Module DQT203

IMK

Introduction toDIFFERENTIAL EQUATIONS

Institut Matematik KejuruteraanUniversiti Malaysia Perlis

DQT203

Content

Chapter RevisionDifferentiation and Integration ……………… 2

Chapter 1First Order Differential Equations ………………… 17

Chapter Overview and Learning Objectives …………………… 171.0 Introduction to Differential Equations ……………………… 181.1 Solution to Differential Equations ……………………… 20

1.1.1 Direct Integration ………… 201.1.2 Separable Equations ……… 211.1.3 Homogeneous Equations ………… 251.1.4 Linear Equations ………… 281.1.5 Exact Equations ………… 34

Chapter 2Second Order Linear Differential Equation ………… 42

Chapter Overview and Learning Objectives ………… 422.0 Introduction to Second Order Linear Differential

Equations … 432.1 Homogeneous Differential Equations ………… 43

2.1.1 Real and Different Roots to Auxiliary Equations … 46

2.1.2 Real and Equal Roots to Auxiliary Equations … 472.1.3 Complex Root to Auxiliary Equations … 48

2.1.4 Equation of the form of … 50

2.2 Non-Homogeneous Differential Equations … 522.2.1 Method of Undetermined Coefficient … 52

2.2.1.1

Case of a polynomial of degree n … 53

2.2.1.2

Case of … 57

2.2.1.3

Case of … 60

2.2.1.4

Case of a Linear Combination of 2.2.11- 2.2.3 … 61

2.3 Method of Variation of Parameters … 64

1

CHAPTER REVISION

Differentiation & Integration

Chapter Outline

1.0 Table Differentiation of Common Functions 1.2 Methods of Differentiation

1.2.1 Functions of a Function ( The Chain Rule)1.2.2 Application of The Chain Rule1.2.3 Differentiation of a Product1.2.4 Differentiation of a Quotient

1.3 Integration1.3.1 The Basic Standard Integrals1.3.2 Integration by Substitution1.3.3 Integration by Parts1.3.4 Integration by Partial Fractions

Chapter Overview and Learning Objectives

The aim for this chapter is to provide a short self assessment for students who would like to acquire a basic understanding of elementary differentiation and integration. It is also a prerequisite technique to students in order to understand and solve the differential equations problems. Thus, I urge every student who is weak at differentiation and integration to revise this chapter, and make sure you did all the tutorial questions provided at the end of the chapter.

After careful study of this chapter, you should be able to do the following:

1. Understand the concept of differentiation and integration.2. Able to differentiate and integrate any given function.

2

1.1 Table of Differentiation of Common Functions

constant 0

1

-

* k is a constant

Source: Lecture Notes EQT102 by Aishah Mohd Noor

3

Examples

Differentiate with respect to x

Solutions

4

1.2 Methods of Differentiation

1.2.1 Functions of a Function ( The Chain Rule)

If y is a function of u and u is a function of x, then

Another important result:

This rule is very important in differentiation. It is used in implicit differentiation, finding rates of changes, as well as parametric differentiation.

1.2.2 Application of The Chain Rule

Examples

Differentiate the following

a) b) c)

Solutions

a) Let , and y= cos u , - sin u

b) Let , and

c) Let , and

5

Some extended differentiation involves chain rule:

y (u is a function of x)

un nun-1u’

kun nkun-1u’

eu eu u’

au au ln (a) u’

ln (u) u’

sin (u) u’

cos (u) u’

tan (u) u’

sec (u) u’

cosec (u) u’

sinh (u) u’

cosh (u) u’

Properties of ln

a)

b)

c)


1.2.3 Differentiation of a Product

The product rule is used to differentiate the product of two functions. Given y = u.v, where u and v are functions of x, then

Examples

6

Differentiate, with respect to x,

Solutions

1.2.4 Differentiation of a Quotient

The quotient rule is used to differentiate functions in the form of

quotient such as where u and v are functions of x, then

Examples

Differentiate the following with respect to x:

Solutions

7

8

1.3 Integration

1.3.1 The Basic Standard Integrals

ln x + C

ex + C

tan x + C

cosh x + C

sinh x + C


9


1.3.2 Integration by Substitution

Example

Find

Solution

Let , or ; and =

Hence, = =

1.3.3 Integration by Parts

Why use parts?

