. (2012). Introduction to classical and variational partial differential
equations
Quezon City: The University of the Philippines Press
Zill, Dennis G.(2009). A first course in differential equations with modeling applications.
9th Australia: Brooks/Cole Cengage Learning
LET’S CHECK! (PRACTICE PROBLEMS)
Problem 1
Identify the differential equation type that will the given equation, and explain how.
𝐲′ + 𝐲𝐱 = 𝐲 𝟐
𝐱𝐲′ = 𝐲 + 𝟐𝐱 𝟑
(𝐱 𝟑 + 𝐱𝐲 𝟐 )𝐲′ = 𝐲 𝟑
(𝟔𝐱 𝟐 − 𝐲 + 𝟑)𝐝𝐱 + (𝟑𝐲 𝟐 − 𝐱 − 𝟐)𝐝𝐲 = 𝟎
(𝐱 𝟐 + 𝟒)𝐲 ′ = 𝟐𝐱𝐲
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Problem 2
Solve the particular solution of the DE (𝟏 + 𝐞𝐱)𝐲 ′ = 𝐞𝐱 if 𝐲(𝟎) = 𝟎.
Problem 3
Solve the differential equation 𝐲 ′ =
𝐱−𝐲+𝟑
𝐱−𝐲
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Problem 4
Solve the DE (𝟐𝐱𝐲 − 𝐬𝐢𝐧𝐱)𝐝𝐱 + (𝐱 𝟐 − 𝐜𝐨𝐬𝐲)𝐝𝐲 = 𝟎.
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Problem 5
Solve the DE 𝐱 𝟐 𝐲′ + 𝐱𝐲 + 𝟐 = 𝟎.
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Problem 6
Determine the solution of 𝐲 ′ +
𝟐𝐲
𝐱
= 𝟐𝐱√𝐲.
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LET’S ANALYZE! (PRACTICE PROBLEMS)
Problem 1
𝐲
For 𝐲 ′ 𝐥𝐧 𝐱 − 𝐱 = 𝟎 what will be its solution?
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Problem 2
𝑑𝑦
Solve the differential equation 𝑑𝑥 + 3𝑥 2 𝑦 = 6𝑥 2
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Problem 3
Solve the equation 𝑥 2 𝑦’ + 𝑥𝑦 = 1 if 𝑦(1) = 2
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Problem 4
Solve 𝑦’ + 2𝑥𝑦 = 1
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Problem 5
y
Solve y ′ + x − √y = 0 if y(1) = 0
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Problem 6
Solve the equation 6y ′ − 2y = xy 4 if y(0) = −2
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IN A NUTSHELL!
Activity 1
From our previous lesson, we learned about what is order and degree. Now how does
the topic related at the types of differential equations?
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Activity 2
We learned all about the different types of differential equation and their different
approaches to an equation. Now can you share your knowledge on how to a differential
equation? Explain what are the indications for each types of differential equation.
Big Picture
Week 4-5: Unit Learning Outcomes (ULO): At the end of the unit, you are expected to
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a. apply the solutions of first order differential equation to engineering problems.
Big Picture in Focus ULO a. apply the solutions of first order differential equation to
engineering problems.
Metalanguage
In this section, equation modeling is a big task that you must undertake and master,
by this you can solve situational problems in growth and decay, half-life, cooling and
heating of an object, and mixtures. Being an engineer, you must contain this basic
knowledge to practice your field. Please refer to these definitions in case you will
encounter difficulty in the in understanding educational concepts.
Please proceed immediately to the “Essential Knowledge” part since the first
lesson is also definition of essential terms.
Essential Knowledge
The knowledge that you had gained from the previous topics will play a big role at this
part of the subject. Solving the general solution and particular solution, and deciding
what first order differential equation must be utilized to model/generate the solution.
FIRST ORDER DIFFERENTIAL EQUATION (APPLICATIONS)
1. Growth and Decay
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2. Newtons Law of Cooling and Heating
3. Mixture of two substances
4. Electric Circuit
Growth and Decay
The term growth and decay is not limited to the growth of a population and its decay,
it can be the increase and decrease of a certain substance, loss and gain of a capacity,
and so on.
We have this differential equation that will describe any growth and decay.
𝐝𝐏
= 𝐤𝐏
𝐝𝐭
Where,
𝐏 = 𝐏𝐨𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧 𝐚𝐭 𝐚𝐧𝐲 𝐭𝐢𝐦𝐞 (𝐭)
𝐝𝐏
𝐝𝐭
= 𝐆𝐫𝐨𝐰𝐭𝐡/𝐃𝐞𝐜𝐚𝐲 𝐫𝐚𝐭𝐞 𝐚𝐭 𝐚𝐧𝐭 𝐭𝐢𝐦𝐞 (𝐭)
𝐤 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐨𝐟 𝐩𝐫𝐨𝐩𝐨𝐫𝐭𝐢𝐨𝐧𝐚𝐥𝐢𝐭𝐲
Here, we will solve for the general solution of the given differential equation.
Applying variable separable, we will have
∫
𝐝𝐏
= ∫ 𝐤𝐝𝐭
𝐏
𝐥𝐧 𝐏 = 𝐤𝐭 + 𝐂
(1)
Example 1
A bacteria culture starts with 500 bacteria and after 3 hours there are 8000 bacteria.
a) Find the number of bacteria after 4 hours, and c) When will the population reach
30,000?
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Primarily, we should analyze the time and its situation. And, we will be utilizing
equation (1).
Initially, the population of bacteria is 500 units (take not that is zero (0) time). In here
we can solve the value of 𝐂.
𝐥𝐧 𝟓𝟎𝟎 = 𝐤(𝟎) + 𝐂
→
𝐂 = 𝐥𝐧 𝟓𝟎𝟎
Then after 3 hours bacteria grown 8000 units, here we can solve for the value of 𝐤
with 𝐂 = 𝐥𝐧 𝟓𝟎𝟎.
𝐥𝐧 𝟖𝟎𝟎𝟎 = 𝐤(𝟑) + 𝐥𝐧 𝟓𝟎𝟎
𝐤=
𝐥𝐧 𝟏𝟔
𝟑
At this point, we can observe that all the constant variable from equation (1) has its
respective values, so we can solve any situation in the problem.
For the first question, what will be the population after 4 hours?
𝟏
𝐥𝐧 𝐏 = ( 𝐥𝐧 𝟏𝟔) (𝟒) + 𝐥𝐧 𝟓𝟎𝟎
𝟑
𝐏 = 𝟐𝟎, 𝟏𝟓𝟖. 𝟕𝟑𝟕 𝐮𝐧𝐢𝐭𝐬 ≈ 𝟐𝟎, 𝟏𝟔𝟎 𝐮𝐧𝐢𝐭𝐬
Then for the next question, what will be the time needed to attain 30,000 units?
𝐥𝐧 𝟑𝟎, 𝟎𝟎𝟎 =
𝟏
𝐥𝐧 𝟏𝟔 (𝐭) + 𝐥𝐧 𝟓𝟎𝟎
𝟑
𝐭 = 𝟒. 𝟒𝟑 𝐡𝐨𝐮𝐫𝐬
Problem 2
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Suppose a population of insects increases according to the law of exponential growth.
There were 130 insects at the third day of the experiment and 380 insects at the
seventh day. Approximately how many insects were in the original population?
As we can observe there is no initial value at the problem, but still we can solve it by
using what is given. We will use the most recent event which is the 130 insects at the
third day
𝐥𝐧 𝟏𝟑𝟎 = 𝐤(𝟑) + 𝐂
→
𝐂 = 𝐥𝐧 𝟏𝟑𝟎 − 𝐤(𝟑)
Then for the next event is, 380 insects at the seventh day,
𝐥𝐧 𝟑𝟖𝟎 = 𝐤(𝟕) + 𝐂
→
𝐂 = 𝐥𝐧 𝟑𝟖𝟎 − 𝐤(𝟕)
Here, we have two equations and two unknowns, so we can solve for 𝐂 and 𝐤.
𝐂 = 𝟒. 𝟎𝟔𝟑𝟓
𝐤 = 𝟎. 𝟐𝟔𝟖𝟐
So now, we can determine what is the initial population of insects.
𝐥𝐧 𝐏 = 𝟎. 𝟐𝟔𝟖𝟐(𝟎) + 𝟒. 𝟎𝟔𝟑𝟓
𝐏 = 𝟓𝟖 𝐢𝐧𝐬𝐞𝐜𝐭𝐬
Newtons Law of Cooling and Heating
In the late of 17th century British scientist Isaac Newton studied cooling of bodies.
Experiments showed that the cooling rate approximately proportional to the difference
of temperatures between the heated body and the environment. This fact can be
written as the differential relationship:
𝐝𝐐
= 𝛂𝐀(𝐓 − 𝐓𝐒),
𝐝𝐭
where 𝐐 is the heat, 𝐀 is the surface area of the body through which the heat is
transferred, 𝐓 is the temperature of the body, 𝐓𝐬 is the temperature of the surrounding
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environment, α is the heat transfer coefficient depending on the geometry of the body,
state of the surface, heat transfer mode, and other factors.
As 𝐐 = 𝐂𝐓, where 𝐂 is the heat capacity of the body, we can write:
𝐝𝐓
= 𝛂𝐀𝐂(𝐓𝐒 − 𝐓) = 𝐤(𝐓 − 𝐓𝐒)
𝐝𝐭
The given differential equation has the solution in the form:
𝐝𝐓
= 𝐤(𝐓 − 𝐓𝐒)
𝐝𝐭
∫
𝐝𝐓
= ∫ 𝐤𝐝𝐭
(𝐓 − 𝐓𝐒)
𝐥𝐧 (𝐓 − 𝐓𝐒) = 𝐤𝐭 + 𝐂
→
𝐂𝐨𝐨𝐥𝐢𝐧𝐠
𝐥𝐧 (𝐓𝐒 − 𝐓) = 𝐤𝐭 + 𝐂
→
𝐇𝐞𝐚𝐭𝐢𝐧𝐠
Example 3
The temperature of a body dropped from 𝟐𝟎𝟎℉ to 𝟏𝟎𝟎℉ for the first hour. Determine
how many degrees the body cooled in one hour more if the environment temperature
is 𝟎℉?
Let the initial temperature of the heated body be 𝐓 = 𝟐𝟎𝟎℉. The further temperature
dynamics is described by the formula
𝐥𝐧( 𝟐𝟎𝟎 − 𝟎) = 𝐤(𝟎) + 𝐂
𝐂 = 𝐥𝐧 𝟐𝟎𝟎
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At the end of the first hour the body has cooled to 100∘. Therefore, we can write the
following relationship
𝐥𝐧( 𝟏𝟎𝟎 − 𝟎) = 𝐤(𝟏) + 𝐥𝐧 𝟐𝟎𝟎
𝟏𝟎𝟎
𝟏
) = 𝐥𝐧
𝐤 = 𝐥𝐧 (
𝟐𝟎𝟎
𝟐
Finally, for another hour
𝐥𝐧 𝐓 − 𝟎 = 𝐥𝐧
𝟏
(𝟐) + 𝐥𝐧 𝟐𝟎𝟎
𝟐
𝐓 = 𝟓𝟎℉
Example 4
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.
Here, we have 𝐓𝐒 = 𝟐𝟎℃ and for the initial state we have a 𝐓 = 𝟒𝟎℃
𝐥𝐧 𝟒𝟎 − 𝟐𝟎 = 𝐤(𝟎) + 𝐂
𝐂 = 𝐥𝐧 𝟐𝟎
Then, after ten (𝟏𝟎) minutes the temperature of the body dropped to 𝟑𝟓℃
𝐥𝐧 𝟑𝟓 − 𝟐𝟎 = 𝐤(𝟏𝟎) + 𝐥𝐧 𝟐𝟎
𝐤=
𝟑
𝐥𝐧 𝟒
𝟏𝟎
Now, for the time required to attain 𝟑𝟎℃
𝟑
𝐥𝐧 𝟑𝟎 − 𝟐𝟎 = 𝟒 (𝐭) + 𝐥𝐧 𝟐𝟎
𝟏𝟎
𝐥𝐧
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𝐭 = 𝟏𝟒 𝐦𝐢𝐧𝐮𝐭𝐞𝐬
Mixture Problems
At this section of the topic we will be modeling an equation to generate a solution to
solve problems. The only formula that we will be utilizing is
𝐝𝐐
= 𝐑𝐎𝐆 − 𝐑𝐎𝐋
𝐝𝐭
Where:
𝐦𝐚𝐬𝐬
𝐑𝐎𝐆 = 𝐫𝐚𝐭𝐞 𝐨𝐟 𝐠𝐚𝐢𝐧 ( 𝐭𝐢𝐦𝐞 )
𝐦𝐚𝐬𝐬
𝐑𝐎𝐋 = 𝐫𝐚𝐭𝐞 𝐨𝐟 𝐥𝐨𝐬𝐬 ( 𝐭𝐢𝐦𝐞 )
Example 5
A tank originally contains 100 gal of fresh water. Then water containing 1/2 lb of salt
per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to
leave at the same rate. What is the amount of salt at any instant? 10 minutes? 30
minutes?
First, we need to analyze what is the given
At point A which is the intake, has a flowrate of 𝟐 𝐠𝐚𝐥/𝐦𝐢𝐧 with a concentration of
𝟏 𝐥𝐛𝐬𝐚𝐥𝐭
𝟐 𝐠𝐚𝐥𝐰𝐚𝐭𝐞𝐫
. Then at point B, initially it contains 𝟏𝟎𝟎 𝐠𝐚𝐥𝐥𝐨𝐧𝐬 𝐨𝐟 𝐟𝐫𝐞𝐬𝐡 𝐰𝐚𝐭𝐞𝐫. At point C, it
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is said that the flowrate is the same with the intake which is 𝟐 𝐠𝐚𝐥/𝐦𝐢𝐧 but with
unknown concentration.
Now we will use our formula,
𝐝𝐐
= 𝐑𝐎𝐆 − 𝐑𝐎𝐋
𝐝𝐭
For ROG it will be easy to determine, since the volume flow rate and the density of the
intake is given
𝛒𝐢𝐧 =
𝐦𝐚𝐬𝐬𝐢𝐧
𝐯𝐨𝐥𝐮𝐦𝐞𝐢𝐧
→
𝟏 𝐥𝐛𝐬𝐚𝐥𝐭
𝐦𝐚𝐬𝐬𝐢𝐧
=
𝐠𝐚𝐥
𝟐 𝐠𝐚𝐥𝐰𝐚𝐭𝐞𝐫
𝟐 𝐦𝐢𝐧
𝐦𝐚𝐬𝐬𝐢𝐧 = 𝐑𝐎𝐆 = 𝟏
𝐥𝐛
𝐦𝐢𝐧
For ROL, we need to analyze the whole system running. We go back to the formula of
density,
𝛒𝐨𝐮𝐭 =
𝐦𝐚𝐬𝐬𝐨𝐮𝐭
𝐯𝐨𝐥𝐮𝐦𝐞𝐨𝐮𝐭
Here, the only data that we have is the volume flow rate which is equal to the intake
𝐠𝐚𝐥
𝟐 𝐦𝐢𝐧
𝛒𝐨𝐮𝐭 =
𝐦𝐚𝐬𝐬𝐨𝐮𝐭
𝟐
→
𝟐𝛒𝐨𝐮𝐭 = 𝐦𝐚𝐬𝐬𝐨𝐮𝐭
(1)
Now, we will analyze the density in a different way, which involves equation modeling.
