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DIFFERENTIAL EQUATIONS
WEEK 1
JUNE 23, 2021
ABDULLAH H. EDRIS
INSTRUCTOR
O947 – 827 - 4051
abdullahedris150@g
mail.com
Prepared by: Mr. Abdullah H. Edris Differential Equations
SYLLABUS IN MATH 211 (ONLINE LEARNING)
Name: MATH 211
Course Description: Differential Equation
Credit: 3 units
I. COURSE OUTLINE
Prelim Coverage
Day 1 Orientation
FIRST-ORDER DIFFERENTIAL EQUATION
Week 1 Day 2, Definition of Basic Terms
and Day 3, Families of Solutions
Week 2 and Elimination of arbitrary constants
Day 4 Families of Curves
EQUATIONS OF ORDER ONE
Separation of Variables
Day 5,
Homogeneous functions and equations
Week 3 Day 6, with homogenous coefficients
And Day 7,
Exact Equations
Week 4 And
Linear Equations of order one
Day 8
Integrating factors and determination of
integrating factors
Week 5 Day 9 Long Quiz
Day 10 Prelim Exam
Midterm Coverage
ADDITION TOPICS ON EQUATIONS OF
ORDER ONE
Day 11,
Substitution suggested by the Equation
Week 6 Day 12,
Bernoulli’s Equation
And Day 13,
Week 7 And Coefficient linear in two variables
Day 14 The General linear differential equation of
order n
Some properties of differential operators
Day 15 LINEAR EQUATIONS WITH CONSTANTS
Week 8 and COEFFICIENTS
Day 16 The auxiliary equation; distinct roots
The auxiliary equation; repeated roots
Day 19 Long Quiz
Week 9
Day 20 Midterm Examinations
Final Coverage
Day 21 ELEMENTARY APPLICATIONS
Week 10 and Newton’s law of cooling
Day 22 Velocity of escape from the earth
Week 11 Day 23 ELEMENTARY APPLICATIONS
Prepared by: Mr. Abdullah H. Edris Differential Equations
and Logistic growth and the price of
Day 24 commodities
Orthogonal Tragectories
THE LAPLACE TRANSFORM
The concept and definition of the
Day 25,
Laplace transform
Weeek 12 Day 26,
Transform of elementary functions
And Day 27,
Sectionally continuous functions and
Week 13 And
functions of exponential order
Day 28
Transforms of derivatives
Derivatives of transforms
Day 29 Review
Week 14
Day 30 Consultation
Day 31 Long Quiz
Week 15
Day 32 Final Exam
II. COURSE REQUIREMENTS:
a. Quizzes/Assignments Through Edmodo/Flexiquiz
b. Problem Sets Through Edmodo/Flexiquiz
c. Major Exams Through Edmodo/Flexiquiz
d. Attendance Through Edmodo
III. TEACHING METHODS AND STRATEGIES
a. Instructional Materials Through Edmodo
b. Video Lectures Through Edmodo
IV. EVALUTAION SCHEME
Prelim 20%
Midterm 20%
Finals 30%
Attendance 10%
Quizzes/Assignments 20%
Total 100%
PASSING RATE 50%
V. SPECIAL EXAM
A special exam is only viable iff. The student presents an acceptable reason.
The score that will be agreed upon shall be:
o The highest possible score shall be the score of the top scoring student;
o The student will also be demerit of 25% from his/her incurred score.
Prepared by: Mr. Abdullah H. Edris Differential Equations
Lesson 1: Definition of Basic Terms
Learning Outcome(s): At the end of this lesson, the students will be able to:
1. Define and illustrate some basic terms converted with differential equations;
2. Classify differential equations and determine their degree and order.
Differential Equation
Differential equation is an equation involving any equation which contains derivatives,
either ordinary derivative or partial derivative.
Two Kinds of Differential Equation
1. Ordinary Differential Equation (ODE) is a DE whose dependent variable depends on only
one independent variable.
2. Partial Differential Equation (PDE) is a DE whose dependent variable depends on two or
more independent variables.