The function cannot be solved using any integration methods (i.e.

standard integral, substitution, ) that you

encounter before. Usually used for functions involves product either log, exponent and

trigonometry. Neither of the term in the function is the derivative of the other.

Example:

How to solve?

Using a formula

10

How to determine u and v’?

Example

1.

2. kalau tiada log, jadikan u = (kuasa positif), v’=

3. ada log, jadikan

4. ada trigo, jadikan

5. tiada ,

* From the previous examples, a priority order to choose u is:-

Choose the u:-

Solve the following:-

1.3.4 Integration by Partial Fractions

Why use partial fractions?

The given function is in the form of fraction, but not in the form of

(i.e. the numerator is not the derivative of the denominator).

For such case, firstly express the fractions in terms of its partial fractions (a simpler algebraic fractions which we shall most likely be able to integrate separately without difficulty).

Example:

3

'3

2

2

xdxxv

xv

11

Wrong!

Choose v’ so that you can easily get v

How to solve?

Express the given function in terms of its partial fractions using the following rules:a) The numerator must be of lower degree that the denominator.

Otherwise, divide out by long division.

Example:

Long division:

b) Factorize the denominator into its prime factors. This is important, since the factors obtained determine the shape of the partial fractions.

Example:

Remember

c) A linear factor gives a partial fractions of the form

d) Factor give partial fractions

Example:

e) Factor give partial fractions

Example:

f) A quadratic factor gives a partial fraction

Next, how to determine A, B, C, D, ….

12

Evaluate the following.

13

TUTORIAL 1 Differentiation & Integration

A. DIFFERENTIATION

I. Find the derivative of each of the following

1. 2. 3. 4. 5. 6. 7. 8.

9. 10. 11. 12.

13. 14.

15. 16.

17. Find f’(2) when 18. Find f’(-1) when

19. 20. 21. 22. 23. 24 25. 26. 27. 28.

29. 30. Find for

31. 32.

33. 34.

35. 36.

14

15

B. INTEGRATION

I. Integrate the following

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25.

II. Find the partial fractions of each of the following (26 – 28)

26.

27.

28.

III. Integration by partial fractions

29.

30.

31.

32.

33.

Integration by parts

34.

35.

36.

37.

38.

39.

17

CHAPTER 1First Order

Differential Equations

Chapter Outline

1.0 Introduction to Differential Equations 1.1 Solution to Differential Equations

1.1.1 Direct Integration1.1.2 Separable Equations1.1.3 Homogeneous Equations1.1.4 Linear Equations1.1.5 Exact Equations


This chapter presents four methods of solving first order differential equations. We begin with the easiest method direct integration. In this part, students need to make sure that they are familiar with the integration techniques. It is crucially important to students because the integration methods will be used throughout the chapter. Students who master every integration techniques have the advantages to understand the topic well. Students who are weak at integration will encounter difficulties in problem solving. Thus, it is advisable for these students to do exercise so that problem in integration can be minimized. The following subchapter exposed students to methods of solving three types of differential equation. To make things easier, students are advised to get familiar with the types of differential equations and its solution techniques so that they will not have problems.


1. Clearly understand the types of first order differential equations.2. Able to identify all types of first order differential equations and

solve using appropriate method.

18

3. Clearly understand basic concepts of first order and its applications.

19

1.0 Introduction

Often in physic, engineering, and other technical areas, we need to search for an unknown function. In many cases, this search leads to an equation involving derivatives (or differentials) of the unknown function. Such equations involving derivatives (or differentials) are called differential equations.

In other words, a differential equation is a relationship between an independent variable, x, a dependent variable, y, and one or more differential coefficients of y with respect to x.