We let the mass of salt be equal to 𝐐 at any time 𝐭 over the total volume during the
process which incorporates the initial volume of the tank plus the flowrate. As we can
observe at we subtract the two flowrates and multiplied it by 𝐭, we do this to cancel out
the time unit to the volume flowrate so we can determine the total volume.
𝐐
𝟏𝟎𝟎 + (𝟐 − 𝟐)𝐭
Simplify the equation, and substitute to equation (1)
𝐐
= 𝐦𝐚𝐬𝐬𝐨𝐮𝐭 = 𝐑𝐎𝐋
𝟓𝟎
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At this point we can substitute all what we had solved to our differential equation
𝐝𝐐
𝐐
=𝟏−
𝐝𝐭
𝟓𝟎
Here, we can start solving the solution,
∫
𝐝𝐐
= ∫ 𝐝𝐭
𝐐
𝟏−
𝟓𝟎
(−𝟓𝟎)𝐥𝐧 𝟏 −
𝐐
=𝐭+𝐂
𝟓𝟎
Now that we generated our solution, as we can observe we have one constant variable
which is 𝐂, we can solve its value by utilizing the initial conditions. 𝐐 = 𝟎 and 𝐭 = 𝟎
(𝐐 = 𝟎 due the tank is initially containing fresh water, no salt content),
(−𝟓𝟎)𝐥𝐧 𝟏 −
𝟎
=𝟎+𝐂
𝟓𝟎
𝐂=𝟎
After ten (10) minutes, how much salt does the tank have?
(−𝟓𝟎)𝐥𝐧 𝟏 −
𝐐𝐭=𝟏𝟎𝐦𝐢𝐧𝐬
= 𝟏𝟎 + 𝟎
𝟓𝟎
𝐐 = 𝟗. 𝟎𝟔𝟑 𝐥𝐛𝐬 𝐨𝐟 𝐬𝐚𝐥𝐭
And for thirty (30) minutes has,
(−𝟓𝟎)𝐥𝐧 𝟏 −
𝐐𝐭=𝟑𝟎𝐦𝐢𝐧𝐬
= 𝟑𝟎 + 𝟎
𝟓𝟎
𝐐 = 𝟐𝟐. 𝟓𝟔𝟎 𝐥𝐛𝐬 𝐨𝐟 𝐬𝐚𝐥𝐭
Example 6
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A tank contains 8 liters of water in which is dissolved 32 g (grams) of chemical. A
solution containing 2 grams/liters of the chemical flows into the tank at a rate of 4 liters
/min, and the well-stirred mixture flows out at a rate of 2 liters/min. Determine the
amount of chemical in the tank after 20 minutes. What is the concentration of chemical
in the tank at that time?
Point A is the intake which has a flowrate of 𝟒
𝐥𝐢𝐭𝐞𝐫𝐬
𝐦𝐢𝐧
𝐠𝐫𝐚𝐦𝐬𝐬𝐚𝐥𝐭
and a concentration of 𝟐 𝐥𝐢𝐭𝐞𝐫𝐬
𝐰𝐚𝐭𝐞𝐫
.
Point B initially contains 𝟖 𝐥𝐢𝐭𝐞𝐫𝐬 with 𝟑𝟐 𝐠𝐫𝐚𝐦𝐬 of dissolved chemical. At point C, it is
said that the flowrate is the same with the intake which is 𝟐 𝐥𝐢𝐭𝐞𝐫𝐬/𝐦𝐢𝐧 but with
unknown concentration.
Now we will use our formula,
𝐝𝐐
= 𝐑𝐎𝐆 − 𝐑𝐎𝐋
𝐝𝐭
For ROG it will be easy to determine, since the volume flow rate and the density of the
intake is given
𝛒𝐢𝐧 =
𝐦𝐚𝐬𝐬𝐢𝐧
𝐯𝐨𝐥𝐮𝐦𝐞𝐢𝐧
→
𝟐
𝐠𝐫𝐚𝐦𝐬𝐬𝐚𝐥𝐭
𝐦𝐚𝐬𝐬𝐢𝐧
=
𝐥𝐢𝐭𝐞𝐫𝐬𝐰𝐚𝐭𝐞𝐫 𝟒 𝐥𝐢𝐭𝐞𝐫𝐬
𝐦𝐢𝐧
𝐦𝐚𝐬𝐬𝐢𝐧 = 𝐑𝐎𝐆 = 𝟖
𝐠𝐫𝐚𝐦𝐬
𝐦𝐢𝐧
For ROL, we need to analyze the whole system running. We go back to the formula of
density,
𝛒𝐨𝐮𝐭 =
𝐦𝐚𝐬𝐬𝐨𝐮𝐭
𝐯𝐨𝐥𝐮𝐦𝐞𝐨𝐮𝐭
Here, the only data that we have is the volume flow rate which is 𝟐
𝐥𝐢𝐭𝐞𝐫𝐬
𝐦𝐢𝐧
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𝛒𝐨𝐮𝐭 =
𝐦𝐚𝐬𝐬𝐨𝐮𝐭
𝟐
→
𝟐𝛒𝐨𝐮𝐭 = 𝐦𝐚𝐬𝐬𝐨𝐮𝐭
(1)
Now, we will analyze the density in a different way, which involves equation modeling.
We let the mass of salt be equal to 𝐐 at any time 𝐭 over the total volume during the
process which incorporates the initial volume of the tank plus the flowrate. As we can
observe at we subtract the two flowrates and multiplied it by 𝐭, we do this to cancel out
the time unit to the volume flowrate so we can determine the total volume.
𝐐
𝟖 + (𝟒 − 𝟐)𝐭
Simplify the equation, and substitute to equation (1)
𝐐
= 𝐦𝐚𝐬𝐬𝐨𝐮𝐭 = 𝐑𝐎𝐋
𝟒+𝐭
At this point we can substitute all what we had solved to our differential equation
𝐝𝐐
𝐐
=𝟖−
𝐝𝐭
𝟒+𝐭
Here, we can observe that the differential equation that we had come up with is
different from the previous example. The way that we will be solving this is either in
variable separable or linear differential equation. So at this point you probably figured
out that if the given at the initial state is fresh water you will end up solving the
differential equation in separable, and in the other hand if it has a mixture you will be
ending up solving it by linear differential equation.
𝐝𝐐
𝟏
)=𝟖
+ 𝐐(
𝐝𝐭
𝟒+𝐭
Let 𝐏(𝐭) =
𝐝𝐭
𝟏
and 𝐐(𝐭) = 𝟖, and the integrating factor will be ∅ = 𝐞∫𝟒+𝐭 = 𝟒 + 𝐭
𝟒+𝐭
𝐐(𝟒 + 𝐭) = 𝟖 ∫(𝟒 + 𝐭)𝐝𝐭 + 𝐂
𝐭𝟐
𝐐(𝟒 + 𝐭) = 𝟖 (𝟒𝐭 + ) + 𝐂
𝟐
→
𝐐(𝟒 + 𝐭) = 𝟑𝟐𝐭 + 𝟒𝐭 𝟐 + 𝐂
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Now that we generated our solution, as we can observe we have one constant variable
which is 𝐂, we can solve its value by utilizing the initial conditions. 𝐐 = 𝟑𝟐 𝐠𝐫𝐚𝐦𝐬 and
𝐭=𝟎
𝟑𝟐(𝟒 + 𝟎) = 𝟑𝟐(𝟎) + 𝟒(𝟎)𝟐 + 𝐂
𝐂 = 𝟏𝟐𝟖
After ten (20) minutes, how much salt does the tank have?
𝐐(𝟒 + 𝟐𝟎) = 𝟑𝟐(𝟐𝟎) + 𝟒(𝟐𝟎)𝟐 + 𝐂
𝐐 = 𝟗𝟖. 𝟔𝟔𝟕 𝐠𝐫𝐚𝐦𝐬 𝐨𝐟 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥
And for the concentration, the total volume at 20 minutes we will be using
𝐯𝐨𝐥𝐮𝐦𝐞 𝐭=𝟐𝟎𝐦𝐢𝐧 = 𝟐(𝟒 + 𝐭) = 𝟐(𝟒 + 𝟐𝟎) = 𝟒𝟖 𝐥𝐢𝐭𝐞𝐫𝐬. Therefore, our concentration at
𝐭 = 𝟐𝟎 𝐦𝐢𝐧𝐮𝐭𝐞𝐬 is,
𝛒𝐭=𝟐𝟎𝐦𝐢𝐧𝐬 =
Series RL Circuit
𝟗𝟖. 𝟔𝟔𝟕𝐠𝐫𝐚𝐦𝐬 𝐨𝐟 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥
𝐠𝐫𝐚𝐦𝐬 𝐨𝐟 𝐜𝐡𝐞𝐦𝐢𝐜𝐚𝐥
= 𝟐. 𝟎𝟓𝟔
𝟒𝟖 𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫
𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫
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The RL circuit shown above has a resistor and an inductor connected in series. A
constant voltage V is applied when the switch is closed.
The (variable) voltage across the resistor is given by:
𝐕𝐑 = 𝐢𝐑
The (variable) voltage across the inductor is given by:
𝐕𝐋 = 𝐋
𝐝𝐢
𝐝𝐭
Kirchhoff's voltage law says that the directed sum of the voltages around a circuit must
be zero. This results in the following differential equation:
𝐕 = 𝐢𝐑 + 𝐋
𝐝𝐢
𝐝𝐭
And for the solution,
𝐢𝐑 + 𝐋
𝐝𝐢
=𝐕
𝐝𝐭
𝐝𝐢 𝐕 − 𝐢𝐑
=
𝐝𝐭
𝐋
−
→
→
𝐋
∫
𝐝𝐢
= 𝐕 − 𝐢𝐑
𝐝𝐭
𝐝𝐢
𝐝𝐭
= ∫
𝐕 − 𝐢𝐑
𝐋
𝐥𝐧 𝐕 − 𝐢𝐑 𝐭
= +𝐂
𝐑
𝐋
Since the case is always starts with 𝐭 = 𝟎 and 𝐢 = 𝟎 we can further complete our
solution. Now 𝐂 = −
𝐥𝐧 𝐕
𝐑
Plugging in back 𝐂 to our solution, and we will have,
−
Rearranging the equation
𝐥𝐧 𝐕 − 𝐢𝐑 𝐭 𝐥𝐧 𝐕
= −
𝐑
𝐋
𝐑
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𝐥𝐧 𝐕 𝐥𝐧 𝐕 − 𝐢𝐑 𝐭
−
=
𝐑
𝐑
𝐋
→
−𝐥𝐧 𝐕 + 𝐥𝐧 𝐕 − 𝐢𝐑 = −
𝐕 − 𝐢𝐑
𝐑
)= 𝐭
𝐥𝐧 (
𝐕
𝐋
𝟏−
𝐑𝐭
𝐢𝐑
= 𝐞− 𝐋
𝐕
→
→
𝐑𝐭
𝐋
𝐑𝐭
𝐕 − 𝐢𝐑
= 𝐞− 𝐋
𝐕
𝐑𝐭
𝟏 − 𝐞− 𝐋 =
𝐢𝐑
𝐕
Finally, we have,
𝐢=
𝐑𝐭
𝐕
(𝟏 − 𝐞− 𝐋 )
𝐑
Example 7
An RL circuit has an emf of 5 V, a resistance of 50 Ω, an inductance of 1 H, and no
initial current. Find the current in the circuit at time 0.1 sec.
We will just simply use our derived formula
𝐢=
𝐢=
𝐑𝐭
𝐕
(𝟏 − 𝐞− 𝐋 )
𝐑
𝟓𝟎(𝟎.𝟏)
𝟓
(𝟏 − 𝐞− 𝟏 )
𝟓𝟎
𝐢 = 𝟎. 𝟎𝟗𝟗 𝐀
Example 8
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A series RL circuit with R = 50 Ω and L = 10 H has a constant voltage V = 100 V
applied at t = 0 by the closing of a switch. Solve for the current at t = 0.5 s,
𝐢=
𝐢=
𝐑𝐭
𝐕
(𝟏 − 𝐞− 𝐋 )
𝐑
𝟓𝟎(𝟎.𝟓)
𝟏𝟎𝟎
(𝟏 − 𝐞− 𝟏𝟎 )
𝟓𝟎
𝐢 = 𝟏. 𝟖𝟑𝟓𝟖 𝐀
Series RC Circuit
In this section we see how to solve the differential equation arising from a circuit
consisting of a resistor and a capacitor. (See the related section Series RL Circuit in the
previous section.)
In an RC circuit, the capacitor stores energy between a pair of plates. When voltage
is applied to the capacitor, the charge builds up in the capacitor and the current drops
off to zero.
Case 1: Constant Voltage
The voltage across the resistor and capacitor are as follows:
𝐕𝐑 = 𝐢𝐑
And,
𝐕𝐂 =
𝟏
∫ 𝐢𝐝𝐭
𝐂
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Where, 𝐂 = 𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐚𝐧𝐜𝐞
Kirchhoff's voltage law says the total voltages must be zero. So applying this law to a
series RC circuit results in the equation:
𝟏
𝐢𝐑 + ∫ 𝐢𝐝𝐭 = 𝐕
𝐂
One way to solve this equation is to turn it into a differential equation, by differentiating
throughout with respect to t
𝐑
𝐝𝐢 𝐢
+ =𝟎
𝐝𝐭 𝐂
Solving the equation gives us
Divide through by R
𝐝𝐢
𝟏
+ ( )𝐢 = 𝟎
𝐝𝐭
𝐑𝐂
We recognize this as a first order linear differential equation. Where 𝐏(𝐭) =
𝟏
𝐑𝐂
and
𝐭
𝐐(𝐱) = 𝟎, so then ∅ = 𝐞𝐑𝐂
𝐭
𝐭
𝐢𝐞𝐑𝐂 = ∫(𝟎)𝐞𝐑𝐂 𝐝𝐭 + 𝐊
𝐕
→
𝐭
𝐢𝐞𝐑𝐂 = 𝐊
𝐕
At initial conditions 𝐭 = 𝟎 and 𝐢 = 𝐑, then 𝐊 = 𝐑. Plugging in all the values to our
solution, and we will have,
𝐢=
𝐕 −𝐭
𝐞 𝐑𝐂
𝐑
Note: We are assuming that the circuit has a constant voltage source, V. This
equation does not apply if the voltage source is variable.
