COURSE OUTLINE
Department & Faculty :
Dept. of Mathematical Sciences, Faculty of Science Page : 1 of 2
Subject & Code: DIFFERENTIAL EQUATIONS 1 (SSCM 1703)
Total Lecture Hours: 42 hours
Semester: Semester I
Academic Session: 2014/2015
Prepared by:
Name: Pn. Halijah Osman
Signature:
Date: 4 Sept 2014
Certified by:
Name:
Signature:
Date:
Lecturers Tel Room No e-mail
Pn. Halijah Osman
Dr. Anati Ali
34379
34262
C15-320
C22-415
[email protected] [email protected]
Synopsis :
Basic Theory of Linear Differential Equations: Initial Value Problems (IVPs) and Boundary Value Problems (BVPs);
existence and uniqueness, interval of existence. Homogeneous Equations; superposition principle, linear
independence/dependence, Wronskian, general solution of homogeneous equations, general solution of nonhomogeneous
equations. Methods of Solution: reduction of order, method of undetermined coefficients, variation of parameters, Cauchy-
Euler equations. Application of Second Order Differential Equations: mechanical vibrations; damped and undamped free
and forced vibration. Electrical circuits with and without resistance. Laplace Transforms: definition and derivation,
exponential order. Properties of Laplace Transform; linearity, first shifting theorem, multiplication by tn, Laplace transforms of
of the derivatives. Laplace transforms of fundamental functions, the unit step functions (with second shifting property) and
Delta Dirac functions. Inverse Laplace transforms; completing the square, method of partial fractions, convolution theorem.
Solving IVPs and simultaneous system of first order equations. Homogeneous System of First Order ODEs with Constant
Coefficients: reduction of higher order ODEs to a system of first order ODEs. Solution of a system of first order equations (up
to 3x3 matrices) using the eigenvalue method.
Objectives:
On completing the course, students should be able to:
1. Comprehend some basic theories of linear differential equations including those on Laplace transforms and systems
of equations.
2. Determine the solution of second order linear differential equations using the methods of undetermined coefficients,
variation of parameters, reduction of order, and Cauchy-Euler equations.
3. Solve higher order linear equations using the methods of undetermined coefficients and variation of parameters.
4. Obtain Laplace and inverse Laplace transforms and solve initial value problems using the method of Laplace
transform.
5. Solve systems of first order ODEs using the Laplace transforms and eigenvalue methods.
6. Think critically during discussion, tests and in assignments.
Text:
1. W.E. Boyce and C. DiPrima, Elementary Differential Equations and Boundary Value Problems, 9th edition, John
Wiley and Sons; 2009.
References :
1. Nagle, Saff and Snider, Fundamentals Of Differential Equations 5th Edition, Addison Wesley Longman; 2000.
2. Edward and Penney, Elementary Differential Equations Fifth Edition, Pearson Prentice Hall; 2004.
3. Paul, Robert and Glen, Differential Equations Second Edition, Brooks/Cole Thomson Learning; 2002.
COURSE OUTLINE
Department & Faculty :
Dept. of Mathematical Sciences, Faculty of Science Page : 2 of 2
Subject & Code: DIFFERENTIAL EQUATIONS 1 (SSCM 1703)
Total Lecture Hours: 42 hours
Semester: Semester I
Academic Session: 2014/2015
Prepared by:
Name: Pn. Halijah Osman
Signature:
Date: 4 Sept 2014
Certified by:
Name:
Signature:
Date:
Teaching Methodology: Lectures and discussions
Weekly Schedule
Week Lecture Topics Notes
1
7/9/14-13/9/14
Basic Theory of Linear Differential Equations: Initial Value Problems (IVPs) and
Boundary Value Problems (BVPs); existence and uniqueness, interval of existence.
2
14/9/14-20/9/14
Homogeneous Equations; superposition principle, linear independence/dependence,
Wronskian, general solution of homogeneous equations. General solution of
nonhomogeneous equations.
Malaysia Day
16/9/14
3
21/9/14-27/9/14 Solution for Linear Second and Higher Order Equations with Constant
Coefficients: Characteristic method for homogeneous linear equations
4
28/9/14-4/10/14 Method of undetermined coefficients.
5
5/10/14-11/10/14 Solution for Linear Second and Higher Order Equations with Constant or Variable Coefficients: Method of variations of parameters.
Eid Al-Adha
5/10/14
6
12/10/14-18/10/14 Reduction of order method, Cauchy-Euler’s Equations.
Test 1
19/10/14-25/10/14
MID SEMESTER BREAK
Depaavali
23/10/14
1st Muharam 1436
25/10/14
7
26/10/14-1/11/14
Applications: mechanical vibrations - damped and undamped free and forced
vibration. Electrical circuits with and without resistance.
UTM 53rd
Convocation
1/11/14-4/11/14
8
2/11/14-8/11/14
Laplace Transforms: Definition and Laplace transforms for standard elementary
functions. Exponential order.
9
9/11/14-15/11/14 Linearity property. First shift theorem. Multiplication by tn.
10
16/11/14-22/11/14
Laplace transforms of unit step functions and Delta Dirac functions. Second Shift
theorem. Laplace transforms of the derivatives.
Birthday of DYMM
Sultan Johor
22/11/14
11
23/11/14-29/11/14
Inverse Laplace transforms; completing the square, method of partial fractions,
convolution theorem.
Hol Day of Almarhum
Sultan Johor
29/11/14
12
30/11/14-6/12/14 Volterra integral equations, Solving IVPs and system of first order equations.
Test 2
13
7/12/14-13/12/14
Homogeneous System of First Order ODEs with Constant Coefficients: Reduction
of higher order ODE to a system of first order ODEs. Solution of a 2x2 system of first
order equation using eigenvalue method.
14
14/12/14-20/12/14 Continue up to 3x3 systems (for distinct eigenvalues only)
15
21/12/14-27/12/14 REVISION WEEK
Christmas Day
25/12/14
16
28/12/14-17/1/14 FINAL EXAMINATION
Maulidur Rasul
3/1/14
Assessment:
Tests Content
Test I (20%) 1 hr 15 mins Lectures: Weeks 1-5
Test II (20%) I hr 15 mins Lectures: Weeks 6-11
Final Examination (50%) Lectures: Weeks 1-14
Assignments (10%) Lectures: Weeks 1-13
SSCM 1703
INTRODUCTION:
TECHNIQUES
FIRST ORDER DE (SEPARABLE AND LINEAR TYPE)
Differentiation Techniques
Rules Example
Product Rule
y is in this form: y = u(x)v(x) Formula:
Given y = x sin x, find . Solution: From equation, choose u = x and v = sin x
Find : 1
Find : cos
Substitute u, v, and in formula:
cos sin 1 = x cos x + sin x
Quotient Rule
y is in this form: y = Formula:
Given y = , find Solution: From equation, choose u = x and v = sin x
Find : 1
Find : cos
Substitute u, v, and in formula:
sin 1 cos
sin
=
Chain Rule
y is in this form: y = f(u), where u = g(x) Formula:
Given y = sin 2x, find . Solution: From equation, choose u = 2x Now the equation becomes y = sin u Find : cos Find : 2
Substitute and in formula:
cos 2
= 2 cos u = 2 cos 2x
Implicit Differentiation
- Is used when y is written implicitly as functions of x:
- Techniques: differentiate both sides of equation with respect to x:
If y5 – y = 5x, find . Solution:
5
5
5 5
5 1 5
55 1
Apply chain rule to find :
u = , = 5y4
Integration techniques
Techniques Examples
Substitution of u
Integration form:
or dx
sin or
When to use (condition): When derivative of f(x) is a constant multiplication of g(x). or
= cg(x), where c constant How to use? 1. Choose u=f(x), make sure it satisfy the above condition. 2. Differentiate u wrt x:
which implies g(x)dx = du
3. Replace f(x) with u and
g(x)dx with du 3. Integrate. 4. Replace back all u with corresponding x term.
Example 1: Solve sin 2 cos 2 , Solution: Let say f(x) = sin 2x, g(x) = cos 2x Let u = f(x): u = sin 2x
Find : = 2cos 2x hence g(x)dx = cos 2x dx = du Replace and integrate:
sin 2 cos 2 =
Replace back in x: sin 2 cos 2 dx = + A Example 2: Solve sin 3 5 . Let f(x) = 3x + 5, g(x) = 1 Let u = 3x + 5
3
dx = du
= c g(x), with c = 2
Replace all x with corresponding function in u and integrate:
sin 3 5 sin13
sin cos Replace u with corresponding function in x
sin 3 5 13
cos 3 5
By Parts
If , where u = y1 and dv = y2 dx. Formula:
Solve 2 sin Solution: Choose u = 2x and dv = sin x dx
Find : 2 du = 2 dx Find v: Integrate both sides of dv = sin x dx , gives v = -cos x Substitute u, v and du in the formula:
2 sin 2 cos cos 2
2 cos 2 cos = - 2x cos x – (- 2sin x + A) = - 2x cos x + 2 sin x + A
Tabular Method Solving using integration by parts also could be set up using table. Case 1 (one function, either y1 or y2 can becomes zero after differentiating a few times) Sign Differentiate Integrate
1
A1
B1
-1
1
1
1
1
1
.