Example 1

The order of a differential equation is given by the highest derivative involved in the equation.

Function Order

1st order

2nd order

3rd order

2nd order

Exercise 1

State the order of the following functions:

Function Order

20

Reminder

A 1st order differential equation is derived from a function having 1 arbitrary constant.A 2nd order differential equation is derived from a function having 2 arbitrary constants.An nth order differential equation is derived from a function having n arbitrary constants.

Example 2

Function Order

is derived from a function having 1 arbitrary constantis derived from a function having 1 arbitrary constantis derived from a function having 2 arbitrary constants

21

1.1 Solution of Differential Equations

To solve a differential equation, we have to find the function for which the equation is true. This means that we have to manipulate the equation so as to eliminate all the differential coefficients and leave a relationship between y and x. The rest of this particular program is devoted to the various methods of solving first order differential equations.

1.1.1 Direct Integration

If the equation can be arranged in the form of , then the

equation can be solved by simple integration.

Example 1

Find the function y of

Solution

As always, of course, the constant of integration must be included. Here it provides the one arbitrary constant which we always get when solving a first order differential equation.

Example 2

Solve

Solution

Leave on the LHS,

Example 3

22

Find the particular solution of the equation , given that y=3

when x=0.

Solution

First re-write the equation in the form of

Then, Knowing that when y=3, x=0, we can evaluate c:

Thus,

1.1.2 Separable Equations

If the given equation is of the form , the variable y on

the RHS prevents solving by direct integration. We therefore have to devise some other method of solution.

Let us consider equations of the form . and of the

form , i.e. equations in which the RHS can be expressed as

products or quotients of functions of x or of y.

Example 1

Solve

Solution

We can re-write this as

Now, integrate both sides with respect to x

23

Example 2

Solve

Solution

Reminder

Separable equations:

Example 3

Solve

Solution

Exercise

1. Solve ans:

2. Solve ans:

24

3. Solve ans:

Revision Exercise

Find the general solution of the following equations:

25

TUTORIAL 2SEPARABLE EQUATIONS

I. Solve each differential equation.1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12. 13.

II. Find the particular solution of each differential equation subject to the given conditions.

14. 15.

16. 17.

18.

III. Solve the given differential equation by separation variables.

19. 20.

21. 22.

23. 24.

25. 26.

27.

IV. Solve the given initial value problem.

28. 29.

30.

26

1.1.3 Homogeneous Equations

Consider . This look simple enough, but we find that we

cannot express the RHS in the form of “x-factor”, so we cannot solve by the method of separating variables.

In this case, we make the substitution , where v is a function of x. So .

Differentiate with respect to x we get,

Solve using separable method:

Since , substitute

Factorize with respect to x

27

Example 2

Solve .

(Here, all the terms in the RHS are of degree 2, i.e. the equation is homogeneous.)

Solution

Exercise

1. Solve ans:

2. Solve ans:

Reminder

Homogeneous equations:

Revision Exercise

Solve the following:

28

TUTORIAL3HOMOGENEOUS EQUATIONS

I. Solve each differential equation.

1.

2.

3.

4.

5. given that y=3 when x=2.

6.

7.

8.

9.

10.

29

1.1.4 Linear Equations

Consider the equation

This clearly an equation of the first order, but different from those we have dealt with so far. In fact, none of our previous methods could be used to solve this one, so we have to find a further method to attack.

How do we know that an equation is a linear equation? An equation is said to be linear if it has the form of

Where a, b and c are constants or functions of x or a combination of both. To solve a linear equation, we need to transform any given equation to a general form of

Where and and leave the coefficient of .