The time constant in the case of an RC circuit is
𝛕 = 𝐑𝐂
The function
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𝐢=
𝐕 −𝐭
𝐞 𝐑𝐂
𝐑
Will have an exponential decay shape due to the current stops flowing as the capacitor
becomes fully charged.
Applying our expressions from above, we have the following expressions for the
voltage across the resistor and the capacitor
𝐭
𝐕𝐑 = 𝐢𝐑 = 𝐕𝐞−𝐑𝐂
𝐕𝐂 =
𝐭
𝟏
∫ 𝐢𝐝𝐭 = 𝐕(𝟏 − 𝐞−𝐑𝐂 )
𝐂
While the voltage over the resistor drops, the voltage over the capacitor rises as it is
charged
Case 2: Variable Voltage and 2 – mesh Circuit
We need to solve variable voltage cases in q, rather than in i, since we have an integral
to deal with if we use i.
So, we will make the substitutions
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𝐢=
𝐝𝐪
𝐝𝐭
And
𝐪 = ∫ 𝐢𝐝𝐭
And so, the equation 𝐢 involving an integral
𝟏
𝐢𝐑 + ∫ 𝐢𝐝𝐭 = 𝐕
𝐂
Becomes the differential equation in 𝐪
𝐑
𝐝𝐪 𝟏
+ 𝐪=𝐕
𝐝𝐭 𝐂
Example 9
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 voltage across the capacitor is zero. Obtain the subsequent voltage
across the capacitor.
𝟏
From the formula 𝐢𝐑 + 𝐂 ∫ 𝐢𝐝𝐭 = 𝐕, we will obtain
𝐑
𝐝𝐪 𝟏
+ 𝐪=𝐕
𝐝𝐭 𝐂
Plugging in the values
𝟓
𝐝𝐪
𝟏
+
𝐪 = 𝟏𝟎𝟎
𝐝𝐭 𝟎. 𝟎𝟐
→
𝟓
𝐝𝐪
+ 𝟓𝟎𝐪 = 𝟏𝟎𝟎
𝐝𝐭
𝐝𝐪
+ 𝟏𝟎𝐪 = 𝟐𝟎
𝐝𝐭
Now we solve it by linear DE where 𝐏(𝐭) = 𝟏𝟎 and 𝐐(𝐱) = 𝟐𝟎, then the value for ∅ =
𝐞∫ 𝟏𝟎𝐝𝐭 = 𝐞𝟏𝟎𝐭
𝐪𝐞𝟏𝟎𝐭 = 𝟐𝟎 ∫(𝐞𝟏𝟎𝐭 )𝐝𝐭 + 𝐂
→
𝐪 = 𝟐 + 𝐂𝐞−𝟏𝟎𝐭
𝐪𝐞𝟏𝟎𝐭 = 𝟐𝐞𝟏𝟎𝐭 + 𝐂
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Plugging in our initial values we will have the value for 𝐂 = −𝟐 and we will have a
solution of
𝐪 = 𝟐(𝟏 − 𝐞−𝟏𝟎𝐭 )
LET’S CHECK! (PRACTICE PROBLEMS)
Problem 1
Carbon-14 is a radioactive isotope of carbon that has a half – life of 5600 years. It is
used extensively in dating organic material that is tens of thousands of years old. What
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fraction of the original amount of Carbon-14 in a sample would be present after 10,000
years?
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Problem 2
A herd of llamas has 1000 llamas in it, and the population is growing exponentially. In
four (4) years the population increase to 2000 llamas. How long will it double its recent
population?
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Problem 3
A population of a small city had 3000 people in the year 2000 and has grown at a rate
proportional to its size. In the year 2005 the population was 3700. Estimate the
population of the city in 2006 and in 2010.
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Problem 4
A cheese cake taken out of the oven with an ideal internal temperature of 165 F, and
is placed into a 35 F refrigerator. After 10 minutes, the cheese cake has cooled 150
F. How long will it take to cool the cheese cake to 70 F?
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Problem 5
A tank has pure water flowing into it at 10 liters/min. The contents of the tank are kept
thoroughly mixed, and the contents flow out at 10 liters/min. Initially, the tank contains
10 kg of salt in 100 liters of water. How much salt will there be in the tank after 30
minutes?
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Problem 6
A tank initially contains 100 liters of water with 25 grams of salt. The tank is rinsed with
fresh water flowing in at a rate of 5 liters per minute and leaving the tank at the same
rate. The water in the tank is well-stirred. Find the time such that the amount the salt
in the tank is 5 grams.
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Problem 7
An electrical circuit consist of a series connected resistor (100 ohms) and a coil with
inductance of 50 H. If the initial DC source is 200 volts, solve for the current charge at
2 seconds.
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Problem 8
An electrical circuit consist of a series connected resistor (100 ohms) and a capacitor
C = 0.01µF. At the initial moments a DC source with a 200 voltage is connected to the
circuit. Solve for the current charge at 2 seconds.
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LET’S ANALYZE! (PRACTICE PROBLEMS)
Problem 1
The number of bacteria in a liquid culture is observed to grow at a rate proportional to
the number of cells present. At the beginning of the experiment there are 10,000 cells
and after three hours there are 500,000. How many will there be after one day of
growth if this unlimited growth continues? What is the doubling time of the bacteria?
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Problem 2
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A zircon sample contains 4000 atoms of the radioactive element 235U. Given that 23
5U has a half‐ life of 700 million years, how long would it take to decay to 125 atoms
?
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Problem 3
A slow economy caused a company’s annual revenues to drop from $530, 000 at 2008
to $386,000 in 2010. If the revenue is following an exponential pattern of decline, what
is the expected revenue in 2012?
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Problem 4
The brewing pot temperature of coffee is 180 F. and the room temperature is 76 F.
After 5 minutes, the temperature of the coffee is 168 F. How long will it take for the
coffee to reach a serving temperature pf 155 F?
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Problem 5
A tank initially contains 40 gal of sugar water having a concentration of 3 lbs of sugar
for each gallon of water. At time zero, sugar water with a concentration of 4 lbs of
sugar per gal begins pouring into the tank at a rate of 2 gal per minute. Simultaneously,
a drain is opened at the bottom of the tank so that the volume of the sugar-water
solution in the tank remains constant. How much sugar is in the tank after 15 minutes?
How long will it take the sugar content in the tank to reach 150 lb? 170 lb?
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Problem 6
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A tank has pure water flowing into it at 10 liters/min. The contents of the tank are kept
thoroughly mixed, and the contents flow out at 10 liters/min. Initially, the tank contains
10 kg of salt in 100 liters of water. How much salt will there be in the tank after 30
minutes?
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Problem 7
A coil which has an inductance of 40mH and a resistance of 2Ω is connected together
to form a LR series circuit. If they are connected to a 20V DC supply. What will be the
value of the induced emf after 10 ms.
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Problem 8
A series RC circuit with R = 10 W and C = 0.01 F is connected with a battery of E =
200 V. At t = 0, the voltage across the capacitor is zero. Obtain the subsequent voltage
across the capacitor.
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IN A NUTSHELL!
ACTIVITY 1
Learning all the applications under the first order differential equation, can you share
your thoughts what is the importance in understanding first the first order differential
equation types?
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ACTIVITY 2
How can you tell that a given set of data is exponentially increasing or decreasing?
What are the possible methods on determining the situation?
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Big Picture
Week 6-7: Unit Learning Outcomes (ULO): At the end of the unit, you are expected to
a. recall and solve the linear differential equation of nth order
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Big Picture in Focus ULO. recall and solve the linear differential equation of nth order
Metalanguage
In this section, equation modeling is a big task that you must undertake and master,
by this you can solve situational problems in growth and decay, half-life, cooling and
heating of an object, and mixtures. Being an engineer, you must contain this basic
knowledge to practice your field. Please refer to these definitions in case you will
encounter difficulty in the in understanding educational concepts. In addition, the
theoretical structure and methods of solution developed in the preceding chapter for
second order linear equations extend directly to linear equations of third and higher
order.
The general solution of the nonhomogeneous equation can be written in the form
𝑦 = 𝜑(𝑡) = 𝑐1 𝑦1 (𝑡) + 𝑐2 𝑦2 (𝑡) + 𝑌(𝑡) → (𝑎)
where 𝑦1 and 𝑦2 are a fundamental set of solutions of the corresponding homogeneous
equation, 𝑐1 and 𝑐2 are arbitrary constants, and 𝑦 is some specific solution of the
nonhomogeneous equation (a).
Method of Undetermined Coefficients. The method of undetermined coefficients
requires us to make an initial assumption about the form of the particular solution 𝑦(𝑡),
but with the coefficients left unspecified. We then substitute the assumed expression
into Eq. (a) and attempt to determine the coefficients so as to satisfy that equation. If
we are successful, then we have found a solution of the differential equation (a) and
can use it for the particular solution Y(t). If we cannot determine the coefficients, then
this means that there is no solution of the form that we assumed. In this case we may
modify the initial assumption and try again
.
Essential Knowledge
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The knowledge that you had gained from the previous topics will play a big role at this
part of the subject. Solving the general solution and particular solution, and deciding
what first order differential equation must be utilized to model/generate the solution.
Also, in this chapter we briefly review this generalization, taking particular note of those
instances where new phenomena may appear, because of the greater variety of
situations that can occur for equations of higher order
For you to be able to understand the concepts under this topic, here are steps involved
in finding the solution of an initial value problem consisting of a nonhomogeneous
equation of the form
𝑎𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 𝑔(𝑡) → (𝑏)
where the coefficients 𝑎, 𝑏, and 𝑐 are constants, together with a given set of initial
conditions.
1. Find the general solution of the corresponding homogeneous equation.
2. Make sure that the function 𝑔(𝑡) in Eq. (b) belongs to the class of functions
discussed in this section; that is, be sure it involves nothing more than exponential
functions, sines, cosines, polynomials, or sums or products of such functions. If this is
not the case, use the method of variation of parameters (discussed in the next section).
3. If 𝑔(𝑡) = 𝑔1 (𝑡) +··· + 𝑔𝑛(𝑡) that is, if 𝑔(𝑡) is a sum of n terms—then form n
subproblems, each of which contains only one of the terms 𝑔1 (𝑡), ... , 𝑔𝑛 (𝑡). The ith
subproblem consists of the equation
𝑎𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 𝑔𝑖 (𝑡),
where i runs from 1 𝑡𝑜 𝑛.
4. For the 𝑖 𝑡ℎ subproblem assume a particular solution 𝑌𝑖 (𝑡) consisting of the
appropriate exponential function, sine, cosine, polynomial, or combination thereof. If
there is any duplication in the assumed form of 𝑌𝑖 (𝑡) with the solutions of the
homogeneous equation (found in step 1), then multiply 𝑌𝑖 (𝑡) by 𝑡, or (if necessary) by
𝑡 2 , so as to remove the duplication.
5. Find a particular solution 𝑌𝑖 (𝑡) for each of the subproblems. Then the sum 𝑌1 (𝑡) +
··· + 𝑌𝑛 (𝑡) is a particular solution of the full nonhomogeneous equation (b).
6. Form the sum of the general solution of the homogeneous equation (step 1) and the
particular solution of the nonhomogeneous equation (step 5). This is the general
solution of the nonhomogeneous equation.
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7. Use the initial conditions to determine the values of the arbitrary constants
remaining in the general solution. For some problems this entire procedure is easy to
carry out by hand, but often the algebraic calculations are lengthy. Once you
understand clearly how the method works, a computer algebra system can be of great
assistance in executing the details.
LINEAR DIFFERENTIAL OPERATORS
Linear differential equations. The general linear ODE of order n is
𝐲 𝐧 + 𝐩𝟏 (𝐱)𝐲 𝐧−𝟏 + . . . + 𝐩𝐧 (𝐱)𝐲 = 𝐪(𝐱)
(1)
If 𝐪(𝐱) ≠ 𝟎, the equation is inhomogeneous. We then call
𝐲 𝐧 + 𝐩𝟏 (𝐱)𝐲 𝐧−𝟏 + . . . + 𝐩𝐧 (𝐱)𝐲 = 𝟎
(2)
The associated homogeneous equation or the reduced equation.
The theory of the n-th order linear ODE runs parallel to that of the second order
equation. In particular, the general solution to the associated homogeneous equation
(2) is called the complementary function or solution, and it has the form
𝐲𝐂 = 𝐂𝟏 𝐲𝟏 + . . . + 𝐂𝐧 𝐲𝐧
(3)
Where, 𝐂𝐢 are constants,
where the 𝐲𝐢 are n solutions to (2) which are linearly independent, meaning that none
of
them can be expressed as a linear combination of the others, i.e., by a relation of the
form
(the left side could also be any of the other 𝐲𝐢 )
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𝐲𝐧 = 𝐚𝟏 𝐲𝟏 + . . . + 𝐚𝐧 − 𝟏𝐲𝐧 − 𝟏
Where, 𝐚𝐢 are constants,
Once the associated homogeneous equation (2) has been solved by finding 𝐧
independent solutions, the solution to the original ODE (1) can be expressed as
𝐲 = 𝐲𝐩 + 𝐲𝐜
(4)
where 𝐲𝐩 is a particular solution to (1), and 𝐲𝐜 is as in (3).
LINEAR DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS
From now on we will consider only the case where (1) has constant coefficients. This
type of ODE can be written as
𝐲 𝐧 + 𝐚𝟏 (𝐱)𝐲 𝐧−𝟏 + . . . + 𝐚𝐧 (𝐱)𝐲 = 𝐪(𝐱)
(5)
using the differentiation operator 𝐃, we can write (5) in the form
(𝐃𝐧 + 𝐚𝟏 𝐃𝐧−𝟏 + . . . + 𝐚𝐧) 𝐲 = 𝐪(𝐱)
𝐩(𝐃)𝐲 = 𝐪(𝐱) ,
(6)
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Where
𝐩(𝐃) = 𝐃𝐧 + 𝐚𝟏 𝐃𝐧−𝟏 + . . . + 𝐚𝐧
(7)
We call 𝐩(𝐃) a polynomial differential operator with constant coefficients. We think of
the formal polynomial p(D) as operating on a function 𝐲(𝐱), converting it into another
function; it is like a black box, in which the function 𝐲(𝐱) goes in, and 𝐩(𝐃)𝐲 (i.e., the
left side of (5)) comes out.