. . .
.
. 0
Note: 1. Fill in the shaded box as follows:
- Sign column: Constant equal to 1 with alternating sign beginning with positive sign.
- A1: • choose either y1 or y2 • Must be a function that
can become zero after differentiating a few times.
- B1: • The other function that
not chosen as A1. 2. Fill in the non-shaded box following the formula written in the box.
3. Multiply following the arrows. Each arrow will have its own answer.
4. To write final answer for integration: sum answers from all arrows and add with a constant, c.
Example for Case 1: To solve 2 sin , 2x can be differentiated twice to be zero, fill 2x as A1. Choose sin x as B1. Sign Differentiate Integrate
1
2x
sin x
-1
2 = 2
-cos x
1 2
= 0
-sin x
Arrow 1: 1(2x)(-cos x) = -2x cos x (1) Arrow 2: (-1)(2)(-sin x) = 2 sin x (2) Final answer:
sin = (1) + (2) + c = -2x cos x + 2sin x + c
Case 2 (both y1 and y2 can’t be zero after differentiating a few times) Sign Differentiate Integrate
1
A1
B1
-1
1
1
1
1
1
Note: - Choose A1 and B1. - Steps 2 as in Case 1 are
applied. - In step 3, the same as done in
Case 1 is applied for the first two arrows.
- For the third arrow, multiply and put the integration sign.
- Sum up answers from all arrows and simplify to get final answer.
Example for Case 2: To solve e sin , Both and sin x not leads to zero when differentiated. Sign Differentiate Integrate
1
sin x
-1
=
-cos x
1 =
-sin x
Arrow 1: 1( )(-cos x) = - cos x (1) Arrow 2: (-1)( )(-sin x) = sin x (2) Arrow 3: 1 sin x = sin x (3) Final answer:
e sin = (1) + (2) + (3) = - cos x + sin x + sin x Look! The integration in the RHS is same as the one in LHS but multiplied with -1. We can group them on the LHS to be: 2 e sin = - cos x + sin x Therefore,
sin = (- cos x + sin x)
Partial Fraction
- Is used to transform a complicated functions to a simpler functions before doing integration.
Function Partial fraction Linear denominator:
Repeated denominator:
Quadratic denominator:
Determine the suitable partial
fraction based on its denominator.
Arrange the equation Find the value for constants A, B.
• To find A, eliminate B (to find B, eliminate A) by choosing the suitable value for x and substitute it in equation obtained in step 2.
Substitute A and B in partial fraction, and integrate.
Solve . Solution: We may write
Arrange: , gives
1 = A(x - 2) + B(x + 2) Find constants:- Eliminate B: substitute x = -2 in equation, we have: 1 = - 4A A = - 1/4 Eliminate A: substitute x = 2 in equation, we have: 1 = 4B B = 1/4
= - ln(x+2)/4 + ln(x+2)/4 + C
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
1
- Ordinary Differential Equations (ODE)
Contains one or more dependent variables
with respect to one independent variable
- Partial Differential Equations (PDE)
involve one or more dependent variables
and two or more independent variables
is the dependent variable
while is the independent
variable is the dependent variable
while is the independent
variable
Can you determine which one is the DEPENDENT VARIABLE and which
one is the INDEPENDENT VARIABLES from the following equations ???
Classification by type
Dependent Variable: w
Independent Variable: x, t
Dependent Variable: u
Independent Variable: x, y
Dependent Variable: u
Independent Variable: t
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
2
- Order of Differential Equation
Determined by the highest derivative
- Degree of Differential Equation
Exponent of the highest derivative
Examples:
a)
b)
c)
d)
- Linear Differential Equations
Dependent variables and their derivative are of degree 1
Each coefficient depends only on the independent variable
A DE is linear if it has the form
Examples:
1)
2)
3)
Order : 1 Degree: 2
Order : 2 Degree: 1
Order : 2 Degree: 1
Order : 3 Degree: 4
Classification by order / degree
Classification as linear / nonlinear
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
3
- Nonlinear Differential Equations
Dependent variables and their derivatives are not of degree 1
Examples:
1)
2)
3)
Initial conditions : will be given on specified given point
Boundary conditions : will be given on some points
Examples :
1) Initial condition
2) Boundary condition
Initial Value Problems (IVP)
Initial Conditions:
Boundary Value Problems (BVP)
Boundary Conditions:
Order : 1 Degree: 1
Order : 1 Degree: 2
Order : 3 Degree: 2
Initial & Boundary Value Problems
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
4
- General Solutions
Solution with arbitrary constant depending on the order
of the equation
- Particular Solutions
Solution that satisfies given boundary or initial conditions
Examples:
(1)
Show that the above equation is a solution of the following DE
(2)
Solutions:
(3)
(4)
Insert (1) and (4) into (2)
Proven that is the solution for the given DE.
EXERCISE:
Show that is the solution of the
following DE
Solution of a Differential Equation
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
5
Example 1:
Find the differential equation for
Solution:
( 1 )
( 2 )
Try to eliminate A by,
a) Divide (1) with :
( 3 )
b) (2)+(3) :
( 4 )
Example 2:
Form a suitable DE using
Solutions:
Forming a Differential Equation
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
6
Exercise:
Form a suitable Differential Equation using
Hints:
1. Since there are two constants in the general solution, has to
be differentiated twice.
2. Try to eliminate constant A and B.
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
7
1.2 First Order Ordinary Differential Equations (ODE)
Types of first order ODE
- Separable equation
- Homogenous equation
- Exact equation
- Linear equation
- Bernoulli Equation
1.2.1 Separable Equation
How to identify?
Suppose
Hence this become a SEPARABLE EQUATION if it can be written as
Method of Solution : integrate both sides of equation
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
8
Example 1:
Solve the initial value problem
Solution:
i) Separate the functions
ii) Integrate both sides
Answer:
iii) Use the initial condition given,
iv) Final answer
Note: Some DE may not appear separable initially
but through appropriate substitutions, the DE can be
separable.
Use your Calculus
knowledge to
solve this
problem !
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
9
Example 2:
Show that the DE can be reduced to a separable
equation by using substitution . Then obtain the solution
for the original DE.
Solutions:
i) Differentiate both sides of the substitution wrt
(1)
(2)
ii) Insert (2) and (1) into the DE
(3)
iii) Write (3) into separable form
iv) Integrate the separable equation
Final answer :
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
10
Exercise
1) Solve the following equations
a.
b.
c.
d.
2) Using substitution , convert
to a separable equation. Hence solve the original equation.
1.2.2 Homogenous Equation
How to identify?
Suppose , is homogenous if
for every real value of
Method of Solution :
i) Determine whether the equation homogenous or not
ii) Use substitution and in the original DE
iii) Separate the variable and
iv) Integrate both sides
v) Use initial condition (if given) to find the constant value
Separable
equation
method
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
11
Example 1:
Determine whether the DE is homogenous or not
a)
b)
Solutions:
a)
this differential equation is homogenous
b)
this differential equation is non-homogenous
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
12
Example 2:
Solve the homogenous equation
Solutions:
i) Rearrange the DE
(1)
ii) Test for homogeneity
this differential equation is homogenous
iii) Substitute and into (1)
iv) Solve the problem using the separable equation method
Final answer :
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
13
Example 3:
Find the solution for this non-homogenous equation
(1)
by using the following substitutions (2),(3)
Solutions: i) Differentiate (2) and (3)
and substitute them into (1),
ii) Test for homogeneity,
iii) Use the substitutions and
iv) Use the separable equation method to solve the problem
Ans:
Note:
Non-homogenous can be reduced to a homogenous
equation by using the right substitution.
REMEMBER!
Now we use
instead of
LASTLY, do not forget to
replace with
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
14
1.2.3 Exact Equation
How to identify?
Suppose ,
Therefore the first order DE is given by
Condition for an exact equation.