Solution steps:

1. Find the integrating factor (pronounce as rho)

2. Multiply both sides with , we get

You will now see that the LHS is the differential coefficient of

Simplify (integrate both sides with respect to x), to get

3. Solve for . Finally solve for y.

30

Example 1

Identify P, Q and from

Solution

List down the following

Example 2

Solve

Solution

1. P=-1, Q=x, the integrating factor is

2. Multiply both sides with

we get

3. Integrate to solve for y. Integrating by part

Example 3

Solve

31

Solution

Rewrite to its general form

then solve

Example 4

Solve

Solution

Using substitution

32

Example 5

Solve

Solution

Rewrite your general form of linear equation

Solve for y

Example 6

Solve

Solution

Rewrite your general form of linear equation

33

Solve for y

Exercise

1. Solve ans:

2. Solve the equation given that y=10 when x=4.

ans:

Revision Exercise

Solve the following

1. ans:

2. ans:

3. ans:

34

TUTORIAL 4LINEAR EQUATIONS

I. Solve each differential equation.

1.

2.

3.

4.

5.

6. , y(0)=4

7. , y(0)=2

8. ,

9.

10.

11.

35

1.1.5 Exact Equations

Unlike homogeneous equations, many first order differential equations cannot be reduced to separable differential equations by a suitable substitution. For example, the first order differential equation

is neither separable nor can be reduced to a separable equation by an appropriate substitution. In this section, we deal with equations whose solution are given by formulas of the form .

A first order differential equation that can be written in the form

where

for some function is called an exact differential equation. In calculus, we know that the total differential of a function is

.

Therefore, the equation is exact if there exists a function such that is the total differential of

. Thus, we can prove that an equation is an exact equation if and only if

.

Solution of an exact differential equations:

1. Write down your M and N

and

2. Test for exactness

Yes, exact!

3. Let be the solution, and integrate M with respect to x.

36

4. Differentiate F with respect to y, and equate the result to N to find

5. Integrate with respect to y to find (There is no need to include an arbitrary constant)

6. The solution of the equation is

Example 1

Solve

Solution

1. Write down your M and N

2. Test for exactness

Yes, exact! Therefore, there exist a function such that

and

3. Let be the solution, and integrate M with respect to x.

4. Differentiate F with respect to y, and equate the result to N to find

37

5. Integrate with respect to y to find

6. The solution of the equation is

thus,

Thus, the solutions are given implicitly by .

Example 2

Solve

Solution

Rewrite the equation in the form

Yes, exact!

Find F

Differentiate F with respect to y, equate to N

38

Find

Thus, the solution is .

Example 3

Solve

Solution

Rewrite to which can be simplified to .

Keep on working the rest…

Yes? No?

Find F


Find

39


Example 4

Solve

Solution

Yes? No?

Find F


Find


Example 5

Solve

Solution

40

Yes? No?

Find F


Find


41

Exercise

1. Solve ans:

2. Solve ans:

42

TUTORIAL 5EXACT EQUATIONS

Solve each differential equation.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

43

CHAPTER 2Second Order Linear

Differential Equations

Chapter Outline

2.0 Introduction2.1 Homogeneous Differential Equations

2.1.1 Real and Different Roots to the Auxiliary Equation2.1.2 Real and Equal Roots to the Auxiliary Equation2.1.3 Complex Roots to the Auxiliary Equation

2.1.4 Equations of the form

2.2 Non-Homogeneous Differential Equations2.2.1 The Method of Undetermined Coefficient2.2.1.1 Case 2.2.1.2 Case 2.2.1.3 Case 2.2.1.4 Case A linear combination of i, ii, and iii.

2.3 Method of Variation of Parameters (Kaedah Ubahan Parameter)


This chapter deals with second order linear differential equations (SOLDE), as an extension from the previous chapter. Students will be exposed to two types of SOLDE namely homogeneous and non-homogeneous differential equations and its solution methods. It is very important for every student to clearly understand each type of equation and its characteristics so that the learning process is easy.