Our main goal in this section of the Notes is to develop methods for finding particular
solutions to the ODE (5) when 𝐪(𝐱) has a special form: an exponential, 𝐬𝐢𝐧𝐞 or 𝐜𝐨𝐬𝐢𝐧𝐞,
𝐱 𝐤 , or a product of these. (The function 𝐪(𝐱) can also be a sum of such special
functions.) These are the most important functions for the standard applications.
The reason for introducing the polynomial operator p(D) is that this allows us to use
polynomial algebra to help find the particular solutions. The rest of this chapter of the
Notes will illustrate this. Throughout, we let
𝐩(𝐃) = 𝐃𝐧 + 𝐚𝟏 𝐃𝐧−𝟏 + . . . + 𝐚𝐧
(7)
OPERATOR RULES
Our work with these differential operators will be based on several rules they satisfy.
In stating these rules, we will always assume that the functions involved are sufficiently
differentiable, so that the operators can be applied to them.
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Sum rule. If 𝐩(𝐃) and 𝐪(𝐃) are polynomial operators, then for any (sufficiently
differentiable) function 𝐮,
[𝐩(𝐃) + 𝐪(𝐃)]𝐮 = 𝐩(𝐃)𝐮 + 𝐪(𝐃)𝐮
(8)
Linearity rule. If 𝐮𝟏 and 𝐮𝟐 are functions, and 𝐜𝐢 constants,
𝐩(𝐃)(𝐜𝟏 𝐮𝟏 + 𝐜𝟐 𝐮𝟐 ) = 𝐜𝟏 𝐩(𝐃)𝐮𝟏 + 𝐜𝟐 𝐩(𝐃)𝐮𝟐
(9)
The linearity rule is a familiar property of the operator 𝐃𝐤 ; it extends to sums of these
operators, using the sum rule above, thus it is true for operators which are polynomials
in D. (It is still true if the coefficients ai in (7) are not constant, but functions of 𝐱.)
Multiplication rule. If 𝐩(𝐃) = 𝐠(𝐃) 𝐡(𝐃), as polynomials in 𝐃, then
𝐩(𝐃)𝐮 = 𝐠(𝐃)(𝐡(𝐃)𝐮)
(10)
The picture illustrates the meaning of the right side of (10). The property is true when
h(D) is the simple operator 𝐚𝐃𝐤 , essentially because
𝐃𝐦 (𝐚𝐃𝐤 𝐮) = 𝐚𝐃𝐦+𝐤 𝐮
it extends to general polynomial operators h(D) by linearity. Note that a must be a
constant; it’s false otherwise.
An important corollary of the multiplication property is that polynomial operators with
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constant coefficients commute; i.e., for every function 𝐮(𝐱),
𝐠(𝐃)(𝐡(𝐃)𝐮) = 𝐡(𝐃)(𝐠(𝐃)𝐮)
(11)
For as polynomials, 𝐠(𝐃)𝐡(𝐃) = 𝐡(𝐃)𝐠(𝐃) = 𝐩(𝐃), say; therefore, by the
multiplication rule, both sides of (11) are equal to 𝐩(𝐃)𝐮, and therefore, equal to each
other.
The remaining two rules are of a different type, and more concrete: they tell us how
polynomial operators behave when applied to exponential functions and products
involving exponential functions.
Substitution rule.
𝐩(𝐃)𝐞𝐚𝐱 = 𝐩(𝐚)𝐞𝐚𝐱
Proof. We have, by repeated differentiation,
𝐃𝐞𝐚𝐱 = 𝐚𝐞𝐚𝐱 , 𝐃𝟐 𝐞𝐚𝐱 = 𝐚𝟐 𝐞𝐚𝐱 , . . . , 𝐃𝐤 𝐞𝐚𝐱 = 𝐚𝐤 𝐞𝐚𝐱
Therefore,
(𝐃𝐧 + 𝐂𝟏 𝐃𝐧−𝟏 + . . . + 𝐜𝐧 )𝐞𝐚𝐱 = (𝐚𝐧 + 𝐂𝟏 𝐚𝐧−𝟏 + . . . + 𝐜𝐧 ) 𝐞𝐚𝐱
which is the substitution rule (12)
The exponential-shift rule. This handles expressions such as 𝐱 𝐤 𝐞𝐚𝐱 and 𝐱 𝐤 𝐬𝐢𝐧 𝐚𝐱.
𝐩(𝐃)𝐞𝐚𝐱 𝐮 = 𝐞𝐚𝐱 𝐩(𝐃 + 𝐚)𝐮
(13)
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Proof. We prove it in successive stages. First, it is true when 𝐩(𝐃) = 𝐃, since by the
product rule for differentiation,
𝐃𝐞𝐚𝐱 𝐮(𝐱) = 𝐞𝐚𝐱 𝐃𝐮(𝐱) + 𝐚𝐞𝐚𝐱 𝐮(𝐱) = 𝐞𝐚𝐱 (𝐃 + 𝐚)𝐮(𝐱)
(14)
To show the rule is true for 𝐃𝐤 , we apply (14) to 𝐃 repeatedly:
𝐃𝐞𝐚𝐱 𝐮(𝐱) = 𝐞𝐚𝐱 𝐃𝐮(𝐱) + 𝐚𝐞𝐚𝐱 𝐮(𝐱) = 𝐞𝐚𝐱 (𝐃 + 𝐚)𝐮(𝐱)
by
(14)
= 𝐞𝐚𝐱 (𝐃 + 𝐚)((𝐃 + 𝐚)𝐮)
by
= 𝐞𝐚𝐱 (𝐃 + 𝐚)𝟐 𝐮
by
(14)
(10)
In the same way,
𝐃𝟑 𝐞𝐚𝐱 𝐮 = 𝐃(𝐃𝟐 𝐞𝐚𝐱 𝐮) = 𝐃(𝐞𝐚𝐱 (𝐃 + 𝐚)𝟐 𝐮)
by the previous
= 𝐞𝐚𝐱 (𝐃 + 𝐚)((𝐃 + 𝐚)𝟐 𝐮) by (14)
= 𝐞𝐚𝐱 (𝐃 + 𝐚)𝟑 𝐮
by (10)
and so on. This shows that (13) is true for an operator of the form 𝐃𝐤 . To show it is
true
for a general operator
𝐩(𝐃) = 𝐃𝐧 + 𝐚𝟏 𝐃𝐧−𝟏 + . . . + 𝐚𝐧 ,
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we write (13) for each 𝐃𝐤 (𝐞𝐚𝐱 𝐮), multiply both sides by the coefficient 𝐚𝐤 , and add up
the resulting equations for the different values of 𝐤.
Example 1
Evaluate (𝐃 − 𝟏)𝟑 𝐲 = 𝟎
We can expand (𝐃 − 𝟏)𝟑 = 𝐃𝟑 − 𝟑𝐃𝟐 + 𝟑𝐃 − 𝟏 that denotes a differential operator of
order 3. When applied to a function y (which we assume to be thrice differentiable) it
yields
(𝐃𝟑 − 𝟑𝐃𝟐 + 𝟑𝐃 − 𝟏)𝐲 = 𝐃𝟑 𝐲 − 𝟑𝐃𝟐 𝐲 + 𝟑𝐃𝐲 − 𝐲
𝐝𝟑 𝐲
𝐝𝟐 𝐲
𝐝𝐲
−𝟑 𝟐+𝟑
−𝐲=𝟎
𝟑
𝐝𝐱
𝐝𝐱
𝐝𝐱
Example 2
Find 𝐃𝟑 𝐞−𝐱 𝐬𝐢𝐧 𝐱
Using exponential shifting.
𝐃𝟑 𝐞−𝐱 𝐬𝐢𝐧 𝐱 = 𝐞−𝐱 (𝐃 − 𝟏)𝟑 𝐬𝐢𝐧 𝐱 = 𝐞−𝐱 (𝐃𝟑 − 𝟑𝐃𝟐 + 𝟑𝐃 − 𝟏) 𝐬𝐢𝐧 𝐱
= 𝐞−𝐱 (𝟐 𝐜𝐨𝐬 𝐱 + 𝟐 𝐬𝐢𝐧 𝐱)
Since, 𝐃𝟐 𝐬𝐢𝐧 𝐱 = − 𝐬𝐢𝐧 𝐱 , and 𝐃𝟑 𝐬𝐢𝐧 𝐱 = − 𝐜𝐨𝐬 𝐱
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Example 3
𝟑
Perform the operation (𝐃𝟐 + 𝟓𝐃 − 𝟏)(𝐭𝐚𝐧 𝟐𝐱 − 𝐱)
First, we need to distribute,
𝟑 𝟑
𝐃𝟐 𝐭𝐚𝐧 𝟐𝐱 + 𝟓𝐃 𝐭𝐚𝐧 𝟐𝐱 − 𝐭𝐚𝐧 𝟐𝐱 − 𝐃𝟐 − 𝟓𝐃 +
𝐱 𝐱
Then differential accordingly,
𝟖 𝐬𝐞𝐜 𝟐 𝟐𝐱 (𝐭𝐚𝐧𝟐𝐱) –
𝟔
𝟏𝟓
𝟑
+ 𝟏𝟎 𝐬𝐞𝐜 𝟐 𝟐𝐱 + 𝟐 – 𝐭𝐚𝐧𝟐𝐱 +
𝟑
𝐱
𝐱
𝐱
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT
COEFFICIENTS
Solve 𝐧𝐭𝐡 order homogeneous linear equations
𝐚𝐧 𝐲 (𝐧) + 𝐚𝐧−𝟏 𝐲 (𝐧−𝟏) + . . . +𝐚𝟏 𝐲 ′ + 𝐚𝐨 𝐲 = 𝟎
Where 𝐚𝐧 , … , 𝐚𝟏 , 𝐚𝐨 are constant with 𝐚𝐧 ≠ 𝟎
Solution method:
1. Find the roots of the characteristic polynomial
𝐚𝐧 𝐲 (𝐧) + 𝐚𝐧−𝟏 𝐲 (𝐧−𝟏) + . . . +𝐚𝟏 𝐲 ′ + 𝐚𝐨 𝐲 = 𝟎
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2. Evaluate the roots.
3. Use the proper formula and substitute.
Case I
If the roots from the auxiliary equation are real and distinct
𝐲 = 𝐂𝟏 𝐞𝐫𝟏 + 𝐂𝟐 𝐞𝐫𝟐 + 𝐂𝐧 𝐞𝐫𝐧 𝐱 + 𝐂𝐧−𝟏 𝐞𝐫𝐧−𝟏 𝐱
(1)
Example 4
Solve the IVP of the differential equation 𝐲 ′′ + 𝟏𝟏𝐲 ′ + 𝟐𝟒𝐲 = 𝟎, but 𝐲(𝟎) = 𝟎 and
𝐲 ′ (𝟎) = −𝟕
Transform the equation
(𝐃𝟐 + 𝟏𝟏𝐃 + 𝟐𝟒)𝐲 = 𝟎
Now, take the isolated equation, and we will have an auxiliary equation of
𝐦𝟐 + 𝟏𝟏𝐦 + 𝟐𝟒 = 𝟎
Solve for the roots, 𝐌𝟏 = −𝟖 and 𝐌𝟏 = −𝟑. As we can observe the roots are under
CASE I, Using (1)
𝐲 = 𝐂𝟏 𝐞−𝟖𝐱 + 𝐂𝟐 𝐞−𝟑𝐱
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You can either start with −𝟑 and −𝟖. The answer will be the same
For the particular solution, It requires second derivative. So, we will differentiate our
general solution, which results
𝐲 ′ = −𝟖𝐂𝟏 𝐞−𝟖𝐱 − 𝟑𝐂𝟐 𝐞−𝟑𝐱
Now, plug in the initial conditions to get the following system of equations.
𝟎 = 𝐂𝟏 + 𝐂𝟐
−𝟕 = −𝟖𝐂𝟏 − 𝟑𝐂𝟐
𝟕
𝟕
Solving this system gives 𝐂𝟏 = 𝟓 and 𝐂𝟐 = − 𝟓 . The actual solution to the differential
equation is then
𝐲𝐩 =
𝟕 −𝟑𝐱 𝟕 −𝟖𝐱
𝐞
− 𝐞
𝟓
𝟓
Example 5
𝐝𝟑 𝐲
𝐝𝟐 𝐲
𝐝𝐲
Solve for the general solution of 𝐝𝐱𝟑 + 𝟒 𝐝𝐱𝟐 − 𝟕 𝐝𝐱 − 𝟏𝟎𝐲 = 𝟎
The equation will be transformed into
(𝐃𝟑 + 𝟒𝐃𝟐 − 𝟕𝐃 − 𝟏𝟎)𝐲 = 𝟎
And we will have an auxiliary equation of
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𝐦𝟑 + 𝟒𝐦𝟐 − 𝟕𝐦 − 𝟏𝟎 = 𝟎
For the roots, 𝐦𝟏 = −𝟏. 𝐦𝟐 = 𝟐, and 𝐦𝟑 = −𝟓 which are real and distinct, so
therefore, our general solution will be
𝐲 = 𝐂𝟏 𝐞−𝐱 + 𝐂𝟐 𝐞𝟐𝐱 + 𝐂𝟑 𝐞−𝟓𝐱
Case II
If the roots from the auxiliary equation are real and repeating
𝐲 = 𝐞𝐫𝐱 (𝐂𝟏 + 𝐂𝟐 𝐱 + ⋯ + 𝐂𝐧 𝐱 𝐧−𝟏 )
(2)
Example 6
Solve the IVP 𝐲 ′′ − 𝟒𝐲 ′ + 𝟒𝐲 = 𝟎 when 𝐲(𝟎) = 𝟏𝟐 and 𝐲 ′ (𝟎) = 𝟑
Solve for the roots, for the equation 𝐦𝟐 − 𝟒𝐦 + 𝟒 = 𝟎
𝐦𝟏 = 𝟐 = 𝐦𝟐
We can observe that the roots are equal and real, so we will be using (2) for the general
solution
𝐲 = 𝐞𝟐𝐱 (𝐂𝟏 + 𝐂𝟐 𝐱)
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Now, for the particular solution we need to differentiate our general solution once
𝐲 = 𝐞𝟐𝐱 (𝐂𝟏 + 𝐂𝟐 𝐱)
𝐲 ′ = 𝟐𝐞𝟐𝐱 (𝐂𝟏 + 𝐂𝟐 𝐱) + 𝐂𝟐 𝐞𝟐 𝐱
Plugging in the initial conditions gives the following system
𝟏𝟐 = 𝐂𝟏
−𝟑 = 𝟐𝐂𝟏 + 𝐂𝟐
So, we have values for 𝐂𝟏 = 𝟏𝟐 and 𝐂𝟐 = −𝟐𝟕
Therefore, our particular solution will be
𝐲𝐩 = 𝟏𝟐𝐞𝟐𝐱 − 𝟐𝟕𝐱𝐞𝟐𝐱
Problem 7
Solve the differential equation 𝟏𝟔𝐲 ′′ − 𝟒𝟎𝐲 + 𝟐𝟓 = 𝟎
𝟓
Solve for the roots, for the equation 𝐦𝟐 − 𝟏𝟒𝐦 + 𝟐𝟓 = 𝟎. 𝐦𝟏 = 𝐦𝟐 = 𝟒
𝟓
𝐲 = 𝐞𝟒𝐱 (𝐂𝟏 + 𝐂𝟐 𝐱)
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CASE III
If the roots of the auxiliary equation are distinct and conjugate
𝐲 = 𝐞𝐚𝐱 (𝐂𝟏 𝐬𝐢𝐧 𝐛𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝐛𝐱)
(3)
Example 8
Solve the differential equation 𝐲 ′′ − 𝟒𝐲 ′ + 𝟏𝟑𝐲 = 𝟎
We have 𝐦𝟏 = 𝟐 + 𝟑𝐢 and 𝐦𝟐 = 𝟐 − 𝟑𝐢, clearly this is under CASE III, using (3). Our
values for 𝐚 = 𝟐 and 𝐛 = 𝟑. Plugging in all the values and we will have
𝐲 = 𝐞𝟐𝐱 (𝐂𝟏 𝐬𝐢𝐧 𝟑𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝟑𝐱)
Example 9
Solve the differential equation 𝐲 ′′ − 𝟏𝟎𝐲 ′ + 𝟐𝟗 = 𝟎 if 𝐲(𝟎) = 𝟏 and 𝐲 ′ (𝟎) = 𝟑.