Method of Solution (Method 1):
i) Write the DE in the form
And test for the exactness
ii) If the DE is exact, then
(1), (2)
To find , integrate (1) wrt to get
(3)
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
15
iii) To determine , differentiate (3) wrt to get
iv) Integrate to get
v) Replace into (3). If there is any initial conditions given,
substitute the condition into the solution.
vi) Write down the solution in the form
, where A is a constant
Method of Solution (Method 2):
i) Write the DE in the form
And test for the exactness
ii) If the DE is exact, then
(1), (2)
iii) To find from , integrate (1) wrt to get
(3)
iv) To find from , integrate (2) wrt to get
(4)
v) Compare and to get value for and .
vii) Replace into (3) OR into (4).
viii) If there are any initial conditions given, substitute the conditions into
the solution.
ix) Write down the solution in the form
, where A is a constant
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
16
Example 1:
Solve
Solution (Method 1):
i) Check the exactness
Since , this equation is exact.
ii) Find
(1)
(2)
To find , integrate either (1) or (2), let’s say we take (1)
(3)
iii) Now we differentiate (3) wrt to compare with
(4)
Now, let’s compare (4) with (2)
iv) Find
v) Now that we found , our new should looks like this
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
17
vi) Write the solution in the form
, where is a constant
Exercise :
Try to solve Example 1 by using Method 2
Answer:
i) Check the exactness
Since , this equation is exact.
ii) Find
(1)
(2)
To find , integrate both (1) and (2),
(3)
(4)
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
18
iii) Compare to determine the value of and
Hence,
and
iv) Replace into OR into (4)
v) Write the solution in the form
Example 2:
Show that the following equation is not exact. By using integrating
factor, , solve the equation.
Solution:
i) Show that it is not exact
Since , this equation is not exact.
Note:
Some non-exact equation can be turned into exact
equation by multiplying it with an integrating factor.
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
19
ii) Multiply into the DE to make the equation exact
iii) Check the exactness again
Since , this equation is exact.
iv) Find
(1)
(2)
To find , integrate either (1) or (2), let’s say we take (2)
(3)
v) Find
(4)
Now, let’s compare (4) with (1)
vi) Write
, where
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
20
Exercises :
1. Try solving Example 2 by using method 2.
2. Determine whether the following equation is exact. If it is,
then solve it.
a.
b.
c.
3. Given the differential equation
i. Show that the differential equation is exact. Hence, solve
the differential equation by the method of exact equation.
ii. Show that the differential equation is homogeneous.
Hence, solve the differential equation by the method of
homogeneous equation. Check the answer with 3i.
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
21
1.2.4 Linear First Order Differential Equation
How to identify?
The general form of the first order linear DE is given by
When the above equation is divided by ,
( 1 )
Where and
Method of Solution :
i) Determine the value of dan such the the coefficient
of is 1.
ii) Calculate the integrating factor,
iii) Write the equation in the form of
iv) The general solution is given by
NOTE:
Must be here!!
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
22
Example 1:
Solve this first order DE
Solution:
i) Determine and
ii) Find integrating factor,
iii) Write down the equation
iv) Final answer
Note:
Non-linear DE can be converted into linear DE by
using the right substitution.
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
23
Example 2:
Using , convert the following non-linear DE into linear DE.
Solve the linear equation.
Solutions:
i) Differentiate to get and replace into the non-
linear equation.
ii) Change the equation into the general form of linear
equation & determine and
iii) Find the integrating factor,
iv) Find
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
24
Since ,
v) Use the initial condition given, .
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
25
1.2.5 Equations of the form
How to identify?
When the DE is in the form
( 1 )
use substitution
to turn the DE into a separable equation
( 2 )
Method of Solution :
i) Differentiate ( 2 ) wrt ( to get )
( 3 )
ii) Replace ( 3 ) into ( 1 )
iii) Solve using the separable equation solution
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
26
Example 1:
Solve
Solution: i) Write the equation as a function of
( 1 )
ii) Let and differentiate it to get
( 2 )
iii) Replace ( 2 ) into ( 1 )
iv) Solve using the separable equation solution
Substitution Method
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
27
Since ,
Since ,
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
28
1.2.6 Bernoulli Equation
How to identify?
The general form of the Bernoulli equation is given by
( 1 )
where
To reduce the equation to a linear equation, use substitution
( 2 )
Method of Solution :
iv) Divide ( 1 ) with
( 3 )
v) Differentiate ( 2 ) wrt ( to get )
( 4 )
vi) Replace ( 4 ) into ( 3 )
vii) Solve using the linear equation solution
Find integrating factor,
Solve
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
29
Example 1:
Solve
( 1 )
Solutions:
i) Determine
ii) Divide ( 1 ) with
( 2 )
iii) Using substitution,
( 3 )
iv) Replace ( 3 ) into ( 2 ) and write into linear equation form
v) Find the integrating factor
vi) Solve the problem
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
30
vii) Since
Exercise: Answer:
1.
2.
3.
4.
5.
6. Given the differential equation,
4 3 2 32 (4 ) 0.x ydy x y x dx
Show that the equation is exact. Hence solve it.
7. The equation in Question 6 can be rewritten as a Bernoulli equation,
12 4 .
dyx y
ydx
By using the substitution 2,z y solve this equation. Check the
answer with Question 6.
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
31
1.3 Applications of the First Order ODE
The Newton’s Law of Cooling is given by the following equation
Where is a constant of proportionality is the constant temperature of the surrounding medium
General Solution
Q1: Find the solution for
It is a separable equation. Therefore
Q2: Find when
Q3: Find . Given that
The Newton’s Law of Cooling
Do you know
what type of DE
is this?
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
32
Example 1:
According to Newton’s Law of Cooling, the rate of change of the temperature
satisfies
Where is the ambient temperature, is a constant and is time in minutes.
When object is placed in room with temperature C, it was found that the
temperature of the object drops from 9 C to C in 30 minutes. Then
determine the temperature of an object after 20 minutes.
Solution:
i) Determine all the information given.
Room temperature = C
When
When
Question: Temperature after 20 minutes,
ii) Find the solution for
iii) Use the conditions given to find and
When , C
When
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
33
iv)
Exercise: 1. A pitcher of buttermilk initially at 25⁰C is to be cooled
by setting it on the front porch, where the temperature is . Suppose that the temperature of the buttermilk has dropped to after 20 minutes. When will it be at
?
2. Just before midday the body of an apparent homicide victim is found in a room that is kept at a constant temperature of 70⁰F. At 12 noon, the temperature of the body is 80⁰F and at 1pm it is 75⁰F. Assume that the temperature of the body at the time of death was 98.6⁰F and that it has cooled in accord with Newton’s Law. What was the time of death?
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
34
The differential equation
Where is a constant
is the size of population / number of dollars / amount of radioactive The problems:
1. Population Growth 2. Compound Interest 3. Radioactive Decay 4. Drug Elimination
Example 1: A certain city had a population of 25000 in 1960 and a population of 30000 in 1970. Assume that its population will continue to grow exponentially at a constant rate. What populations can its city planners expect in the year 2000? Solution:
1) Extract the information
2) Solve the DE
Separate the equation and integrate
Natural Growth and Decay
Do you know
what type of DE
is this?
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
35
3) Use the initial & boundary conditions
In the year 2000, the population size is expected to be
Exercise: 1) (Compounded Interest) Upon the birth of their first child, a couple
deposited RM5000 in an account that pays 8% interest compounded continuously. The interest payments are allowed to accumulate. How much will the account contain on the child’s eighteenth birthday? (ANS: RM21103.48)
2) (Drug elimination) Suppose that sodium pentobarbitol is used to
anesthetize a dog. The dog is anesthetized when its bloodstream contains at least 45mg of sodium pentobarbitol per kg of the dog’s body weight. Suppose also that sodium pentobarbitol is eliminated exponentially from the dog’s bloodstream, with a half-life of 5 hours. What single dose should be administered in order to anesthetize a 50-kg dog for 1 hour? (ANS: 2585 mg)
3) (Half-life Radioactive Decay) A breeder reactor converts relatively stable uranium 238 into the isotope plutonium 239. After 15 years, it is determined that 0.043% of the initial amount of plutonium has disintegrated. Find the half-life of this isotope if the rate of disintegration is proportional to the amount remaining. (ANS: 24180 years)
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
36
Given that the DE for an RL-circuit is
Where
is the voltage source
is the inductance is the resistance
CASE 1 : (constant)
(1)
i) Write in the linear equation form
ii) Find the integrating factor,
iii) Multiply the DE with the integrating factor
iv) Integrate the equation to find
Electric Circuits - RC
Do you know
what type of DE
is this?