Example of Differential Equation
1. 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 𝑔(𝑡) ODE
𝑑2 𝑦 𝑑𝑦
2. 𝑠𝑖𝑛(𝑦) = (1 − 𝑦 ) + 𝑦 2 𝑒 −5𝑦 ODE
𝑑𝑥 2 𝑑𝑥
3. 𝑦 4 + 10𝑦 ′′′ − 4𝑦 ′ + 2𝑦 = 𝑐𝑜𝑠(𝑡) ODE
𝜕2 𝑢 𝜕2 𝑢 𝜕2 𝑢
4. + 𝜕𝑦2 + 𝜕𝑧 2 = 0 PDE
𝜕𝑥 2
Order of Differential Equation
Order of DE – the order of differential equation is the largest derivative of the present in the DE.
Degree of Differential Equation
Degree of DE – the degree of the order of the differential equation.
Example. State whether the differential is Ordinary Differential Equation or Partial Differential
Equation and state its Order and Degree.
Prepared by: Mr. Abdullah H. Edris Differential Equations
Types of
Given Order Degree
DE
𝒅𝟐 𝒙
1. 𝒅𝒚𝟐 + 𝒌𝟐 𝒙 = 𝟎 ODE 2nd 1st
2. [(𝒙𝟐 + 𝒚𝟐 )𝒅𝒙 + 𝟐𝒙𝒚𝒅𝒚 = 𝟎]
[(𝒙𝟐 + 𝒚𝟐 )𝒅𝒙 + 𝟐𝒙𝒚𝒅𝒚
𝟏 ODE 1st 1st
= 𝟎]
𝒅𝒙
𝒅𝒚
(𝒙𝟐 + 𝒚𝟐 ) + 𝟐𝒙𝒚
𝒅𝒙
3. 𝒚′′′ − 𝟑𝒚′ + 𝟐𝒚 = 𝟎 ODE 3rd 1st
𝝏𝟐 𝒖 𝝏𝒖
4.𝝏𝒙𝟐 + 𝝏𝒚 = 𝟎 PDE 2nd 1st
𝒅𝟐 𝒚 𝒅𝒚
5. 𝒙 𝒅𝒙𝟐 + 𝒚 𝒅𝒙 + 𝟒𝒚𝟐 = 𝟏 ODE 2nd 1st
𝒅𝟑 𝒚 𝒅𝒚
6. 𝒔𝒊𝒏 (𝒅𝒙𝟑 ) = 𝒅𝒙 + 𝒙 ODE 1st 1st
𝒅𝒚 𝒅𝟐 𝒚
7. √(𝒅𝒙)𝟐 + 𝟑𝒚 = 𝒅𝒙𝟐
𝟐
𝟐 𝟐
𝒅𝒚 𝒅𝟐 𝒚
(√( ) + 𝟑𝒚) = ( 𝟐 )
𝒅𝒙 𝒅𝒙 ODE 2nd 2nd
𝟐
𝒅𝒚 𝟐 𝒅𝟐 𝒚
( ) + 𝟑𝒚 = ( 𝟐 )
𝒅𝒙 𝒅𝒙
𝒅𝒚 𝟑 𝒅𝟐 𝒚
8. 𝟑𝒚𝟐 (𝒅𝒙) − 𝒅𝒙𝟐 = 𝒔𝒊𝒏(𝒙𝟐 ) ODE 2nd 1st
Prepared by: Mr. Abdullah H. Edris Differential Equations
Exercise 1
State whether the differential is Ordinary Differential Equation or Partial Differential
Equation and state its Order and Degree. Write your answers on the space provided.