We have 𝐦𝟏 = 𝟓 + 𝟐𝐢 and 𝐦𝟐 = 𝟓 − 𝟐𝐢, clearly this is under CASE III, using (3). Our
values for 𝐚 = 𝟓 and 𝐛 = 𝟐. Plugging in all the values and we will have
𝐲 = 𝐞𝟓𝐱 (𝐂𝟏 𝐬𝐢𝐧 𝟐𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝟐𝐱)
CASE IV
If the roots of the auxiliary equation are repeated and conjugate
𝐂𝟐 𝐱 𝐜𝐨𝐬 𝐛𝐱)
𝐲 = 𝐞𝐚𝐱 (𝐂𝟏 𝐬𝐢𝐧 𝐛𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝐛𝐱) + 𝐞𝐚𝐱 (𝐂𝟏 𝐱 𝐬𝐢𝐧 𝐛𝐱 +
(4)
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Example 10
Solve the general solution of the differential equation 𝐲 (𝟒) + 𝟖𝐲 ′′ + 𝟏𝟔𝐲 = 𝟎
For the roots of the equation we have, 𝐦𝟏 = 𝐦𝟑 = 𝟐𝐢 and 𝐦𝟐 = 𝐦𝟒 = −𝟐𝐢
Clearly, this is under CASE IV. We will use (4) and substitute all the values 𝐚 = 𝟎 and
𝐛=𝟐
𝐲 = 𝐞𝟎𝐱 (𝐂𝟏 𝐬𝐢𝐧 𝟐𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝟐𝐱) + 𝐞𝟎𝐱 (𝐂𝟑 𝐱 𝐬𝐢𝐧 𝟐𝐱 + 𝐂𝟒 𝐱 𝐜𝐨𝐬 𝟐𝐱)
𝐲 = (𝐂𝟏 𝐬𝐢𝐧 𝟐𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝟐𝐱) + (𝐂𝟑 𝐱 𝐬𝐢𝐧 𝟐𝐱 + 𝐂𝟒 𝐱 𝐜𝐨𝐬 𝟐𝐱)
Example 11
Solve the general solution of the differential equation 𝐲 (𝟒) + 𝟒𝐲 (𝟑) + 𝟖𝐲 ′′ + 𝟖𝐲 ′ + 𝟒𝐲 =
𝟎
For the roots of the equation we have, 𝐦𝟏 = 𝐦𝟑 = −𝟏 + 𝐢 and 𝐦𝟐 = 𝐦𝟒 = −𝟏 − 𝐢
Substitute the values, where 𝐚 = −𝟏 and 𝐛 = 𝟏, and we will have an answer of
𝐲 = 𝐞−𝐱 (𝐂𝟏 𝐬𝐢𝐧 𝐱 + 𝐂𝟐 𝐜𝐨𝐬 𝐱) + 𝐞−𝐱 (𝐂𝟑 𝐱 𝐬𝐢𝐧 𝐱 + 𝐂𝟒 𝐱 𝐜𝐨𝐬 𝐱)
NON-HOMOGENEOUS EQUATIONS: UNDETERMINED COEFFICIENTS
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Unlike the homogeneous equation, non-homogeneous equation has a function at the
right side of its equation
𝐚𝐧 𝐲 𝐧 + 𝐚𝐧−𝟏 𝐲 𝐧−𝟏 + ⋯ + 𝐚𝟏 𝐲 ′ + 𝐚𝟎 𝐲 = 𝐟(𝐭)
The way that we will be solving the non-homogeneous equation is one step different
to the homogeneous equation. The formula of the solution is
𝐲 = 𝐲𝐜 + 𝐲𝐩
Which 𝐲𝐜 is the complementary equation, we can solve 𝐲𝐜 by utilizing what we have
learned from the previous lecture.
And 𝐲𝐩 is the particular solution of the given non – homogeneous equation.
Equivalent value for the functions at the right side of the non – homogeneous equation.
𝐚𝐱 𝟐 + 𝐛𝐱 + 𝐜
𝐀𝐱 𝟐 + 𝐁𝐱 + 𝐂
𝐚𝐞𝐛𝐱
𝐀𝐞𝐛𝐱
𝐚 𝐬𝐢𝐧 𝐛𝐱 or 𝐚 𝐜𝐨𝐬 𝐛𝐱
𝐚𝐱 𝐧 𝐞𝐛𝐱
𝐚𝐱 𝐧 𝐬𝐢𝐧 𝐛𝐱 or 𝐚𝐱 𝐧 𝐜𝐨𝐬 𝐛𝐱
𝐀 𝐬𝐢𝐧 𝐛𝐱 + 𝐁 𝐜𝐨𝐬 𝐛𝐱
𝐀𝐱 𝐧 𝐞𝐛𝐱
𝐀𝐱 𝐧 𝐬𝐢𝐧 𝐛𝐱 + 𝐁𝐱 𝐧 𝐜𝐨𝐬 𝐛𝐱
We can observe all the coefficients at the left side (𝐚, 𝐛, 𝐜) are disregarded at the left
side, in which they are replaced by our undetermined coefficients (𝐀, 𝐁, 𝐂), hence, the
functions of 𝐞, 𝐬𝐢𝐧𝐞, and 𝐜𝐨𝐬𝐢𝐧𝐞 remains.
But this method has limitations, it wont work to linear equation which contains variable
coefficients. In addition, the term 𝐟(𝐭) is should not be with these forms:
𝐚 𝐩𝐨𝐥𝐲𝐧𝐦𝐢𝐚𝐥 𝐩(𝐭) = 𝐚𝐨 + 𝐚𝟏 𝐭 + ⋯ + 𝐚𝐍 𝐭 𝐍 ,
𝐚𝐧 𝐞𝐱𝐩𝐨𝐭𝐞𝐧𝐭𝐢𝐚𝐥 𝐞𝐚𝐭
𝐜𝐨𝐬𝛃𝐭,
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𝐬𝐢𝐧𝛃𝐭,
(𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥) ∙ 𝐞𝐚𝐭 ,
(𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥) ∙ 𝐜𝐨𝐬𝛃𝐭,
(𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥) ∙ 𝐬𝐢𝐧𝛃𝐭,
(𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥) ∙ 𝐞𝐚𝐭 𝐜𝐨𝐬𝛃𝐭,
(𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥) ∙ 𝐞𝐚𝐭 𝐬𝐢𝐧𝛃𝐭,
𝐚𝐧𝐝 𝐚𝐧𝐲 𝐥𝐢𝐧𝐞𝐚𝐫 𝐜𝐨𝐦𝐛𝐢𝐧𝐚𝐭𝐢𝐨𝐧 𝐚𝐛𝐨𝐯𝐞
So, any equation that has 𝐟(𝐭) we can solve them by using the method, as long as it
doesn’t belong to the equation listed above.
Example 12
Solve the differential equation 𝐲 ′′ − 𝟑𝐲 ′ − 𝟒𝐲 = 𝟑𝐞𝟐𝐱
First, we need to solve for the complementary solution of the equation
𝐲 = 𝐲𝐜 + 𝐲𝐩
At this point we will just consider our equation at the left side with will be transformed
into an auxiliary equation
𝐦𝟐 − 𝟑𝐦 − 𝟒 = 𝟎
Our roots are 𝐦𝟏 = 𝟒 and 𝐦𝟐 = −𝟏
𝐲𝐂 = 𝐂𝟏 𝐞𝟒𝐱 + 𝐂𝟐 𝐞−𝐱
Now we set aside our 𝐲𝐂 and solve for our 𝐲𝐏 . In solving the 𝐲𝐏 we will be utilizing the
function at the right side of our non – homogeneous equation.
𝐲𝐩 = 𝐀𝐞𝟐𝐱
At this point we will be differentiating our 𝐲𝐩 depending what is the highest order to our
non – homogeneous equation. Since the highest order is second, we will be
differentiating 𝐲𝐩 to 𝐲𝐩 ′′ .
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𝐲𝐩 = 𝐀𝐞𝟐𝐱
𝐲𝐩 ′ = 𝟐𝐀𝐞𝟐𝐱
𝐲𝐩 ′′ = 𝟒𝐀𝐞𝟐𝐱
Then we will substitute these values to our non – homogeneous equation
𝟒𝐀𝐞𝟐𝐱 − 𝟑(𝟐𝐀𝐞𝟐𝐱 ) − 𝟒𝐀𝐞𝟐𝐱 = 𝟑𝐞𝟐𝐱
𝟒𝐀𝐞𝟐𝐱 − 𝟔𝐀𝐞𝟐𝐱 − 𝟒𝐀𝐞𝟐𝐱 = 𝟑𝐞𝟐𝐱
→
𝟒𝐀 − 𝟔𝐀 − 𝟒𝐀 = 𝟑
𝐀=−
→
−𝟔𝐀 = 𝟑
𝟏
𝟐
𝟏
Therefore, our 𝐲𝐩 = − 𝟐 𝐞𝟐𝐱 and now our general solution for the DE is
𝟏
𝐲 = 𝐂𝟏 𝐞𝟒𝐱 + 𝐂𝟐 𝐞−𝐱 − 𝐞𝟐𝐱
𝟐
Example 13
Solve the general solution of 𝐲 ′′′ + 𝟒𝐲 ′′ − 𝟕𝐲 ′ − 𝟏𝟎𝐲 = 𝟐 𝐬𝐢𝐧 𝟐𝐱
First, solve for the complementary solution 𝐲𝐂 ,
𝐦𝟑 + 𝟒𝐦𝟐 − 𝟕𝐦 − 𝟏𝟎 = 𝟎
Having the roots of 𝐦𝟏 = −𝟏, 𝐦𝟐 = 𝟐, and 𝐦𝟑 = −𝟓
𝐲𝐂 = 𝐂𝟏 𝐞−𝐱 + 𝐂𝟐 𝐞𝟐𝐱 + 𝐂𝟑 𝐞−𝟓𝐱
Now for the 𝐲𝐏 , observing that the highest order is third
𝐲𝐏 = 𝐀 𝐬𝐢𝐧 𝟐𝐱 + 𝐁 𝐜𝐨𝐬 𝟐𝐱
𝐲𝐏′ = 𝟐𝐀 𝐜𝐨𝐬 𝟐𝐱 − 𝟐𝐁 𝐬𝐢𝐧 𝟐𝐱
𝐲𝐏′′ = −𝟒𝐀 𝐬𝐢𝐧 𝟐𝐱 − 𝟒𝐁 𝐜𝐨𝐬 𝟐𝐱
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𝐲𝐏′′′ = −𝟖𝐀 𝐜𝐨𝐬 𝟐𝐱 + 𝟖𝐁 𝐬𝐢𝐧 𝟐𝐱
Substitute to our non – homogeneous equation
−𝟖𝐀 𝐜𝐨𝐬 𝟐𝐱 + 𝟖𝐁 𝐬𝐢𝐧 𝟐𝐱 − 𝟒(−𝟒𝐀 𝐬𝐢𝐧 𝟐𝐱 − 𝟒𝐁 𝐜𝐨𝐬 𝟐𝐱) − 𝟕(𝟐𝐀 𝐜𝐨𝐬 𝟐𝐱 − 𝟐𝐁 𝐬𝐢𝐧 𝟐𝐱) −
𝟏𝟎(𝐀 𝐬𝐢𝐧 𝟐𝐱 + 𝐁 𝐜𝐨𝐬 𝟐𝐱) = 𝟐 𝐬𝐢𝐧 𝟐𝐱
−𝟖𝐀 𝐜𝐨𝐬 𝟐𝐱 + 𝟖𝐁 𝐬𝐢𝐧 𝟐𝐱 + 𝟏𝟔𝐀 𝐬𝐢𝐧 𝟐𝐱 + 𝟏𝟔𝐁 𝐜𝐨𝐬 𝟐𝐱 − 𝟏𝟒𝐀 𝐜𝐨𝐬 𝟐𝐱 + 𝟏𝟒𝐁 𝐬𝐢𝐧 𝟐𝐱 −
𝟏𝟎𝐀 𝐬𝐢𝐧 𝟐𝐱 − 𝟏𝟎𝐁 𝐜𝐨𝐬 𝟐𝐱 = 𝟐 𝐬𝐢𝐧 𝟐𝐱
−𝟐𝟐𝐀 𝐜𝐨𝐬 𝟐𝐱 + 𝟔𝐁 𝐜𝐨𝐬 𝟐𝐱 + 𝟔𝐀 𝐬𝐢𝐧 𝟐𝐱 + 𝟐𝟐𝐁 𝐬𝐢𝐧 𝟐𝐱 = 𝟐 𝐬𝐢𝐧 𝟐𝐱
@ 𝐜𝐨𝐬 𝟐𝐱
@ 𝐬𝐢𝐧 𝟐𝐱
−𝟐𝟐𝐀 + 𝟔𝐁 = 𝟎
−𝟔𝐀 + 𝟐𝟐(−
𝐁=−
𝟏𝟏
𝟑
𝐀
𝟏𝟏
𝟑
𝟑
𝐀) = 𝟐
𝟏𝟏
𝐀 = − 𝟏𝟑𝟎 and 𝐁 = 𝟏𝟑𝟎
𝟑
𝟏𝟏
Therefore, our 𝐲𝐏 = − 𝟏𝟑𝟎 𝐬𝐢𝐧 𝟐𝐱 + 𝟏𝟑𝟎 𝐜𝐨𝐬 𝟐𝐱. Now our general solution will be
𝐲 = 𝐂𝟏 𝐞−𝐱 + 𝐂𝟐 𝐞𝟐𝐱 + 𝐂𝟑 𝐞−𝟓𝐱 −
𝟑
𝟏𝟏
𝐬𝐢𝐧 𝟐𝐱 +
𝐜𝐨𝐬 𝟐𝐱
𝟏𝟑𝟎
𝟏𝟑𝟎
Example 14
Solve
𝐭 ′′′ + 𝟒𝐭 ′′ − 𝟕𝐭 ′ − 𝟏𝟎𝐭 = 𝟏𝟎𝟎𝐱 𝟐 − 𝟔𝟒𝐞𝟑𝐱 if the initial values are 𝐭(𝟎) =
−𝟐𝟎, 𝐭 ′ (𝟎) = −
𝟐𝟗
𝟓
, and 𝐭 ′′ (𝟎) = −
𝟏𝟗
𝟓
First, solve the complementary solution 𝐲𝐂
𝐦𝟑 + 𝟒𝐦𝟐 − 𝟕𝐦 − 𝟏𝟎 = 𝟎
The roots will be, 𝐦𝟏 = −𝟏, 𝐦𝟐 = 𝟐, and 𝐦𝟑 = −𝟓. Therefore, the solution is
𝐭 𝐂 = 𝐂𝟏 𝐞−𝟓𝐱 + 𝐂𝟐 𝐞𝟐𝐱 + 𝐂𝟑 𝐞−𝐱
Next will be the particular solution 𝐲𝐏 . As we can observe the function at the right side
of the non – homogenous equation are two different terms, we have 𝟏𝟎𝟎𝐱 𝟐 and 𝟔𝟒𝐞𝟑𝐱 .