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
37
CASE 2 : or
Consider , the DE can be written as
i) Write into the linear equation form and determine and
ii) Find integrating factor,
iii) Multiply the DE with the integrating factor
iv) Integrate the equation to find
(1)
Tabular Method
Differentiate Sign Integrate
+
+
-
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
38
(2)
From (1)
(3)
Replace (3) into (2)
Exercise: 1) A 30-volt electromotive force is applied to an LR series circuit in which
the inductance is 0.1 henry and the resistance is 15 ohms. Find the curve if . Determine the current as .
2) An electromotive force
is applied to an LR series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current if .
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
39
The Newton’s Second Law of Motion is given by
Where is the external force is the mass of the body
is the velocity of the body with the same direction with is the time
Example 1:
A particle moves vertically under the force of gravity against air resistance ,
where is a constant. The velocity at any time is given by the differential
equation
If the particle starts off from rest, show that
Such that . Then find the velocity as the time approaches
infinity.
Solution:
i) Extract the information from the question
Initial Condition
ii) Separate the DE
Vertical Motion – Newton’s Second Law of Motion
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
40
Let ,
iii) Integrate the above equation
iv) Use the initial condition,
v) Rearrange the equation
Using Partial Fraction
CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSE1793
41
vi) Find the velocity as the time approaches infinity.
When , =>
1
1 Basic Theory of Linear Differential
Equations
1.1 Introduction
The fourth order DE has the following form:
( ) ( ) ( ) ( ) ( ) ( ) ( ) (1)
1.2 Existence and Uniqueness
Two items of great interest while solving a DE
Existence (“Does this problem have a solution?”)
Uniqueness (“Is the solution that is found the only solution to this problem?”)
Referring to Equation (1), if
( ) homogeneous equation
( ) non-homogeneous equation
( ) ( ), … ( ) are constants the equation is said to have constant coefficient
( ) ( ), … ( ) are variables the equation is said to have variable coefficient
2
Theorem 1.1 (Existence and Uniqueness Theorem)
Let ( ) ( ) ( ) and ( ) all be continuous
functions in an interval (a,b) containing the point and let
( ) be nonzero in (a,b). Then the initial value problem
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
(2)
has a unique solution for all x in (a,b).
1.3 Interval of Existence
Referring to Theorem 1.1, the interval of existence for
Equation (2) is (a,b) where the initial point is within (a,b).
How to find the interval of existence?
1. Divide the DE with the coefficient of the highest
derivative to be of this form:
( ) ( ) ( ) ( ) ( ) ( )
2. Find x values that makes ( ) ( ) ( ) and ( )
discontinuous.
3. Write the corresponding intervals.
4. If given the initial condition, choose its correct interval.
3
Example 1a, Tutorial 1
4
Example 2a, Tutorial 1
5
1.4 Linear Dependent and Linear Independent
Definition (Linear Dependent and Linear Independent)
A set of functions * + is linearly dependent on
the interval (a,b) if there exist a set of constants * +,
not all zero, such that for all in
(a,b).
If no such set of constants exists, the functions are linearly
independent on (a,b).
In other words, the functions are linearly independent if
implies that
. (or all constants are zero)
If given a set of function , we may check whether
these functions are linearly dependent or not:
1. Write .
2. Simplify the equation if possible.
3. Check whether all constants are zero or not (equate
coefficient)
4. If all constants are zero, linearly independent.
If not all constants are zero, linearly dependent, and
there will be a linear relation among them.
6
Example 1.4.1
Given and . Show
that they are linearly dependent.
Solution
Write ,
Hence, ( ) ( )
( ) ( ) ( )
Equate the coefficient, we obtain:
and are not zero. Therefore, they are linearly
dependent.
Its linear relation is .
Example 1.4.2
Given and . Show that they are linearly
independent.
Solution
Write
Hence,
( )
7
is never be zero, therefore
Equate the coefficients, we obtain:
and .
Therefore, they are linearly independent.
Example 1.4.3 (No. 4a, Tutorial 1)
8
1.5 Solutions of a DE
For a fourth order linear homogeneous DE
( ) ( ) ( ) ( ) ( ) ( ) (3)
there will be four solutions, ( ), ( ) ( ) and ( ).
Suppose that ( ), ( ),… ( ) are solutions to the above
DE and that ( ) , then the four solutions are
called a fundamental set of solutions.
Note: The number of solutions is depending on the order of
the DE.
1.5.1 Verifying a solution
Example 1.5.1
Given the differential equation
.
Show that is a solution to the above differential
equation.
How?
Find :
Find :
9
Substitute , and in the given differential equation
(DE):
Since the function satisfies the DE, it is verified that
it is a solution to the given DE.
1.5.2 Wronskian, ( )
The Wronskian for a set of functions * + is given
by the following determinant:
( ) |
|
1.5.3 Linear Combination of Solutions
Given the fourth order linear differential equations
( ) ( ) ( ) ( ) ( ) ( ) (4)
If ( ), ( ) ( ) and ( ) are solutions of the above
homogeneous linear DE, then any linear combination
( ) ( ) ( ) ( ) is also a solution.
10
Proof
Consider a second order linear homogeneous DE
( ) ( ) ( ) (5)
where and are its solutions. Since is a solution we
may substitute it in Eq. (5) and obtain
( ) ( ) ( ) . (6)
Since is a solution we similarly obtain
( ) ( ) ( ) . (7)
The linear combination of solutions for this case is
( ) ( ). Substitute it in Eq. (5) to verify whether it
is also a solution, gives
(
) (
) ( )
Rearrange will gives
(
) (
)
(
) (
)
= 0 = 0
(Eq. (6)) (Eq. (7))
Therefore, it is verified that ( ) ( ) is also a
solution.
11
1.5.4 General Solution
The general solution for a problem with more than one
solution could be written based on the linear combination of
solutions.
Example 1.5 (No. 5a, Tutorial 1)
For a DE with two solutions, and , its
general solution could be written as
For a DE with n solutions, , , … and ,
its general solution could be written as
12
1
2 Second Order and Higher Order Linear ODEs 2.1 Solving a Second Order Homogenous Linear ODE
How to find the solution of this equation:
2
20.
d y dya b cy
dx dx (i)
1) Assume the solution in the form of mty e . Thus, we have
22; ;
2
dy d ymt mt mty e me m edt dt
(ii)
2) The solution mty e must satisfy (i). Use (ii) in (i) gives the
characteristic (auxiliary) equation:
2 0am bm c (iii)
3) Find the roots 2
1 2
4,
2
b b acm m
a
For 2 4 0b ac roots are real and distinct (m1 m2)
For 2 4 0b ac roots are real and equal (m1 = m2)
For 2 4 0b ac roots are complex conjugate numbers
4) The solution of (i) depends on the type of roots.
2
Case 1: Roots are real and distinct, 21
mm ,
General solution: 1 2m x m xy Ae Be .
Example 2.2.1:
Solve '' 5 ' 6 0.y y y
Solution:
1) Use the characteristic equation: 2 5 6 0.m m
2) Find the roots : (m+3) (m+2) = 0 1m = -3; 2m = -2.
3) The general solution is 3 2x xy Ae Be
Example 2.2.2:
Solve 11 0.y y y
Solution:
1) Use the characteristic equation:
2 11 0.m m
2) Find the roots : 2
45
2
1m
1
1 45,
2m
2
1 45.
2m
3) The general solution is 1 2( ) .
m x m xy x Ae Be
( √
)
( √
)
3
Case 2: Roots are real and equal, 1 2m m
General solution: xm
eBxAy 1)( Example 2.2.3:
Solve 6 9 0y y y
Solution:
1) Use the characteristic equation:
0962 mm 2) Find the roots : 0)3( 2 m ⇒ 3,3 m
3) The general solution is 3( ) ( ) xy x A Bx e
Case 3: Roots are complex numbers; im
General solution: )sincos()( xBxAexy x
Example 2.2.4:
Solve 10 26 0.y y y
Solution:
1) Use the characteristic equation: 2 10 26 0.m m
2) Find the roots : 2
)26(410010 m
210 4 10 45
2 2
ii
5 ; 1 3) The general solution is
5( ) ( cos sin )xy x e A x B x
4
Example 2.2.5: Solve the initial value problem: 6 16 0y y y ;
(0) 1, (0) 0.y y
Solution:
1)Use the characteristic equation:
2 6 16 0.m m
2) Find the roots : 6 36 4(16) 6 28
2 2m
=2
283
2i = 3 7 .i
3)The general solution is
3( ) cos 7 sin 7ty t e A t B t
4) Using IC: i) y(0) = 1 ; A = 1 ;
ii) (0) 0y
' 3 37 sin 7 7 cos 7 cos 7 sin 7 ( 3 )t ty e A t B t A t B t e
0 7 3B A 37 B 3
.7
B
5) The solution is
3 3( ) cos 7 sin 7
7
ty t e t t
Initial Value Problem
These are the
initial conditions
5
1. 2 7 3 0y y y (Ans: /2 3
1 2( ) x xy x c e c e )
2. 2 0, (0) 5, (0) 3y y y y y (Ans: ( ) 5 2x xy x e xe )
3. 4 5 0y y y (Ans: 2( ) ( cos sin )xy x e A x B x )
2.2 Higher Order Homogeneous Linear ODE
Recall in second order homogeneous DE, three steps are required in
solving it;
Step 1 Write its characteristic equation
Step 2 Find the roots
Step 3 Write the solution based on the roots
These steps are also applied for higher order homogeneous DE, but
with a bit modification due to the following reasons:
2nd
order: quadratic characteristic equation, 2 roots
3rd
order: cubic characteristic equation, 3 roots
4th
order: quartic characteristic equation, 4 roots
Consider the following fourth order linear homogeneous DE:
+
+
+
+
The following steps are required to solve it:
Step 1: Write characteristic equation
+
+ + +
Step 2: Find its roots
Factorize (if possible).