1. Clearly understand the types of second order linear differential equations.

2. Able to identify all types of second order linear differential equations and solve using appropriate method.

44

3. Clearly understand basic concepts of second order and its applications.

45

2.0 Introduction

In previous chapter, we have studied first-order differential equations. However, knowing only the first-order differential equations is not enough because many engineering problems such as mechanical vibrations, buoyancy, and electric circuits require the use of second-order differential equations, discussed in this chapter. A second-order differential equations can be written as

Where a, b, and c are constant coefficients and f(x) is a given function of x. To make our process of finding the solution simple, we classify our second-order differential equations into two types:

1. Homogeneous differential equations

2. Non-homogeneous differential equations

2.1 Homogeneous Differential Equations

A second-order differential equations are said to homogeneous if , that is,

……….. (1)

The solution of the equation is given by , where A and B are two arbitrary constants and and are the roots of the quadratic equation . This quadratic equation is called the auxiliary equation and is obtained directly from the equation (1), by writing

46

Example 1

Find the auxiliary equations from the given functions

Solution

Since the auxiliary equation is always a quadratic equation, the values (roots) of m can be determined in the usual way, i.e. factorizes.

Example 2

Find the roots of the auxiliary equations

47

Solution

Remarks: The types of solution we get for homogenous differential equation depends on the roots of the auxiliary equation. There are three (3) types of roots:

1. Real and different roots

2. Real and equal roots

3. Complex roots

48

2.1.1 Real and Different Roots to the Auxiliary Equation

Example 3

Find the solution for

Solution

Auxiliary equation:

Example 4

Solve

Solution

Auxiliary equation:

Example 5

Solve

Solution

Auxiliary equation:

Example 6

Solve

Solution

Auxiliary equation:

49

2.1.2 Real and Equal Roots to the Auxiliary Equation

Let us take . The auxiliary equation is: ,

. If and , then these would give the solution and their two terms would combine to give . But every second-order differential equation has two arbitrary constants, so there must be another term containing a second constant. In fact, it can be shown that also satisfy the equation, so that the complete general solution id of the form

, i.e. .

In general, if the auxiliary equation has real and equal roots, giving , the solution of the differential equation is .

Example 7

Solve

Solution

Auxiliary equation:

Example 8

Find the solution for

Solution

Auxiliary equation:

Example 9

Solve

Solution

Auxiliary equation:

50

2.1.3 Complex Roots to the Auxiliary Equation

Now let us see what we get when the roots of the auxiliary equation are complex. Suppose , i.e. and . Then, the solution would be of the form:

From our previous understanding on complex number, we know that:

Our solution above can therefore be written:

Thus, if , the solution can be written in the form:

For example, if , then

For example, if , then

51

Example 10

Solve

Solution

Auxiliary equation:

In this case,

Solution:

Example 11

Solve

Solution

Auxiliary equation:

In this case,

Solution:

52

2.1.4 Equations of the form

Let us now consider the special case of the equation

when ,

i.e

and this can be written as to cover the two cases when the

coefficient of y is positive or negative.

i) If

(This is like , when and )

ii) If

or

Example 12

Solve

Solution

Example 13

Solve

Solution

53

Example 14

Solve

Solution

Example 15

Solve

Solution

54

2.2 Non-Homogeneous Differential Equations

So far we have considered equations of the form for

the case . If , then giving and and the solution is in general .

In the non-homogeneous differential equation , the

substitution would make the left-hand side zero. Therefore, there must be a further term in the solution which will make the LHS equal to and zero. The complete solution which will therefore be of the form , where X is the extra function yet to be found.

is called the complementary function

(a function of x) is called the particular integral

Note that the complete general solution for non-homogeneous equation is given by

2.2.1 The Method of Undetermined Coefficient

At this stage, you have already know how to find the . Next, we are trying to find the . In order to ease our process of finding , lets look at different types of that you might encounter in your equations.

i.

Given

Hint: Look at the highest order of the given polynomial

55

ii. , where P and k are constants. The is given by

iii. , where P and k are constants

Given

Hint: The take the same form if either sine or cosine or both exist.

iv. A linear combination of i, ii, and iii.

Given

To find the particular integral, there are four steps involve:

1. Determine the type of the given 2. Determine the value of r (r=0,1,2)3. Find the and substitute them into the given differential

equation.4. Determine the values of the arbitrary constants by equating the

coefficient.