Given Types of DE Order Degree
𝒅𝒙
1. 𝒅𝒚 = 𝒙𝒚 + 𝟏
2. 𝒚′′′ + 𝒚′′ + 𝒚′ + 𝒚 = 𝟏
𝝏𝟐 𝒚 𝝏𝟒 𝒚
3. 𝝏𝒙𝟐 + 𝝏𝒚𝟒 = 𝒔𝒊𝒏𝒚
𝒅𝟐 𝒚 𝒅𝟑 𝒚
4. 𝒄𝒐𝒔 ( 𝟐) =
𝒅𝒙 𝒅𝒙𝟑
𝟑 𝒅𝒚 𝟐
𝒅 𝒚 𝟐
5. √(𝒅𝒙) + 𝟑𝒚 = 𝒅𝒙𝟐
𝒅𝟐 𝒚 𝒅𝒚 𝟑
6. 𝒅𝒙𝟐 + (𝒅𝒙) = 𝒙𝒚𝟐
𝟐
𝒅𝟐 𝒚 𝒅𝟐 𝒚
7. √𝟐 + (𝒅𝒙𝟐 ) = 𝟐𝒚 (𝒅𝒙𝟐 )
8. (𝒚′)𝟐 − 𝟑𝒚𝒚′ + 𝒙𝒚 = 𝟎
𝟏𝟐
𝒅𝟐 𝒚
9. ( 𝟐) +𝒚=𝟎
𝒅𝒙
𝒅𝟑 𝒚 𝒅𝒚
10. 𝒙 𝒅𝒙𝟑 + 𝟕 𝒅𝒙 − 𝟕𝒚 = 𝟔
Prepared by: Mr. Abdullah H. Edris Differential Equations
Lesson 2: Solutions of Differential Equations
Learning Outcome(s): At the end of this lesson, the students will be able to:
1. Verify functions that serve as solutions to differential equations;
2. Determine which functions satisfy differential equations
A solution of a differential equation in the unknown function y and the independent variable
x on the interval I is a function y(x) that satisfies the differential equation identically for all x in
I.
Example 1. Verify that 3𝑒 −2𝑥 + 4𝑒 𝑥 is a solution of
𝑦 ′′′ − 3𝑦 ′ + 2𝑦 = 𝑜.
Solution:
𝑦 = 3𝑒 −2𝑥 + 4𝑒 𝑥
𝑦′ = −6𝑒 −2𝑥 + 4𝑒 𝑥
𝑦′′ = 12𝑒 −2𝑥 + 4𝑒 𝑥
𝑦′′′ = −24𝑒 −2𝑥 + 4𝑒 𝑥
(−24𝑒 −2𝑥 + 4𝑒 𝑥 ) − 3(−6𝑒 −2𝑥 + 4𝑒 𝑥 ) + 2(3𝑒 −2𝑥 + 4𝑒 𝑥 ) = 0
−24𝑒 −2𝑥 + 4𝑒 𝑥 + 18𝑒 −2𝑥 − 12𝑒 𝑥 + 6𝑒 −2𝑥 + 8𝑒 𝑥 = 0
−24𝑒 −2𝑥 + 18𝑒 −2𝑥 + 6𝑒 −2𝑥 + 4𝑒 𝑥 − 12𝑒 𝑥 + 8𝑒 𝑥 = 0
0= 0
Hence, 𝟑𝒆−𝟐𝒙 + 𝟒𝒆𝒙 is a solution of 𝒚′′′ − 𝟑𝒚′ + 𝟐𝒚 = 𝒐.
𝑑2 𝑥
Example 2. Verify that 𝑥 = 𝑠𝑖𝑛𝑘𝑡 is a solution of 𝑑𝑡 2 + 𝑘 2 𝑥 = 0.
Solution:
𝑥 = 𝑠𝑖𝑛𝑘𝑡
𝑥 ′ = 𝑘𝑐𝑜𝑠𝑘𝑡
𝑥 ′′ = −𝑘 2 𝑠𝑖𝑛𝑘𝑡
−𝑘 2 𝑠𝑖𝑛𝑘𝑡 + 𝑘 2 (𝑠𝑖𝑛𝑘𝑡) = 0
Prepared by: Mr. Abdullah H. Edris Differential Equations
−𝑘 2 𝑠𝑖𝑛𝑘𝑡 + 𝑘 2 𝑠𝑖𝑛𝑘𝑡 = 0
0=0
𝒅𝟐 𝒙
Therefore, 𝒙 = 𝒔𝒊𝒏𝒌𝒕 is a solution of + 𝒌𝟐 𝒙 = 𝟎.
𝒅𝒕𝟐
𝟑
−
Example 3. Show that 𝒚(𝒙) = 𝒙 𝟐 is a solution to 𝟒𝒙𝟐 𝒚′′ + 𝟏𝟐𝒙𝒚′ + 𝟑𝒚 = 𝟎 for x > 0.