or
Guess and long division
Exercise 2.2
6
Note:
For 3rd
order DE,
For 4th
order DE,
Step 3: Write the solution
Based on the roots, there are
4 types of solution for 3rd
order DE
9 types of solution for 4th
order DE
(See the given table below)
Guess one of
the roots, m1
Write its
factor, m – m1
Divide the
characteristic equation
(long division) with
m – m1
Find the roots
from the
answer.
Cubic
characteristic
equation
Quadratic
characteristic
equation
Long division
Roots Factorize/Formula
Quartic
characteristic
equation
Cubic
characteristic
equation
Long division Factorize / Long division
7
Solutions for Third order DE
Roots Solution
+ B + C
, gives , gives
= real number
= gives
gives
Remember
Case 2, 2nd
order?
8
Solutions for Fourth Order DE
Roots Solution
+ B + C +
, gives
, gives
+
, gives
gives
C +
C +
9
, gives
, gives
gives
gives
+
gives
= gives
10
(Two different complex roots)
= gives
= gives
(Two same complex roots)
Both and gives
2.1.2 Solving Nonhomogeneous Higher Order DE using Method of Undetermined
Coefficient.
Example 2.1.2a:
Find the general solution of ''' '' '3 3 .y y y y x+ + + =
Solution:
1. Find ( )hy x :
The characteristic is given by 3 23 3 1 0.m m m+ + + =
After Long Division, then we have
( )3 1 2 31 0 1.m m m m+ = ⇒ = = = −
Hence, the homogeneous equation is given by
( ) ( )2 .xhy x A Bx Cx e−= + +
2. Find ( )py x :
yp x( ) = Fx + E.
3. If Yes, do some modification on py by taking different values of r . Then compare again.
If No, proceed to find ' '' ''', & .p p py y y
( )( )( )
'
''
'''
0
0
p
p
p
y x F
y x
y x
=
=
=
Substitute ' '' ''', , &p p p py y y y into
''' '' '3 3p p p py y y y x+ + + =
and equate coefficient power x . Then, we have
33, 1
F E Fx xE F+ + =
⇒ = − =
So, the integral solution is given by
Note: Compare and . Any similar term? Yes/No?
yp x( ) = x −3.
4. Therefore, the general solution is
y x( ) = yh x( )+ yp x( )= A+ Bx +Cx2( )e−x + x −3
Example 2.1.2b:
Solve y 4( ) +8y '' +16y = 64sin2t.
Solution:
1. Find hy :
The characteristic equation have Two Same Complex Roots which is Case 9. Then,
( ) cos2 sin 2 cos2 sin 2 .hy t A t B t Ct t Dt t= + + +
2. Find py :
Subtitute 𝑦! and its derivatives into nonhomogeneous equation:
Compare the coefficients of the two sides:
Then, the particular solution is given by
3. Therefore, the general solutions of the nonhomogeneous equation are
Example 2.1.2c:
Solve the following differential equation
d 3ydx3
+ 4 d2 ydx2
−7 dydx−10y = 64e3t .
Solution:
1. Solve the corresponding homogeneous equation (see Example 2.1.1e), hence we have 5 2( ) x x x
hy x Ae Be Ce− −= + + .
2. Find ( )py x ,
3. The general solutions to the nonhomogeneous equation are;
Example 2.1.2d:
Determine the form of a particular solution of
y 4( ) + y ''' = −x2 + e−x .
Solution:
1. Find ( )hy x :
The characteristic is given by 4 3 0.m m+ =
Using Factorization, then we have
( )31 2 3 41 0 0, 1.m m m m m m+ = ⇒ = = = = −
Hence, the homogeneous equation is given by
( ) ( )2 0
2 .
x xh
x
y x A Bx Cx e De
A Bx Cx De
−
−
= + + +
= + + +
2. Find py :
After that, substitute into ( )4 ''' 2 .xp py y x e−+ = −
Then, we have
3. Therefore, particular solution is given by
The general solution is : __________________________________
Example 2.1.2e:
Solve the equation the following differential equation
d 3ydx3
+d 2 ydx2
= ex cos x .
Solution:
1. Find ( )hy x :
The characteristic is given by 3 2 0.m m+ =
Using Factorization, then we have
( )21 2 31 0 0, 1.m m m m m+ = ⇒ = = = −
Hence, the homogeneous equation is given by
( ) ( ) 0
.
x xh
x
y x A Bx e Ce
A Bx Ce
−
−
= + +
= + +
2. Find py :
r = 0 : yp = x0ex asin x +bcos x( ) = aex sin x +bex cos x .
Since there are no function in py that duplicate functions in the complementary solution, we proceed in the usual manner. Then, we have
( ) ( )
( ) ( )
( ) ( )
'
''
'''
cos sin ,
sin cos ,
2 cos 2 sin ,
2 2 sin 2 2 cos .
x xp
x xp
x xp
x xp
y ae x be x
y a b e x a b e x
y b e x a e x
y a b e x a b e x
= +
= − + + +
= −
= − + + − +
Product of Exponential and Trigonometry function
Compare to and this is correct!!
After that, substitute into ''' '' cos .xp py y e x+ =
Then, we have
( ) ( )4 2 sin 2 4 cos cos1 1, .10 5
x x xa b e x a b e x e x
a b
− − + − + =
⇒ = − =
Hence, particular solution is given by
( ) 1 1cos sin .10 5
x xpy x e x e x= − +
3. Therefore, the general equation is
( ) 1 1cos sin .10 5
x x xy x A Bx Ce e x e x−= + + − +
Exercise 2.1.2
1. Find the general solution of the differential equation
2. Solve the differential equation
3. Solve the differential equation
4. Solve the equation
5. Find the general solution of the following differential equation
with the initial conditions
6. Solve
Answer
1.
2.
3.
4.
5.
6.
Second order ODE : Applications 2.4.1 MECHANICAL SYSTEM
The governing equation is
( )my by ky f t
where
m = inertia b = damping k=stiffness
( )f t = force vibration.
Undamped; 0b Damped; 0b
Free vibration;
0)( tf
->Homogeneous
0my ky
0my by ky
Force vibration;0)( tf
->Non-
Homogeneous
( )my ky f t
( )my by ky f t
Do you know what
type of equations
are these?
y
Case 1: UNDAMPED FREE VIBRATIONS (When b = 0 ; f(t) = 0) The equation is given by:
02
2
kydt
ydm (i)
Divide by m; 2
20
d y ky
mdt
Rewrite,
22
20
d yy
dt where
m
k .
