Sem3 de note

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


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


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


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


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


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.


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


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 !


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 :


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


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


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 :


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


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)


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


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


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


ii) Find

(1)

(2)

To find , integrate both (1) and (2),

(3)

(4)


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.


19

ii) Multiply into the DE to make the equation exact

iii) Check the exactness again


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


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.


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


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!!


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.


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


24

Since ,

v) Use the initial condition given, .


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 )


i) Differentiate ( 2 ) wrt ( to get )

( 3 )

ii) Replace ( 3 ) into ( 1 )

iii) Solve using the separable equation solution


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


27

Since ,

Since ,


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 )


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


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


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.


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?


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


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?


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?


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)


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?


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

+

+

-


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 .


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


40

Let ,

iii) Integrate the above equation

iv) Use the initial condition,

v) Rearrange the equation

Using Partial Fraction


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+ = ⇒ = = = = −


( ) ( )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+ = ⇒ = = = −


( ) ( ) 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


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


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:


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


2 xxy A Bx Cx e .

Example 2.1.1c:

Solve the equation ''' '' '7 11 5 0.y y y y

Solution:


3 27 11 5 0.m m m



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


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


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


5x xy x A Bx e Ce .

Example 2.1.1d:

Solve ''' '' '4 4 0y y y y

Solution:


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




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


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:


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




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


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:


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:


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 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:


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


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 .



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


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


cos sinx xy x A Be e C x D x .

Example 2.1.2g:


4 ''2 0y y .

Solution:


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


cos 2 sin 2 .y x A Bx C x D x

Example 2.1.2h:


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


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

Education

Sem3 de note