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In this case, we have two particular solutions
𝐭 𝐏 = 𝐭 𝐏𝟏 + 𝐭 𝐏𝟐
We let 𝐲𝐏𝟏 = 𝐀𝐱 𝟐 + 𝐁𝐱 + 𝐂 and 𝐲𝐏𝟐 = 𝐃𝐞𝟑𝐱 and we will have
𝐭 𝐏 = 𝐀𝐱 𝟐 + 𝐁𝐱 + 𝐂 + 𝐃𝐞𝟑𝐱
From the non – homogenous equation, we can see that the highest order is third, so
we need to differentiate our particular solution up to third derivative
𝐭 𝐏 = 𝐀𝐱 𝟐 + 𝐁𝐱 + 𝐂 + 𝐃𝐞𝟑𝐱
𝐭 𝐏 ′ = 𝟐𝐀𝐱 + 𝐁 + 𝟑𝐃𝐞𝟑𝐱
𝐭 𝐏 ′′ = 𝟐𝐀 + 𝟗𝐃𝐞𝟑𝐱
𝐭 𝐏 ′′′ = 𝟐𝟕𝐃𝐞𝟑𝐱
We now substitute what we have solved to our non – homogenous equation
𝟐𝟕𝐃𝐞𝟑𝐱 + 𝟒(𝟐𝐀 + 𝟗𝐃𝐞𝟑𝐱 ) − 𝟕(𝟐𝐀𝐱 + 𝐁 + 𝟑𝐃𝐞𝟑𝐱 ) − 𝟏𝟎(𝐀𝐱 𝟐 + 𝐁𝐱 + 𝐂 + 𝐃𝐞𝟑𝐱 )
= 𝟏𝟎𝟎𝐱 𝟐 − 𝟔𝟒𝐞𝟑𝐱
𝟐𝟕𝐃𝐞𝟑𝐱 + 𝟖𝐀 + 𝟑𝟔𝐃𝐞𝟑𝐱 − 𝟏𝟒𝐀𝐱 − 𝟕𝐁 − 𝟐𝟏𝐃𝐞𝟑𝐱 − 𝟏𝟎𝐀𝐱 𝟐 − 𝟏𝟎𝐁𝐱 − 𝟏𝟎𝐂 − 𝟏𝟎𝐃𝐞𝟑𝐱
= 𝟏𝟎𝟎𝐱 𝟐 − 𝟔𝟒𝐞𝟑𝐱
− 𝟏𝟎𝐀𝐱 𝟐 − (𝟏𝟒𝐀 − 𝟏𝟎𝐁)𝐱 + 𝟖𝐀 − 𝟕𝐁 − 𝟏𝟎𝐂 + (𝟐𝟕𝐃 + 𝟑𝟔𝐃 − 𝟐𝟏𝐃 − 𝟏𝟎𝐃)𝐞𝟑𝐱
= 𝟏𝟎𝟎𝐱 𝟐 − 𝟔𝟒𝐞𝟑𝐱
𝟖𝐀 − 𝟕𝐁 − 𝟏𝟎𝐂 + (−𝟏𝟒𝐀 − 𝟏𝟎𝐁)𝐱 − 𝟏𝟎𝐀𝐱 𝟐 + 𝟑𝟐𝐃𝐞𝟑𝐱 = 𝟏𝟎𝟎𝐱 𝟐 − 𝟔𝟒𝐞𝟑𝐱
Now we will solve the undetermined coefficients:
@𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭𝐬
@𝐱
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𝟖𝐀 − 𝟕𝐁 − 𝟏𝟎𝐂 = 𝟎 → (a)
−𝟏𝟒𝐀 − 𝟏𝟎𝐁 = 𝟎 → (b)
@𝐱 𝟐
@𝐞𝟑𝐱
− 𝟏𝟎𝐀 = 𝟏𝟎𝟎
𝟑𝟐𝐃 = −𝟔𝟒
𝐀 = −𝟏𝟎
𝐃 = −𝟐
Using (b) for 𝐁
Using (a) for 𝐂
−𝟏𝟒(−𝟏𝟎) − 𝟏𝟎𝐁 = 𝟎
𝟖(−𝟏𝟎) − 𝟕(𝟏𝟒) − 𝟏𝟎𝐂 = 𝟎
𝐁 = 𝟏𝟒
𝐂=−
So, now our 𝐭 𝐏 = −𝟏𝟎𝐱 𝟐 + 𝟏𝟒𝐱 + −
𝟖𝟗
𝟓
𝟖𝟗
𝟓
− 𝟐𝐞𝟑𝐱 . Then finally, our general solution is
𝐭 = 𝐂𝟏 𝐞−𝟓𝐱 + 𝐂𝟐 𝐞𝟐𝐱 + 𝐂𝟑 𝐞−𝐱 − 𝟏𝟎𝐱 𝟐 + 𝟏𝟒𝐱 + −
𝟖𝟗
− 𝟐𝐞𝟑𝐱
𝟓
Now for the particular solution, we need to differentiate our general solution up to
second derivative.
@𝐭(𝟎) = −𝟐𝟎
−𝟐𝟎 = 𝐂𝟏 + 𝐂𝟐 + 𝐂𝟑 −
@𝐭 ′ (𝟎) = −
−
𝟐𝟗
𝟓
𝟏𝟗
𝟓
𝟓
−𝟐
𝟐𝟗
𝟓
= −𝟓𝐂𝟏 + 𝟐𝐂𝟐 − 𝐂𝟑 + 𝟏𝟒 − 𝟔
@𝐭 ′′ (𝟎) = −
−
𝟖𝟗
𝟏𝟗
𝟓
= 𝟐𝟓𝐂𝟏 + 𝟒𝐂𝟐 − 𝐂𝟑 − 𝟐𝟎 − 𝟏𝟖
We have three equation three unknowns, so the value for 𝐂𝟏 , 𝐂𝟐 , and 𝐂𝟑 are 𝟐, −𝟐, and
𝟏
− 𝟓 respectively.
Finally, our solution will be
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𝟏
𝟖𝟗
𝐭 = 𝟐𝐞−𝟓𝐱 + −𝟐𝐞𝟐𝐱 − 𝐞−𝐱 − 𝟏𝟎𝐱 𝟐 + 𝟏𝟒𝐱 + −
− 𝟐𝐞𝟑𝐱
𝟓
𝟓
LET’S CHECK! (PRACTICE PROBLEMS)
Problem 1
Find 𝐃𝟑 (𝟐𝐱 𝟐 − 𝟏)𝟑
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Problem 2
Check whether the commutative law of multiplication is valid for the operators 𝐋 =
𝐃𝟐 + 𝟏 and 𝐌 = 𝟐𝐃 + 𝟑.
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Problem 3
Let 𝐀 = 𝐃 + 𝟑 and 𝐁 = 𝟐𝐃 − 𝟏. Compute 𝐀𝐁𝐲
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Problem 4
Solve the IVP, where 𝐱 = 𝐟(𝐭) and 𝐱" − 𝟒𝐱 = 𝟎 subject to the conditions that when
𝐭 = 𝟎, 𝐱 = 𝟎, and 𝐱′ = 𝟑
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Problem 5
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Solve y" − 5y = 0
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Problem 6
Solve y" – y' – 2y = 0
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Problem 7
(𝐝𝟐 𝐲)
(𝐝𝐲)
Solve [(𝐝𝐱𝟐 )] + 𝟒 [(𝐝𝐱)] + 𝟒𝐲 = 𝟎
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Problem 8
(𝐝𝟐𝐱 )
Solve [(𝐝𝐭 𝟐 )] = 𝟎
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Problem 9
(𝐝𝟐 𝐲)
(𝐝𝐲)
Solve the equation [(𝐝𝐱𝟐 )] + 𝟐 [(𝐝𝐱)] + 𝐲 = 𝟎
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Problem 10
Solve 𝐲′′ + 𝐲 = 𝟓𝐬𝐢𝐧𝐭
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Problem 11
Solve 𝐲 ′′ − 𝟖𝐲 ′ + 𝟏𝟔𝐲 = 𝟐𝟒𝐱 + 𝟐
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Problem 12
Solve 𝐲′′ − 𝟑𝐲′ − 𝟒𝐲 = −𝟐𝟓𝐜𝐨𝐬(𝟐𝐭)
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Problem 13
Solve for 𝐲 ′′ − 𝟐𝐲 ′ + 𝟓𝐲 = 𝟒𝐞𝟑𝐭
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LET’S ANALYZE! (PRACTICE PROBLEMS)
Problem 1
If A = D + 2 and B = 3D – 1 evaluate AB and BA
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Problem 2
Let 𝐆 = 𝐱𝐃 + 𝟐 and 𝐇 = 𝐃 – 𝟏; Evaluate 𝐆(𝐇𝐲) and 𝐇(𝐆𝐲).
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Problem 3
Check whether the commutative law of multiplication is valid for the operators 𝐋 =
𝐱𝐃 − 𝟏 and 𝐌 = 𝐃𝟐 + 𝐱 𝟐
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Problem 4
Solve y" + 4y = 0
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Problem 5
Solve the initial value problem 𝐲" − 𝟔𝐲′ + 𝟗𝐲 = 𝟎 with 𝐲(𝟎) = 𝟐, 𝐲′ (𝟎) = 𝟏.
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Problem 6
Find the particular solution of the differential equation 𝐲" + 𝐲′ − 𝟔𝐲 = 𝟎, subject to
the initial conditions 𝐲 = 𝟒 and 𝐲′ = 𝟑 when 𝐱 = 𝟎.
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Problem 7
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𝐝𝟑 𝐲
𝐝𝟐 𝐲
𝐝𝐲
Solve the equation 𝐝𝐱𝟑 – 𝟒 𝐝𝐱𝟐 + 𝐝𝐱 + 𝟔𝐲 = 𝟎.
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Problem 8
Solve 𝐲′′′ − 𝟔𝐲′′ + 𝟏𝟏𝐲′ − 𝟔𝐲 = 𝟎 if 𝐲(𝛑) = 𝟎, 𝐲′ (𝛑) = 𝟎, 𝐲" (𝛑) = 𝟏.
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Problem 9
Suppose that a 14-th order homogeneous linear differential equation with constant
coefficients have characteristic roots: −𝟑, 𝟏, 𝟎, 𝟎, 𝟐, 𝟐, 𝟐, 𝟐, 𝟑 + 𝟒𝐢, 𝟑 + 𝟒𝐢, 𝟑 + 𝟒𝐢, 𝟑 −
𝟒𝐢, 𝟑 − 𝟒𝐢, 𝟑 − 𝟒𝐢
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Problem 10
Solve 𝐲′′′ − 𝟓𝐲′′ − 𝐲′ + 𝟓𝐲 = 𝟏𝟎𝐭 − 𝟔𝟑𝐞−𝟐𝐭 + 𝟐𝟗 𝐬𝐢𝐧(𝟐𝐭)
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Problem 11
Solve the differential equation 𝐲″ − 𝟐𝐲′ − 𝟑𝐲 = 𝐞 𝟐𝐭 + 𝟑𝐭 𝟐 + 𝟒𝐭 − 𝟓 + 𝟓𝐜𝐨𝐬(𝟐𝐭)
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Problem 12
Solve 𝐲″ − 𝟐𝐲′ − 𝟑𝐲 = 𝟑𝐭 𝟐 + 𝟒𝐭 – 𝟓 if the initial conditions are 𝐲(𝟎) = 𝟗, 𝐲′(𝟎) =
−𝟒
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Problem 13
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Solve
𝐲 ″ − 𝟐𝐲 ′ + 𝟒𝐲 = 𝟖𝐭 − 𝟏𝟐𝐬𝐢𝐧 𝟐𝐭
−𝟐, 𝐲′(𝟎) = 𝟖
𝐢𝐟 𝐭𝐡𝐞 𝐢𝐧𝐢𝐭𝐢𝐚𝐥 𝐜𝐨𝐧𝐝𝐢𝐭𝐢𝐨𝐧𝐬 𝐚𝐫𝐞 𝐲(𝟎) =
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IN A NUTSHELL!
Activity 1
From what you have learn, summarize linear differential operators on your own brief
words. And the rules and theorems that are important to the next topic.
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Activity 2
There is a complication that occurs under a certain circumstance, look for it and explain
how did it become an error.
𝐲″ − 𝟐𝐲′ − 𝟑𝐲 = 𝟓𝐞𝟑𝐭
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Activity 3
Check whether the commutative law of multiplication is valid for the operators 𝐋 =
𝐱𝐃 − 𝟏 and 𝐌 = 𝐃𝟐 + 𝐱 𝟐
Self-Help: You can also refer to the sources below to help you further
understand the lesson.
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*Hass, J., Weir, M., Thomas Jr, G. (2012). University Calculus: early transcendental.
Philippines : Pearson Education South Asia.
Stewart, J. (2000). Calculus: Concepts and Contexts. Pacipic Grove, CA: Brook/Cole.