(ii) Step 1: Using characteristic equation:
2 2 0r r i
Step 2: General solution to (ii) is
tktkty sincos)( 21
We also can express y(t) in the more convenient form
)sin()( 2
2
2
1 tkkty
is an angular
frequency and
period =
is an
amplitude
The motion of a
mass in case 1 is
called simple
harmonic motion
y
Notes:
Given tktk sincos 21 , we can write this as
t
kk
kt
kk
kkk sincos
2
2
2
1
2
2
2
2
1
12
2
2
1
i.e
2 2 1 2
1 2 1 22 2 2 2
1 2 1 2
cos sin cos sink k
k t k t k k t tk k k k
where
1 2
2 2 2 2
1 2 1 2
sin , cosk k
k k k k
1
2 2
1 2
2
2 2
1 2
sintan
cos
k
k k
k
k k
1
2
arctank
k
tkk
ttkk
tktk
sin
sincoscossin
sincos
2
2
2
1
2
2
2
1
21
Notes: Alternative representation:
2
2
2
1
2
2
2
2
1
1 sin,coskk
k
kk
k
tkk
ttkk
tktk
cos
sinsincoscos
sincos
2
2
2
1
2
2
2
1
21
Case 2: DAMPED FREE VIBRATIONS (When b ≠ 0 ; f(t) = 0) The equation is given by:
02
2
kydt
dyb
dt
ydm (i)
Step 1: Characteristic equation for (i) is
02 kbrmr Its roots are
224 1
42 2 2
b b mk br b mk
m m m
Step 2: There are 3 cases related to these roots:
1) 2 4b mk Underdamped or Oscillatory Motion
The roots are i , where
m
b
2:
242
1: bmk
m
The general solution to (i) is
)sincos()( 21 tktkety t
can express in
)sin()( 2
2
2
1 tekkty t
2) 2 4b mk Overdamped Motion
The roots are
is
damping factor
Complex
roots
Two distinct
roots
1
2
br
m
,
2
2
14
2 2
br b mk
m m
The general solution to (i) is
1 2
1 2( ) r t r ty t k e k e
3) 2 4b mk Critical damped Motion
The roots are
1 2
2
br r r
m
The general solution to (i) is
1 2( ) ( ) rty t k k t e
Two repeated roots
Case 3: UNDAMPED FORCED VIBRATIONS (When b = 0 ; f(t) ≠ 0) Example 2.4.1 Solve the initial boundary problem for undamped mechanical system.
tFky
dt
ydm cos
2
2
(i)
000 yy where mk
Solution: Find homogeneous solution of (i). Rewrite (i) as
tm
Fy
m
k
dt
ydcos
2
2
(ii)
Set mk
mk ,2
Step 1: Using characteristic Equation:
imm 022
tBtAtyh sincos (iii)
Step 2: Find particular solution of (ii)
Try tDtCtyp sincos (iv) a
2sin cospy C t D t (iv) b
2 2cos sinpy C t D t (iv) c
Note: Since , system not at resonance.
To find const and D and C, substitute (iv) a,b,c into (ii). Equate
coefficients of tcos and tsin 2 2
0( cos sin ) ( cos sin ) cosm C t D t k C t D t F t
2 2
0( )cos ( )sin cosCm Ck t Dk Dm t F t
2 20 ,
FD C
m
2 2
cos sin cosF
y x A t B t tm
(v)
Find A & B using initial condition.
00 y
220
m
FA
2 2
FA
m
00 y , Find B?
B=0.
Final Answer:
2 2 2 2
cos cosF F
y t t tm m
Case 4: DAMPED FORCED VIBRATIONS (When b≠0, f(t)≠0)
The equation is given by:
2
2cos
d y dym b ky F t
dt dt
The general solution is
2
( /2 )
2 2 2
4sin sin( )
2 ( )
b m t Fmk by t Ae t t
m k m b
____________________________________________________________________________
__ Exercise 2.4.1: 1. A 4 pound weight is attached to a spring whose spring constant
is 16 lb/ft. what is the period of simple harmonic motion?
[ans : period = 2 k
m
2period
8 ]
2. A 24-pound weight, is attached to the end of a spring, stretches
it 4 inches. Find the equation of motion if the weight is released from rest from a point 3 inches above the equilibrium position.
[ans : 1
( ) cos4 64
y t t ]
3. A 10 - pound weight attached to a spring stretches it 2 feet. The
weight is attached to a dashpot damping device that offers a resistance numerically equal to ( 0) times the
instantaneous velocity. Determine the values of the damping constant so that the subsequent motion is a) overdamped;
b)critically damped; c) underdamped.
[ans : 5 5 5
) ; ) ; )02 2 2
a b c ]
4. A 16 - pound weight stretches a spring 8/3 feet. Initially the
weight starts from rest 2 feet below the equilibrium position and the subsequent motion takes place in a medium that offers a damping force numerically equal to ½ the instantaneous velocity. Find the equation of motion if the weight is driven by
an external force equal to ( ) 10cos3f t t .
[ans : 24 47 64 47 10
( ) cos sin cos3 sin33 2 2 33 47
t
y t e t t t t
]
5. The motion of a mass-spring system with damping governed
by: 2
216 0;
d yby y
dt
(0) 1, (0) 0.y y
Find the equation of motion and sketch its graph for b = 0, 6, 8 and 10.
6. Determine the equation of motion for an undamped system at
resonance governed by: 2
25cos ;
d yy t
dt
(0) 1, (0) 0.y y
Sketch the solution. 7. The response of an overdamped system to constant force is
governed by:
2
28 6 18;
d y dyy
dt dt
(0) 0, (0) 0.y y
Compute and sketch the displacement y(t). What is the limiting value of y(t) at t ? 2.4.2 RLC – Circuit
The equation for RLC circuit is given by:
;dI q
L RI E tdt c
with the initial conditions : IIqq 0,0
where .dq
Idt
Example 2.4.3:
The series RLC circuit has a voltage source given by
100VE t , a resistor 20R , an inductor 10HL , and a
capacitor 1
6260c
. If the initial current and the initial charge
on the capacitor are both zero, determine the current in the circuit for t>0.
Series RLC Circuit notations:
V - the voltage of the power source (measured in volts V)
I - the current in the circuit (measured in amperes A)
R - the resistance of the resistor (measured in ohms = V/A);
L - the inductance of the inductor (measured in henrys = H = V·s/A)
C - the capacitance of the capacitor (measured in farads = F = C/V =
A·s/V)
q - the charge across the capacitor (measured in coulombs C)
Solution:
The equation :
(i)dI q
L RI E tdt c
The initial conditions: 0 0 , 0 0q I
The information:
16260,100,20,10
ctERL
Method 1: Change the equation into homogeneous equation
Step 1: Differentiate eqn (i) with t, we have
tEdt
d
dt
dq
cdt
dIR
dt
IdL
12
2
tEdt
dI
cdt
dIR
dt
IdL
12
2
(ii)
In our case
16260,100,20,10
ctERL
Eqn. (ii) becomes
0626020102
2
Idt
dI
dt
Id
Or 2
22 626 0
d I dII
dt dt
(iii)
Step 2: Find I(t),
2 2 626 0m m
This is a homogenous equation. Can you remember how to solve this?? Use characteristic equation!!!
This is the general solution of
I(t). Use the initial conditions
given to find A and B.
im 251
cos25 sin 25tI t e A t B t
(iv) Step 3: Need to find A & B . Substitute the Initial condition given into (iv):
0 0 0I A
tBetI t 25sin (v)
What is B? How to find B?
From I.C. 0 0,q we have to find ? at 0.dI
tdt
From (i), at t = 0,
1000
0 c
qRI
dt
dIL
ot
10
100100
Ldt
dI
dt
dIL
otot
From (iv),
teteB
dt
dI tt 25cos2525sin
5
2
25
1025100 BB
Final answer:
2
sin 255
tI t e t
Method 2 : Solve the non-homogeneous equation
New initial
condition t = 0,
I = 0.
Step 1: 2
2We know , substitute to equation (i):
dq dI d qI
dt dt dt
2
2 (vi)
d q dq qL R E t
dt dt c
1
In our case 10 , 20 , 100 , 6260 ,L R E t c
equation (vi) become
2
210 20 6260 100
d q dqq
dt dt
Or
2
210 626 10
d q dqq
dt dt
Step 2:
Then find and h pq q
626
10,25sin25cos
p
t
h qtBtAeq
10
cos 25 sin 25626
tq t e A t B t
(vii) Step 3: i)Given initial charge is 0
0at0 tq
626
10
626
100 AA
ii)Initial current is 0
This is a nonhomogenous equation. Can you remember how to solve this?? You have to
find !!!
This is the general
solution of q(t). Use
the initial conditions
given to find A and B.
0at0 tdt
dqI
Differentiate (vii)
t
t
etBtA
tBtAedt
dq
25sin25cos
25sin2525cos25
tBAtBAedt
dq t 25sin2525sin25
(iii)
From I.C: 0a0 ttdt
dq
1505
1
3130
2
25625
10
25
250
AB
AB
The final answer :
tetI t 25sin
5
2
Exercise 2.4.2
1. An RLC series circuit has a voltage source given by E(t)=20 V , a resistor of 100 Ω, an inductor of 4 H and a capacitor of 0.01 F. If the initial current is zero and the initial charge on the capacitor is 4 C, determine the current in the circuit for t>0.
2. An RLC series circuit has a voltage source given by of E(t)=10cos 20t V , a resistor of 120 Ω, an inductor of 4 H and a capacitor of (2200)-1 F. Find the steady state current (solution) for this circuit.
3. An RLC series circuit has a voltage source given by of
E(t)=30sin 50t V , no resistor, an inductor of 2 H and a capacitor of 0.02 F. What is the current in this circuit for t>0 if at t =0, I(0)=q(0)=0?