2.2.1.1Case

For this case, if the given is of the above form, then the is written as:

Where r will be 0, 1, 2 such that no term in the is a solution in .

56

Example 16

Solve

Solution

Find : Repeat what you have done in the previous chapter.

Find : Determine which case that you deal with.

Thus,

The solution is

57

Example 17

Solve .

Solution

58

Example 18

Solve .

Solution

59

2.2.1.2 Case



Example 19

Solve

Solution



Thus,

The solution is

Example 20

60

Solve given that when and

.

Solution



Thus,

The solution is

61

Example 21

Solve .

Solution

62

2.2.1.4 Case


Given

Hint: The take the same form if either sine or cosine or both exist.


Example

Given

Example 22

Solve .

Solution

Find :

Find :

63

Example 23

Solve

Solution

2.2.1.4 Case A linear combination of 2.2.1.1 - 2.2.1.3.

Given

For this question, you may have a combination of cases that we have discussed before. To make it easier, identify which case that you have in your f(x), and find your separately. Consider these examples.

Example 24

Solve .

Solution

64

65

Example 25

Solve .

Solution

Follow previous example to solve this question. Some hints below may help you.

66

2.3 Method of Variation of Parameters (Kaedah Ubahan Parameter)

We have seen that the method of undetermined coefficients is a simple procedure for determining a particular solution when the equation has constant coefficients and the nonhomogeneous term is of a special type. Next, we are going to look at a more general method, called variation of parameters, for finding a particular solution.

This method works for every function of , but it usually more difficult to apply in practice. Let’s consider a differential equation

. If and are the solution of and respectively.

Hence, the general solution of is

Where,

And Wronskian (W)

In a simple word, to solve

1. Determine a and f(x)

2. Determine and . That is the solutions for a homogeneous equation where f(x)=0

3. Find the Wronskian, W

4. Find u and v

5. Solve by using the equation

67

Example 26

Solve using method of variation of parameter .

Solution

1. a=1 and

2. and .

3. Wronskian, W

4. u and v

2y1y

68

5. Solve,

Example 27

Solve given that

Solution

69

Hence,

Initial value problem:

Solve by adding the two equations

70

71

UNIVERSITI MALAYSIA PERLISINSTITUT MATEMATIK KEJURUTERAAN

TUTORIAL 8VARIATION OF PARAMETER

(NON-HOMOGENEOUS EQUATIONS)

Solve each differential equation.

1.

2.

3.

4. ,

5. ,254 22

2xey

dx

dy

dx

yd

72

ANSWERTutorial

73

TUTORIAL 1

A. DIFFERENTIATION

1. 02. 3. 44. -35. 6. 7. 8. 9. 10. 11. 12.

13.

14.

15.

16.

17. -718. 1019. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

31.

32.

33.

34.

35.

36.

B. INTEGRATION

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

74

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31. @

32. @

33.

34. 35.

36.

37.

38.

39.

TUTORIAL 2

1.

2.

3.

4.

5. 6. 7. 8. 9.

10.

11.

12.

13.

14. 15. 16.

17.

18.

19.

20.

21.

22.

23. 24. 25.

26.

27.

28.

29.

30. FIY ..... find it yourself

TUTORIAL 3

1. 2.

3.

4. 5. 6.

7.

8. 9. 10.

TUTORIAL 4

1.

2.

3. xx ceey 3

2

1

4.

5.

6.

7.

8.

9.

10.

11.

TUTORIAL 5

1. Exact2. Exact3. Exact4. 5. 6.

7.

8. 9.

10.

11.

TUTORIAL 6

1. 2. 3.

4.

5. 6. 7. 8. 9.

10.

11. 12.

TUTORIAL 7

1. 2. 3.

4.

5. 6. 7. 8.

9.

10. 11.

12.

13.

14.

15.

TUTORIAL 8

1.

2.

3.

4.

5.

Documents

Module DQT203