Big Picture in Focus:
ULO-4a. learn and solve about Laplace Transforms of Functions
Big Picture
Week 8-9: Unit Learning Outcomes (ULO): At the end of the unit, you are expected
to:
a. learn and solve about Laplace Transforms of Functions,
b. analyze and compute Inverse Laplace Transforms, and
c. Solve for Initial Value Problems using Laplace Transforms
Metalanguage
In this section, Laplace transform will be a tool for you to solve for IVP’s which will be
discussed in the next learning outcomes. Under this section will be the definition,
transforms of Elementary functions, transforms of 𝑒 −𝛼𝑡 𝑓(𝑡) – Theorem, and
transforms of 𝑡 𝑛 𝑓(𝑡) – Derivatives of Transforms. Being an engineer, you must contain
this basic knowledge to practice your field.
Please proceed immediately to the “Essential Knowledge” part since the first
lesson is also definition of essential terms.
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Essential Knowledge
Laplace transforms are important for any engineer’s mathematical toolbox as they
make unraveling linear ODEs and related initial value problems, as well as systems of
linear ODEs, much easier. Such applications include: electrical networks, springs,
mixing problems, signal processing, and other areas of engineering and physics.
Laplace transform reduces the problem of solving a differential equation to an
algebraic problem.
Definition: Let f(t) be a given of function which is defined for all positive values y = t.
We multiply f(t) by 𝑒 −𝑠𝑡 and integrate with respect to t from zero to infinity. Then, if the
resulting integral exists, it is a function of s, say F(s); where s is a parameter which
may be real or complex.
∞
𝐹(𝑠) = ∫ 𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡
0
The function F(s) is the Laplace transform of the original function f(t) and will be
denoted by ℒ(𝑓). The symbol ℒ, which transforms f(t) into F(s) is called the Laplace
transformation operator. Thus,
∞
𝐹(𝑠) = ℒ(𝑓) = ∫ 𝑒 −𝑠𝑡 𝑓(𝑡)𝑑𝑡
0
The original function f(t) is called the inverse transform or inverse of F(s) and will be
denoted by ℒ −1 (𝐹). That is, f(t) = ℒ −1 (𝐹).
Example 1: Find the Laplace transform of f(t) = 1, when t > 0
∞
𝐹(𝑠) = ℒ(𝑓) = ∫ 𝑒 −𝑠𝑡 (1)𝑑𝑡
0
𝑢 = −𝑠𝑡
𝑑𝑢 = −𝑠 𝑑𝑡
∞
= ∫ 𝑒 𝑢 𝑑𝑢
0
1
= − (𝑒 𝑢 )∞
0
𝑠
1
= − (𝑒 −𝑠𝑡 )∞
0
𝑠
1
= − [𝑒 −𝑠(∞) − 𝑒 −𝑠(0) ]
𝑠
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1
= − (0 − 1)
𝑠
1
𝐹(𝑠) = ; 𝑠 > 0
𝑠
Example 1: Find the Laplace transform of 𝑓(𝑡) = 𝑒 𝛼𝑡 , when t > 0
∞
𝐹(𝑠) = ℒ(𝑓) = ∫ 𝑒 −𝑠𝑡 (𝑒 𝛼𝑡 )𝑑𝑡
∞
0
= ∫ 𝑒 𝑡(𝑎−𝑠) 𝑑𝑡
0
𝑢 = 𝑡(𝑎 − 𝑠)
𝑑𝑢 = (𝑎 − 𝑠) 𝑑𝑡
∞
𝑑𝑢
= ∫ 𝑒𝑢
𝑎−𝑠
0
1
(𝑒 𝑢 )∞
=
0
𝑎−𝑠
=
=
1
∞
(𝑒 𝑡(𝑎−𝑠) )0
𝑎−𝑠
1
(𝑒 ∞(𝑎−𝑠) − 𝑒 0(𝑎−𝑠) )
𝑎−𝑠
1
(0 − 1)
𝑎−𝑠
−1
=
𝑎−𝑠
−1
=
−(−𝑎 − 𝑠)
1
𝐹(𝑠) =
; 𝑠>0
𝑠 − 𝑎
=
Table 6.1
Some functions f(t) and their Laplace Transforms
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PROPERTIES OF LAPLACE TRANSFORM
1. LINEARITY PROPERTY:
If a and b are constants while f(t) and g(t) are functions of t, then
Example 1: Find the Laplace transform of 𝑓(𝑡) = 7𝑠𝑖𝑛𝑡.
ℒ(𝑓) = 7 ℒ(𝑠𝑖𝑛𝑡)
𝟕
𝑭(𝒔) = 𝟐
𝒔 +𝟏
Example 2: Find the Laplace transform of 𝑓(𝑡) = 2𝑡 + 3
ℒ(𝑓) = 2 ℒ(𝑡) + 3
𝐹(𝑠) = 2 ℒ(𝑡) + 3
1
1
𝐹(𝑠) = 2( 2 ) + 3( )
𝑠
𝑠
𝟐 + 𝟑𝒔
𝑭(𝒔) =
𝒔𝟐
2. FIRST SHIFTING PROPERTY:
Example 3: Find the Laplace transform of 𝑓(𝑡) = 𝑒 −3𝑡 𝑐𝑜𝑠5𝑡
𝑠
ℒ(𝑓) = 2
; (𝑤ℎ𝑒𝑟𝑒 𝑠 = 𝑠 + 3)
𝑠 + 52
𝒔+𝟑
𝓛(𝒇) =
(𝒔 + 𝟑)𝟐 + 𝟐𝟓
Example 4: Find the Laplace transform of 𝑓(𝑡) = 𝑡 3 𝑒 5𝑡
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ℒ(𝑓) =
3!
𝑠 3+1
; (𝑤ℎ𝑒𝑟𝑒 𝑠 = 𝑠 − 5)
𝟔
(𝒔 − 𝟓)𝟒
𝓛(𝒇) =
3. CHANGE OF SCALE PROPERTY
Example 5: Find the Laplace transform of 𝑓(𝑡) = 𝑐𝑜𝑠3𝑡
𝑠
Remember that, ℒ(cost) = 𝑠2 +12
1
ℒ(cos3t) = 3 (
1
𝑠
3
𝑠 2
( ) +1
3
𝑠
3
𝑠2
ℒ(cos3t) = 3 (
9
+1
)
)
𝑠
1
ℒ(cos3t) = 3 ( 𝑠23+9 )
9
1
𝑠
9
ℒ(cos3t) = 3 ⌊(3)(𝑠2+9)⌋
𝒔
ℒ(cos3t) = 𝒔𝟐 +𝟗
𝑠2 − 𝑠+1
Example 6: Find the Laplace transform of 𝑓(2𝑡)𝑤ℎ𝑒𝑛 ℒ[f(t)]=(2𝑠+1)2(𝑠−1)
𝑠 2 𝑠
2
2
𝑠
𝑠
(2( )+1)2 ( −1)
2
2
𝑠2 𝑠
1
( ) − +1
ℒ[f(2t)]=2 [
1
ℒ[f(2t)]=2 [
4
− +1
2
ℒ[f(2t)]=2 [
1
𝑠2 −2𝑠+4
4
𝑠−2
(𝑠+1)2 (
)
2
𝑠2 −2𝑠+4
ℒ[f(2t)]=2 [(
4
𝟏
]
𝑠−2
)
2
(𝑠+1)2 (
1
]
]
2
) ((𝑠+1)2 (𝑠−2))]
𝒔𝟐 −𝟐𝒔+𝟒
ℒ[f(2t)]=𝟒 ((𝒔+𝟏)𝟐 (𝒔−𝟐))
4. MULTIPLICATION BY POWER OF t
If ℒ[f(t)] = F(s), then
𝑑𝑛
ℒ[𝑡 𝑛 𝑓(𝑡)] = (−1)𝑛 𝑑𝑠𝑛 𝐹(𝑠)
Example 7: Find the Laplace transform of 𝑓(𝑡) = 𝑡𝑠𝑖𝑛(𝑡)
By looking at the exponent of t, we have n=1.
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𝑑𝑛
ℒ[𝑡 𝑛 𝐹(𝑡)] = (−1)𝑛 𝑑𝑠𝑛 𝐹(𝑠)
𝑑1
1
(
)
= (−1)
𝑑𝑠1 𝑠 2 + 12
(𝑠 2 + 1)(0) − (1)(2𝑠)
= (−1) (
)
(𝑠 2 + 1)2
−2𝑠
)
= (−1) ( 2
(𝑠 + 1)2
𝟐𝒔
= 𝟐
(𝒔 + 𝟏)𝟐
Example 8: Find the Laplace transform of 𝑓(𝑡) = 𝑡(3𝑠𝑖𝑛2𝑡 − 2𝑐𝑜𝑠2𝑡)
1
2
𝑠
ℒ[3tsint2t] - ℒ[2tcos2t] = 3(−1)1 𝑑 [𝑠2 +4] − 2(−1)1 𝑑 [𝑠2 +22 ]
= −3 [
(𝑠2 +4)(0) − (2)(2𝑠)
(𝑠2 +4)2
]+ 2[
(𝑠2 +4)(1) − (𝑠)(2𝑠)
(𝑠2 +4)2
−𝑠2 +4
−4𝑠
= −3 [(𝑠2 +4)2] + 2 [(𝑠2 +4)2 ]
12𝑠 − 2𝑠2 +8
=
=
(𝑠2 +4)2
𝟐(𝟔𝒔 − 𝒔𝟐 +𝟒)
(𝒔𝟐 +𝟒)𝟐
5. DIVISION BY t
If ℒ[f(t)] = F(s), then
ℒ[
𝑓(𝑡)
𝑡
∞
] = ∫𝑠 𝐹(𝑢)𝑑𝑢
Provided lim [
𝑓(𝑡)
𝑡
𝑡−0
Example 9: Find the Laplace transform of 𝑓(𝑡) =
By identity: 𝑓(𝑡) =
ℒ[
1 − 𝑐𝑜𝑠2𝑡
2
𝑠𝑖𝑛2 𝑡
∞
=∫
𝑠
𝑠𝑖𝑛2 𝑡
𝑡
𝑠𝑖𝑛2 𝑡
𝑡
1
2𝑠
=
1
1 − 𝑐𝑜𝑠2𝑡
2
𝑠
− 2 (𝑠2 +4)
1 1
𝑠
( − 2
)
2 𝑠 𝑠 +4
=
Thus, ℒ (
]=
]
𝑡
∞1 1
) = ∫𝑠
𝑠
( − 𝑠2 +4) 𝑑𝑠
2 𝑠
∞
1
1
𝑠
𝑑𝑠 − ∫
( 2
) 𝑑𝑠
2𝑠
𝑠 2 𝑠 +4
𝑙𝑒𝑡 𝑢 = 𝑠 2 + 4; 𝑑𝑢 = 2𝑠𝑑𝑠
]
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∞
=∫
𝑠
∞
1
1 𝑑𝑢
( )
𝑑𝑠 − ∫
2𝑠
𝑠 2 2𝑢
∞
∞
1
1
= [ 𝑙𝑛(𝑠)] − [ 𝑙𝑛 𝑠 2 + 4]
2
4
𝑠
𝑠
∞
∞
1
1
2
2
[𝑙𝑛(∞)
[𝑙𝑛
(∞
=[
− 𝑙𝑛(𝑠)]] − [
+ 4) − 𝑙𝑛 (𝑠 + 4)] ]
2
4
0
0
1
1
= [(∞) − 𝑙𝑛(𝑠)] − [ ∞ − 𝑙𝑛 (𝑠 2 + 4)]
2
4
1
1
= ∞ − 𝑙𝑛(𝑠) − ∞ + 𝑙𝑛(𝑠)
2
4
1
= [−2𝑙𝑛(𝑠) + 𝑙𝑛(𝑠 2 + 4)]
4
=
1
[−𝑙𝑛(𝑠 2 ) + 𝑙𝑛(𝑠 2 + 4)]
4
=
𝟏
𝒔𝟐 + 𝟒
𝒍𝒏 ( 𝟐 )
𝟒
𝒔
6. LAPLACE TRANSFORMS OF DERIVATIVES
If f(t) is continuous and f'(t) is piecewise continuous for 0 t ≥, then
ℒ{ f'(t) } = s ℒ{f(t) } − f(0)
[Proof]
∞
ℒ[𝑓′′(𝑡) = ∫ 𝑓′(𝑡)𝑒 −𝑠𝑡 𝑑𝑡
0
Integration by parts by letting
u = e-st
dv = f'(t) dt
-st
du= -se dt
v = f(t)
∞
′
−𝑠𝑡
ℒ[𝑓′′(𝑡) = [𝑒 −𝑠𝑡 𝑓(𝑡)]∞
𝑑𝑡 = −𝑓(0) + 𝑠ℒ[𝑓(𝑡)]
0 + 𝑠 ∫ 𝑓 (𝑡)𝑒
0
= ℒ[𝑓′(𝑡) = 𝑠ℒ[𝑓(𝑡)] − 𝑓(0)
Theorem: f(t), f'(t), . . ., f(n-1)(t) are continuous functions for t ≥ 0, and f(n)(t) is
piecewise continuous function, then
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EXAMPLE 10: Find the Laplace Transform of f(t) = sin at.
EXAMPLE 11: Find the Laplace Transform of f(t) =𝑡 2 .
𝑓(𝑡) = 𝑡 2
𝑓(0) = 0
𝑓′(𝑡) = 2𝑡
𝑓′(0) = 0
𝑓′′(𝑡) = 2
𝑓′′(0) = 2
2
ℒ(𝑓′′) = 𝑠 ℒ(𝑓) − 𝑠𝑓(0) − 𝑓′(0)
ℒ(2) = 𝑠 2 ℒ(𝑓) − 𝑠(0) − 0
2
= 𝑠 2 ℒ(𝑓)
𝑠
2
ℒ(𝑓) = 3
𝑠
Self-Help: You can also refer to the sources below to help you further
understand the lesson.
Source
Kreyszig, E.(2011), Advance Engineering Mathematics.
Let’s Check
Find the Laplace Transform of the following (No specific method).
1. 𝑓(𝑡) = 2𝑡 + 3
2. 𝑓(𝑡) = 𝑐𝑜𝑠5𝑡
3. 𝑓(𝑡) = 𝑡 − 𝑐𝑜𝑠ℎ3𝑡
4. 𝑓(𝑡) = 𝑡 3 𝑒 5𝑡
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5.
6.
7.
8.
𝑓(𝑡)
𝑓(𝑡)
𝑓(𝑡)
𝑓(𝑡)
=
=
=
=
4𝑡 3 𝑒 −𝑡
4𝑠𝑖𝑛ℎ3𝑡 − 18𝑒 −5𝑡
2𝑒 −7𝑡 − 𝑡𝑒 −7𝑡
𝑐𝑜𝑠 2 2𝑡
Let’s Analyze
Using a specific method, find the Laplace Transforms of the following.