2.1 Solving Homogeneous Higher Order DE
2.1.1 Solution of Homogeneous Equation for Third Order DE
Consider the following third order linear homogeneous differential equation:
3 2
3 2 1 03 20.
d y d y dya x a x a x a x y
dx dx dx
We can solve higher order homogeneous differential equations just like we solve second order
differential equations and the procedure is the same.
1. Write Characteristic Equation.
3 2
3 2 1 0 0.a m a m a m a
2. Finding roots:
Factorize Directly or
Guess and Long Division
Example 2.1.1a:
Solve ''' '' '5 22 56 0.y y y y
Solution:
1. Use the characteristic equation:
3 25 22 56 0m m m
Case 1: Roots are real and distinct, 1 2 3m m m ,
General Solution: 31 2 m xm x m xxy Ae Be Ce
Cubic
Characteristic
Equation
Quadratic
Characteristic
Equation
Roots
Long
division
Factorize/
Formula
2. Find the roots (factorize if possible)
1 2 34 2 7 0 4, 2, 7m m m m m m
3. We noticed that 1 2 3m m m , therefore the general solution is;
4 2 7x x xy x Ae Be Ce .
Example 2.1.1b:
Find the general solution of
Solution:
1. Use the characteristic equation:
3 23 3 1 0m m m
2. Use Long Division:
Guess one of the roots, 1 1m , then its factor is 1m .
3. Divide the characteristic equation (long division) with its factor, 1m ;
2
3 2
3 2
2
2
2 1
1 3 3 1
. 2 3
2 2
. 1
1
. .
m m
m m m m
m m
m m
m m
m
m
We get the quadratic equation, 2 2 1m m .
Case 2: Roots are real and equal, 1 2 3m m m ,
General Solution: 12 m xxy A Bx Cx e
''' '' '33 0y y y y
4. Find the roots for 2 2 1m m and we have;
22
3
1 2 3
1 2 1 0 1 1 0
1 0
1
m m m m m
m
m m m
5. Therefore, the general solution is
2 xxy A Bx Cx e .
Example 2.1.1c:
Solve the equation ''' '' '7 11 5 0.y y y y
Solution:
1. Use the characteristic equation:
3 27 11 5 0.m m m
2. Use Long Division:
Guess one of the roots, 1 1m , then its factor is 1m .
Case 3: Combination of real, distinct and equal roots, 1 2 3m m m ,
Note:
1
3
1 2 1
3 2
1 2
,
m x
m x
y
m m y A Bx e
m y Ce
y y
General Solution: 31 .m xm x
xy A Bx e Ce
3. Divide the characteristic equation (long division) with its factor, 1m ;
2
3 2
3 2
2
2
6 5
1 7 11 5
. 6 11
6 6
. 5 5
5 5
. .
m m
m m m m
m m
m m
m m
m
m
We get the quadratic equation, 2 6 5m m .
4. Find the roots for 2 6 5 0m m and we have;
22
1 2 3
1 6 5 0 1 5 0
1, 5
m m m m m
m m m
5. Therefore, the general solution is
5x xy x A Bx e Ce .
Example 2.1.1d:
Solve ''' '' '4 4 0y y y y
Solution:
1. Use the characteristic equation:
3 2 4 4 0.m m m
Case 4: Roots are distinct and complex numbers, 31 2, ,i mm m real number,
Note:
3
1 2 1
3 2
1 2
cos sin
,
x
m x
i
y
m m y e A x B x
m y Ce
y y
General Solution: 3cos sin m xxx Cey e A x B x
2. Use Long Division:
Guess one of the roots, 1 1m , then its factor is 1m .
3. Divide the characteristic equation (long division) with its factor, 1m ;
We get the quadratic equation, 2 4m .
4. Find the roots for 2 4 0m and we have;
2
1 2 3
1 4 0 1 2 2 0
1, 2 , 2
m m m m i m i
m m i m i
Where
1
xy Ae and 0
2 cos2 sin 2y e B x C x
5. Therefore, the general solution is
cos2 sin 2xy x Ae B x C x .
Example 2.1.1e:
Solve the initial value problem
''' '' ' ' ''4 7 10 0, 0 3, 0 12, 0 36y y y y y y y
Solution:
1. Use the characteristic equation:
3 24 7 10 0m m m
2
3 2
3 2
4
1 4 4
. . 4 4
4 4
. .
m
m m m m
m m
m
m
Initial Value Problem
2. Use Long Division:
Guess one of the roots, 1 1m , then its factor is 1m .
3. Divide the characteristic equation (long division) with its factor, 1m ;
4. Then, we have
3 2 2
1 2 3
4 7 10 0 3 10 1 0
5 2 1 0
5, 1, 2
m m m m m m
m m m
m m m
5. The general solution is
5 2( ) x x xy x Ae Be Ce
6. Using the initial conditions ' ''0 3, 0 12, 0 36y y y , yield
3
5 2 12
25 4 36
A B C
A B C
A B C
Solve the linear system for A, B, and C;
3 2, 5 2, 1A B C
7. Thus, the solution to the initial value problem is
5 23 5( )
2 2
x x xy x e e e .
2
3 2
3 2
2
2
3 10
1 4 7 10
. 3 7
3 3
. 10 10
10 10
. .
m m
m m m m
m m
m m
m m
m
m
2.1.2 Solution of Homogeneous Equation for Fourth Order DE
The same procedure is applied as before. In finding roots, we may factorize or use long division
as follows:
Example 2.1.2a:
Solve the following differential equation:
4 3 2
4 3 210 35 50 24 0
d y d y d y dyy
dx dx dx dx .
Solution:
1. The characteristic equation is
4 3 210 35 50 24 0m m m m .
Quartic
Characteristic
Equation
Cubic
Characteristic
Equation
Quadratic
Characteristic
Equation
Long
Division
Long
Division
Factorize/
Formula
Roots
Case 1: Roots are real and distinct, 1 2 3 4m m m m ,
General Solution: 431 2 m xm x m x m xx Dy Ae Be Ce e
2. Guess one of the roots, 1 1m , then its factor is 1m . Divide the characteristic equation
(long division) with its factor ;
3 2
4 3 2
4 3
3 2
3 2
2
2
9 26 24
1 10 35 50 24
. 9 35
9 9
. 26 50
26 26
. 24 24
24 24
. .
m m m
m m m m m
m m
m m
m m
m m
m m
m
m
We get the cubic equation, 3 29 26 24m m m .
3. Guess one of the roots, 2 2m , then its factor is 2m . Apply the long division again
and we have;
2
3 2
3 2
2
2
7 12
2 9 26 24
2
. 7 26
7 14
. 12 24
12 24
. .
m m
m m m m
m m
m m
m m
m
m
We get the quadratic equation, 2 7 12m m .
4. Then, we have
4 3 2 3 2
2
1 2 3 4
10 35 50 24 0 1 9 26 24 0
1 2 7 12 0
1 2 3 4 0
1, 2, 3, 4,
m m m m m m m m
m m m m
m m m m
m m m m
5. Therefore, the general solution is given by;
2 3 4 .x x x xy x Ae Be Ce De
Example 2.1.2b:
Find the general solution for the following differential equation:
4 3 2
4 3 24 6 4 0
d y d y d y dyy
dx dx dx dx .
Solution:
1. The characteristic equation is
4 3 24 6 4 1 0m m m m .
2. Guess one of the roots, 1 1m , then its factor is 1m . Divide the characteristic equation
(long division) with its factor ;
Case 2: Roots are real and equal, 1 2 3 4m m m m ,
General Solution: 12 3 m xx Dxy A Bx Cx e
3 2
4 3 2
4 3
3 2
3 2
2
2
3 3 1
1 4 6 4 1
. 3 6
3 3
. 3 4
3 3
. 1
1
. .
m m m
m m m m m
m m
m m
m m
m m
m m
m
m
We get the cubic equation, 3 23 3 1m m m .
3. Three other roots are from, 3 23 3 1m m m and guess on of the roots is 2 1m .
Use long division again:
2
3 2
3 2
2
2
2 1
1 3 3 1
. 2 3
2 2
. 1
1
. .
m m
m m m m
m m
m m
m m
m
m
4. Then, we have
4 3 2 3 2
2
2 2
1 2 3 4
4 6 4 1 0 1 3 3 1 0
1 1 2 1 0
1 1 0
1
m m m m m m m m
m m m m
m m
m m m m
5. Therefore, the general solution is given by
2 3 .xy x A Bx Cx Dx e
Example 2.1.2c:
Solve the following differential equation
4 2
4 218 81 0
d y d yy
dx dx .
Solution:
1. The characteristic equation is
4 218 81 0m m .