Solution:
𝟑
−
𝒚=𝒙 𝟐
−𝟑 −𝟑−𝟏 −𝟑 −𝟓
𝒚′ = 𝒙 𝟐 = 𝒙 𝟐
𝟐 𝟐
−𝟑 −𝟓 −𝟓−𝟏 𝟏𝟓 −𝟕
𝒚′′ = ( )( )𝒙 𝟐 = 𝒙 𝟐
𝟐 𝟐 𝟒
𝟏𝟓 −𝟕 −𝟑 −𝟓 −
𝟑
𝟒𝒙𝟐 ( 𝒙 𝟐 ) + 𝟏𝟐𝒙 ( 𝒙 𝟐 ) + 𝟑 (𝒙 𝟐 ) = 𝟎
𝟒 𝟐
𝟕 𝟑 𝟑
− − −
𝟏𝟓𝒙𝟐 𝒙 𝟐 − 𝟏𝟖𝒙 𝟐 + 𝟑𝒙 𝟐 =𝟎
𝟑 𝟑
− −
𝟏𝟓𝒙𝟐𝒙 𝟐 − 𝟏𝟓𝒙𝟐 𝒙 𝟐 =𝟎
𝟎=𝟎
𝟑
−
Thus, 𝒚(𝒙) = 𝒙 𝟐 is a solution to 𝟒𝒙𝟐 𝒚′′ + 𝟏𝟐𝒙𝒚′ + 𝟑𝒚 = 𝟎.
Example 4. Show that 𝒚′′′ − 𝟑𝒚′ + 𝟐𝒚 = 𝟎 is a solution to 𝒚 = 𝒆−𝟐𝒙.
Solution:
𝒚 = 𝒆−𝟐𝒙
𝒚′ = −𝟐𝒆−𝟐𝒙
𝒚′′ = 𝟒𝒆−𝟐𝒙
𝒚′′′ = −𝟖𝒆−𝟐𝒙
−𝟖𝒆−𝟐𝒙 − ൫−𝟐𝒆−𝟐𝒙൯ + 𝟐൫𝒆−𝟐𝒙 ൯ = 𝟎
−𝟖𝒆−𝟐𝒙 + 𝟔𝒆−𝟐𝒙 + 𝟐𝒆−𝟐𝒙 = 𝟎
−𝟖𝒆−𝟐𝒙 + 𝟖𝒆−𝟐𝒙 = 𝟎
𝟎=𝟎
Prepared by: Mr. Abdullah H. Edris Differential Equations
Hence, 𝒚′′′ − 𝟑𝒚′ + 𝟐𝒚 = 𝟎 is a solution to 𝒚 = 𝒆−𝟐𝒙.
𝒅𝟐 𝒙
Example 4. Show that + 𝒌𝟐 𝒙 = 𝟎 is a solution to 𝒙 = 𝒔𝒊𝒏𝒌𝒕.
𝒅𝒕𝟐
Solution:
𝒙 = 𝒔𝒊𝒏𝒌𝒕
𝒙′ = 𝒌𝒄𝒐𝒔𝒌𝒕
𝒙′′ = −𝒌𝟐𝒔𝒊𝒏𝒌𝒕
−𝒌𝟐 𝒔𝒊𝒏𝒌𝒕 + 𝒌𝟐 (𝒔𝒊𝒏𝒌𝒕) = 𝟎
−𝒌𝟐 𝒔𝒊𝒏𝒌𝒕 + 𝒌𝟐 𝒔𝒊𝒏𝒌𝒕 = 𝟎
𝟎=𝟎
𝒅𝟐 𝒙
Therefore, + 𝒌𝟐 𝒙 = 𝟎 is a solution to 𝒙 = 𝒔𝒊𝒏𝒌𝒕
𝒅𝒕𝟐
Exercise 2
Answer the following clearly. Write your answers on the space provided.
1. Show that 𝑦 = 2𝑒 −𝑥 + 𝑥𝑒 −𝑥 is a solution of 𝑦 ′′ + 2𝑦 ′ + 𝑦 = 0
2. Show that 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2𝑒 2𝑥 is a solution of 𝑦 ′′ − 3𝑦 ′ + 2𝑦 = 0
−𝑥
3. Show that 𝑥 2 + 𝑦 2 is a solution of 𝑦 ′ = .
𝑦
4. Show that 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2𝑥𝑒 𝑥 is a solution of 𝑦 ′′ − 3𝑦 ′ + 2𝑦 = 0
Prepared by: Mr. Abdullah H. Edris Differential Equations