1. Find the Laplace Transforms using Laplace transform of Derivatives given,
𝑓(𝑡) = 𝑠𝑖𝑛 𝑎𝑡
2. Find the Laplace Transforms using Multiplication by 𝑡 𝑛 given, 𝑓(𝑡) = 𝑡 2 𝑐𝑜𝑠 𝑎𝑡
3. Find the Laplace Transforms using Division by t given, 𝑓(𝑡) =
1 − 𝑒 −𝑡
𝑡
4. Find the Laplace Transforms using First Shifting Property, 𝑓(𝑡) = 𝑒 −2𝑡 𝑠𝑖𝑛4𝑡.
5. Find the Laplace Transforms using Change of Scale Property 𝑓(𝑡) =
2𝑠𝑖𝑛(−5𝑡)
In a Nutshell
The Laplace transform is used for solving differential and integral equations. In
physics and engineering it is used for analysis of linear time-invariant systems such
as electrical circuits, harmonic oscillators, optical devices, and mechanical systems.
From the time domain function f(t) to its frequency domain counterpart ℒ[f(t)](s). Such
transforms can be computed directly from the definition of Laplace transform
∞
𝐹(𝑠) = ∫ 𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡
0
Q&A List
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If you have any questions regarding Laplace Transform, kindly write down on
the table provided.
QUESTIONS
1.
2.
3.
4.
5.
ANSWERS
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Big Picture in Focus:
ULO-4b. analyze and compute Inverse Laplace Transforms
Metalanguage
In this section, you will learn about the Inverse of Laplace Transforms. Being an
engineer, you must contain this basic knowledge to practice your field. Please refer to
these definitions of terms to continue.
The Inverse Laplace transform is the transformation of a Laplace transform into a
function of time. If L{f(t)}=F(s) then f(t) is the inverse Laplace transform of F(s), the
inverse being written as:
Partial Fractions are algebraic technique of writing a quotient of polynomials in
simpler form.
Please proceed immediately to the “Essential Knowledge” part since the next
lesson is also definition of essential terms.
Essential Knowledge
In ULOa, we defined the Laplace transform of f by
We’ll also say that f is an inverse Laplace Transform of F, and write
To solve differential equations with the Laplace transform, we must be able to obtain f
from its transform F. There’s a formula for doing this, but we can’t use it because it
requires the theory of functions of a complex variable. Fortunately, we can use the
table of Laplace transforms to find inverse transforms that we will need. The table in
the next page shows the Inverse Laplace Functions of some Laplace Transforms
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Table 6.2: Inverse Laplace Transform
Image source: https://www.quora.com/What-is-the-inverse-Laplace-transform-of-one
EXAMPLE 1: Find the Inverse Laplace Transform of F(s) =
s2
ℒ −1 [
s2 − 3s +4
s3
− 3s + 4
1
1
1
] = ℒ −1 [ ] − ℒ −1 3 [ 2 ] + ℒ −1 4 [ 3 ]
3
s
s
s
s
3t 2−1
4t 3−1
+
(2 − 1)!
(3 − 1)!
4t 2
= 1 − 3t +
2
𝐟(𝐭) = 𝟏 − 𝟑𝐭 + 𝟐𝐭 𝟐
= 1 −
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12
EXAMPLE 2: Find the Inverse Laplace Transform of F(s) = s − 5 −
ℒ −1 [
2s
s2 +49
12
2s
12
2s
− 2
] = ℒ −1 (
) − ℒ −1 ( 2
)
s − 5
s − 5
s + 49
s + 49
𝐟(𝐭) = 𝟏𝟐𝐞𝟓𝐭 – 𝟐𝐜𝐨𝐬𝟕𝐭
EXAMPLE 3: Find the Inverse Laplace Transform of F(s) =
s+4
s2 − s − 6
Remember: From Algebra, a proper rational fraction can be resolved into a sum of
partial fractions. Please review them as we will not tackle that here.
s+4
Therefore, s2 − s − 6 =
s+4
(s+2)(s−3)
which makes this a distinct/non − linear factor
s+4
A
B
=
+
(s − 3)(s + 2)
s − 3 s + 2
s + 4 = A(s + 2) + B(s − 3)
@s = −2: − 2 + 4 = A(−2 + 2) + B(−2 − 3)
2 = B(−5)
B = − 2⁄5
@s = 3: 3 + 4 = A(3 + 2) + B(3 − 3)
7 = 5A
A = 7⁄5
NOTE: I do not require you to use this method only. You can use different method to solve for the
constants.
−2⁄
7⁄
s+4
5
5
F(s) =
𝓛−𝟏 [
𝐬𝟐
s2 − s − 6
=
s − 3
+
s + 2
𝐬+𝟒
] = 𝐟(𝐭) = 𝟕⁄𝟓 𝐞𝟑𝐭 − 𝟐⁄𝟓 𝐞−𝟐𝐭
− 𝐬 − 𝟔
EXAMPLE 4: Find the Inverse Laplace Transform of F(s) =
4s+5
(s − 1)2 (s+2)
4s + 5
A
B
C
=
+
+
2
2
(s − 1) (s + 2)
s − 1 (s − 1)
s +2
𝟒𝐬 + 𝟓 = 𝐀(𝐬 − 𝟏)(𝐬 + 𝟐) + 𝐁(𝐬 + 𝟐) + 𝐂(𝐬 − 𝟏)𝟐
𝟒𝐬 + 𝟓 = 𝐀𝐬 𝟐 + 𝐀𝐬 − 𝟐𝐀 + 𝐁𝐬 + 𝟐𝐁 + 𝐂𝐬 𝟐 − 𝟐𝐂𝐬 + 𝐂
@s = 1: 4 + 5 = B(3)
B=3
@s = −2: −8 + 5 = 9C
C = −1⁄3
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@s 2 : 0 = A + C
0 = A − 1⁄3
A = 1⁄3
1⁄
3
3 +
=
4s + 5
(s − 1)2 (s + 2)
𝟒𝐬 + 𝟓
−𝟏
𝓛
[
𝟐
(𝐬 − 𝟏) (𝐬 + 𝟐)
s − 1
(s − 1)2
−
1⁄
3
s +2
] = 𝐟(𝐭) = 𝟏⁄𝟑 𝐞𝐭 + 𝟑𝐭𝐞𝐭 − 𝟏⁄𝟑 𝐞−𝟐𝐭
Self-Help: You can also refer to the sources below to help you further
understand the lesson.
Sources:
Fox,H., Bolton, B.,( 2002). in Mathematics for Engineers and Technologists
Kreyszig, E.(2011), Advance Engineering Mathematics.
Let’s Check
Find the Inverse Laplace Transform of the following.
3
1. 𝐹(𝑠) = 2
𝑠 +5
1
2. 𝐹(𝑠) =
3𝑠 − 5
5𝑠
3. 𝐹(𝑠) = 2
𝑠 +9
2
4. 𝐹(𝑠) = 4
3𝑠
3𝑠 + 2
5. 𝐹(𝑠) = 2
𝑠 + 25
Let’s Analyze
Find the Inverse Laplace Transform of the following.
𝑠+2
1. 𝐹(𝑠) = 2
𝑠 − 4𝑠 + 13
3(𝑠 2 − 2)2
2. 𝐹(𝑠) =
2𝑠 5
5𝑠 + 3
3. 𝐹(𝑠) =
(𝑠 − 1)(𝑠 2 + 2𝑠 + 5)
1
4. 𝐹(𝑠) = 2
(𝑠 + 6𝑠 + 13)
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5. 𝐹(𝑠) =
𝑠 − 1
(𝑠 + 3)(𝑠 2 + 2𝑠 + 2)
In a Nutshell
Given F(s) and f(t),
𝐹(𝑠)
= ℒ𝑓(𝑠), 𝑡ℎ𝑒𝑛 𝑓(𝑡) 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑙𝑎𝑝𝑙𝑎𝑐𝑒 ℒ 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝑜𝑓 𝐹(𝑠), 𝑎𝑛𝑑 𝑖𝑠 𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑏𝑦
𝑓(𝑡) = ℒ −1 𝐹(𝑡)
ℒ −1 {𝐹(𝑠)}(𝑡)
𝐹 = ℒ(𝑓)
𝑓 = ℒ −1 (𝐹)
Q&A List
If you have any questions regarding Inverse Laplace Transform, kindly write
down on the table provided.
QUESTIONS
1.
2.
3.
4.
5.
ANSWERS
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Big Picture in Focus:
ULO-4c. Solve for Initial Value Problems using Laplace Transforms
Metalanguage
In this section, you will learn about solving Initial Value Problems. Being an engineer,
you must contain this basic knowledge to practice your field. Now what is an Initial
Value Problem?
An Initial Value Problem (or IVP) is a differential equation along with an appropriate
number of initial conditions. Some examples are given below,
You need to have the skills in solving for Laplace transforms in order to understand
this topic. Please proceed immediately to the “Essential Knowledge” part since
the next lesson is also definition of essential terms.
Essential Knowledge
As we have seen, most differential equations have more than one solution. For a firstorder equation, the general solution usually involves an arbitrary constant C, with one
particular solution corresponding to each value of C.
What this means is that knowing a differential equation that a function y(x) satisfies is
not enough information to determine y(x). To find the formula for y(x) precisely, we
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need one more piece of information, usually called an initial condition. For example,
suppose we know that a function y(x) satisfies the differential equation.
𝑦′ = 𝑦
It follows that
𝑦(𝑥) = 𝐶𝑒 𝑥
for some constant C. If we want to determine C, we need atleast one more piece of
information about the function y(x). For example, if we also know that
𝑦(0) = 3
the value of C must be 3, and hence 𝑦(𝑥) = 3𝑒 𝑥 .
For example, 𝑦′ = 𝑦,
3𝑒 𝑥 .
𝑦(0) = 3 is an initial value problem, whose solution is 𝑦 =
In general, we expect that every initial value problem has exactly one solution. We
can find this solution using the following procedure.
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Remember in your first topic:
Refresher.. Find the solution of the following initial value problem.
𝑦′ = −𝑦 2 ,
𝑦(0) = 5
SOLUTION:
𝑑𝑦
∫
= ∫ 𝑑𝑥
−𝑦 2
1
=𝑥+𝐶
𝑦
1
𝑦=
𝑥+𝐶
Plugging in x = 0 and y = 5 gives the equation
1
5=
0+𝐶
1
1
𝐶=
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑡𝑜 𝑦 =
5
𝑥+𝐶
Which gives us
𝒚=
𝟓
𝟓𝒙 + 𝟏
How about using Laplace transform?
Solving differential equations using ℒ [ ].
➔ Homogeneous IVP.
➔ First, second, higher order equations.
➔ Non-homogeneous IVP.
Very Important!!! Please recall: Partial fraction decompositions.
Solving differential equations using ℒ[ ].
Idea of the method:
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Homogeneous IVP
EXAMPLE 1: Use the Laplace transform to find the solution y(t) to the IVP
y′′−y′−2y= 0,
y(0) = 1,
y′(0) = 0.
Solution: Compute the ℒ[ ] of the differential equation,
ℒ [y′′−y′−2y] = ℒ [0]
⇒
ℒ [y′′−y′−2y] = 0.
The ℒ[ ] is a linear function, so
ℒ[y′′] − ℒ[y′] −2 ℒ [y] = 0
Derivatives are transformed into power functions,
[s2 ℒ [y] − s y(0)− y′(0)] − [s ℒ [y] − y(0)] − 2 ℒ [y] = 0,
We obtain(s2−s−2) ℒ[y] = (s−1) y(0) + y′(0).
Introduce the initial condition,
(s2−s−2) ℒ[y] = (s−1)
We can solve for the unknown ℒ[y] as follows,
(𝑠 − 1)
(𝑠 2 − 𝑠 − 2)
The partial fraction method: Find the zeros of the denominator,
ℒ[𝑦] =
Therefore, we rewrite:
ℒ[𝑦] =
(𝑠 − 1)
(𝑠 − 2)(𝑠 + 1)
Find constants a and b such that
(𝑠 − 1)
𝑎
𝑏
=
+
(𝑠 − 2)(𝑠 + 1) 𝑠 − 2 𝑠 + 1
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Note: You can use
other method to get
the constants
EXAMPLE 2: Use the Laplace transform to find the solution y(t) to the IVP
y′′−4y′+4y= 0,
y(0) = 1,
y′(0) = 1.
Solution: Compute the ℒ[ ] of the differential equation,
ℒ [y′′−4y′+4y] = ℒ [0]
⇒
ℒ [y′′−y′−2y] = 0.
The ℒ[ ] is a linear function, so
ℒ[y′′] − 4ℒ[y′] +4 ℒ [y] = 0
Derivatives are transformed into power functions,
[s2 ℒ [y] − s y(0)− y′(0)] − 4[s ℒ [y] − y(0)] + 4 ℒ [y] = 0,
We obtain(s2−4s+4) ℒ[y] = (s−4)y(0) + y′(0).
Introduce the initial condition,
(s2−4s+4) ℒ[y] = (s−3)
We can solve for the unknown ℒ[y] as follows
(𝑠 − 3)
ℒ[𝑦] = 2
(𝑠 − 4𝑠 + 4)
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First, second, higher order equations
EXAMPLE 3:
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Non-homogeneous IVP
EXAMPLE 4:
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Self-Help: You can also refer to the sources below to help you further
understand the lesson.
Sources:
Fox,H., Bolton, B.,( 2002). in Mathematics for Engineers and Technologists
Kreyszig, E.(2011), Advance Engineering Mathematics.
e- source:
https://users.math.msu.edu/users/gnagy/teaching/11-winter/mth235/lslides/l18-235.pdf
Let’s Check
Determine the following equations if they are Homogeneous IVP, First, second,
higher order equations and Non-homogeneous IVP.
1. y′′ − 10y ′ + 9y = 5t
y(0) and y′(0)=2
2. y'' + y' - 2y = 4
y(0) = 2
3. y′′−3y′+ 2y= 3cos(3t)
y(0) = 2, y′(0) = 3
4. y′′−y′−2y= 0, y(0) = 1
y′(0) = 2
y'(0) = 1
5.
Let’s Analyze
Use Laplace Transform to find the solution of the IVP.
1. y′′ − 10y ′ + 9y = 5t
y(0) and y′(0)=2
2. y'' + y' - 2y = 4
y(0) = 2
3. y′′−3y′+ 2y= 3cos(3t)
y(0) = 2, y′(0) = 3
4. y′′−y′−2y= 0, y(0) = 1
y′(0) = 2
y'(0) = 1
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5.
In a Nutshell
In solving DE using Laplace,
Q&A List
If you have any questions regarding Initial Value Problems using Laplace
Transform, kindly write down on the table provided.
QUESTIONS
1.
2.
3.
4.
ANSWERS
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5.