2. By factorization, then we have
4 2 2 2
1 2 3 4
18 81 0 9 9 0
3, 3
m m m m
m m m m
Then,
3
1 2 1
3 3
3 x
x x
m m y A Bx e
Ae Bxe
and
3
3 4 2
3 3
3 x
x x
m m y C Dx e
Ce Dxe
Case 3: 41 2 3, mm m m ,
Note:
1
34
1 2 1
3 2
1 2
m x
m xm
y
m m y A Bx e
m y C Dx e
y y
General Solution: 31 m xm xxy A Bx e C Dx e
3. Therefore, the general solution is given by
3 3 3 3
1 2 .x x x xy x y y Ae Bxe Ce Dxe
Example 2.1.2d:
Find the general solution for the following differential equation
4 2
4 20.
d y d y
dx dx
Solution:
1. The characteristic equation is
4 2 0m m .
2. Factorize the characteristic equation, then we have ;
4 2 2 2
2 2
1 2 3 4
0 1 0
0, 1 0
, 0, , 1
m m m m
m m
m m m m
3. Then,
0
1 2 1, 0 xm m y A Bx e
A Bx
Case 4: 41 2 3, mm m m ,
Note:
1
3 44
1 2 1
3 2
1 2
m x
m x m xm
y
m m y A Bx e
m y Ce De
y y
General Solution: 31 4m xm x m xxy A Bx e Ce De
and
3 4 2, 1 x xm m y Ce De
4. Therefore, the general solution is given by
x xy x A Bx Ce De .
Example 2.1.2e:
Solve the following differential equation.
4 ''' '' '2 11 18 4 8 0y y y y y .
Solution:
1. The characteristic equation is given by
4 3 22 11 18 4 8 0m m m m .
2. Use Long Division:
Guess one of the roots, 1 2m , then its factor is 2m .
3. Divide the characteristic equation (long division) with its factor 2m ;
Case 5: 41 2 3 mm m m ,
Note:
1
4
2
4
1 2 3 1
2
1 2
m x
m xm
y
m m m y A Bx Cx e
y De
y y
General Solution: 1 42 m x m xxy A Bx Cx e De
3 2
4 3 2
4 3
3 2
3 2
2
2
2 7 4 4
2 2 11 18 4 8
2 4
. 7 18
7 14
. 4 4
4 8
. 4 8
4 8
. .
m m m
m m m m m
m m
m m
m m
m m
m m
m
m
4. Three other roots are from, 3 22 7 4 4m m m and guess on of the roots is 2 2m .
Use long division again:
2
3 2
3 2
2
2
2 3 2
2 2 7 4 4
2 4
. 3 4
3 6
. 2 4
2 4
. .
m m
m m m m
m m
m m
m m
m
m
5. Then, we have
4 3 2 3 2
2 2
3
1 2 3 4
2 11 18 4 8 0 2 2 7 4 4 0
2 2 3 2 0
2 2 1 0
12,
2
m m m m m m m m
m m m
m m
m m m m
6. Therefore, the general solution is
1
2 2 2x
xy x A Bx Cx e De .
Example 2.1.2f:
Solve the equation, 4 ''' '2 0y y y .
Solution:
1. Use the characteristic equation; 4 3 2 0m m m .
2. Factor the left side and find the roots; 3 2 2m m m .
3. Note that, one of the roots of the cubic polynomial is 1 1m .Divide the characteristic
equation (long division) with its factor, 1m ;
4. Find the roots for 2 2 2m m and we have 3 4 1m m i . Thus, the characteristic
equation has four distinct roots, two which are complex;
1 2 3 40, 1, 1m m m m i
Case 6: 41 2 3, ,m im m m ,
Note:
1 2
4
1 2 1
3 2
1 2
, sin
m x m x
x
A B
m i Ccos x D x
y
m m y e e
m y e
y y
General Solution: 1 2 sinm x m x xx A B Ccos x D xy e e e
2
3 2
3 2
2
2
2 2
1 2
. 2 0
2 2
. 2 2
2 2
. .
m m
m m m
m m
m m
m m
m
m
5. Therefore, the general solution is
cos sinx xy x A Be e C x D x .
Example 2.1.2g:
Solve the following differential equation.
4 ''2 0y y .
Solution:
1. The characteristic equation is
4 22 0m m .
2. Factorize the characteristic equation, then we have ;
4 2 2 2
2 2
1 2 3 4
2 0 2 0
0, 2 0
, 0, , 2
m m m m
m m
m m m m i
3. Then,
0
1 2 1, 0 xm m y A Bx e
A Bx
Case 7: 41 2 3, ,m im m m ,
Note:
1
4
1 2 1
3 2
1 2
, sin
m x
xm i Ccos x D x
y
m m y A Bx e
m y e
y y
General Solution: 1 sinm x xx Ccos x D xy A Bx e e
and
0
3 4 2, 2 cos 2 sin 2
cos 2 sin 2
xm m i y e C x D x
C x D x
4. Therefore, the general solution is given by
cos 2 sin 2 .y x A Bx C x D x
Example 2.1.2h:
Solve the following differential equation.
416 0.y y
Solution:
1. The characteristic equation, 4 16 0m .
2. We can only find the roots by using De Moivre’s Theorem and the theorem is given by;
1 1
1
cos sin
2 2cos sin , 0,1,2,3,..., 1
n n
n
z r i
k kr i k n
n n
Case 8: Two different complex roots, 41 2 3, , ,i m im m m ,
Note:
2
4
1 1
3 2
1 2
, sin
, sin
x
x
m i Acos x B x
m i Ccos x D x
y
m y e
m y e
y y
General Solution:
sin sin .x xx Acos x B x Ccos x D xy e e
3. Therefore, the four roots of -16 can be found by evaluating the following;
1
4 4
4 24
16 16
16
2 cos sin , 0,1,2,34 2 4 2
k i
e
k i k k
Here are the four roots of -16:
1 10 : 2 cos sin 2 2 2
4 4 2 2
3 3 1 11: 2 cos sin 2 2 2
4 4 2 2
5 5 1 12 : 2 cos sin 2 2 2
4 4 2 2
7 73: 2 cos sin
4 4
k i i i
k i i i
k i i i
k i
1 12 2 2
2 2i i
So, we have two sets of complex roots:
1 2, 2 2m m i and 3 4, 2 2m m i .
4. Then, the general solution is
2
1 2 1
2
3 4 2
, 2 2 cos 2 sin 2
, 2 2 cos 2 sin 2
x
x
m m i y e A x B x
m m i y e C x D x
2 2
1 2
2 2 2 2
cos 2 sin 2 cos 2 sin 2
cos 2 sin 2 cos 2 sin 2 .
x x
x x x x
y x y y e A x B x e C x D x
Ae x Be x Ce x De x
Example 2.1.2i:
Find the general solution for the following differential equation
4 2
4 218 81 0.
d y d yy
dx dx
Solution:
1. The characteristic is;
4 218 81 0m m .
2. Find the roots by using the factorization;
2
4 2 218 81 0 9 0m m m
Then, we have
2
1 29 0 , 3m m m i
and
2
3 49 0 , 3m m m i
3. Both pairs gives
0
1 cos3 sin 3
cos3 sin 3
xy x e A x B x
A x B x
Case 9: Two same complex roots, 41 2 3, , ,m im m m ,
Note:
1
1
sinx Acos x B x
y C Dx
y e
y
General Solution: x C Dxy sinx Acos x B xe
4. Therefore, the general solution is given by
cos3 sin3 .y x C Dx A x B x
Exercise 2.1
1. Find the Third Order Homogeneous Linear Equation which has the particular solution 2 2( ) 2 5t t ty t e e te
2. Solve the following Initial Value Problem ''' '' ' ' '''4 5 0 (0) 4, (0) 7, (0) 23y y y y y y
3. Find the general solution to the differential equation 416 0y y
4. Solve the following differential equation 4 '' 2 0y y y
5. Solve 4 3 2
4 3 22 2 0
d y d y d y dy
dx dx dx dx
6. a) Find general solutions of 4 2
4 28 16 0.
d y d yy
dx dx
b) Solve
4 2
' '' '''
4 28 16 0, 0 1, 0 5, 0 8, 0 28.
d y d yy y y y y
dx dx
Answer
1. ''' ''3 4 0y y y
2. 55 2 t ty e e
3. 2 2 cos 2 sin 2x xy Ae Be C x D x
4. 2 2 cos sinx xy Ae Be C x D x
5. 2 cos sinxy A Be C x D x
6. a) cos 2 sin 2 cos 2 sin 2 .y x A t B t Ct t Dt t
b) cos 2 2sin 2 cos 2 3 sin 2 .y x t t t t t t