FlipWorkbook DUM30172

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

Example:

The table shows the time used to read a newspaper in a day by a group of Kimpalan students.

Time(minutes) 6 -10 11-15 16 - 20 21- 25 26 - 30

Number Of students 23 5 4 1

Find the variance of the distribution.

Solution: Midpoint, x f x2 fx fx2
Time 8 2 64 16 128
6 – 10 13 3 169 39 507
18 5 324 90 1620
11 - 15 23 4 529 92 2116
16 – 20 28 1 784 28 784
21 – 25 ∑f = 15 ∑fx = 265 ∑fx2 = 5155
26 - 30

Mean, x = 265
15

= 17.667 min

( )∑∑σ 2 =
fx 2 − x 2
f

= 5155 − (17.667)2
15

= 31.54 min

Example:

Mr. Zakuan recorded the time taken by each student to complete an additional mathematics

assignment as shown below.

Time (minutes) 30 - 34 35 – 39 40 – 44 45 – 49 50 - 54

Number of students 49 15 10 2

Find the standard deviation for this set of data.

Solution:

Time Midpoint, x f x2 fx fx2
30 – 34 32 4 1024 128 4096
1369 333 12321
35 – 39 37 9 1764 630 26460
2209 470 22090
40 – 44 42 15 104 5408
2704 ∑fx = 1665 ∑fx2 = 70735
45 – 49 47 10

50 - 54 52 2

∑f = 40

Mean, x = 1665
40

= 41.625

( )σ =∑ fx2 − x 2
∑f

36

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

= 70735 − (41.625)2
40

= 5.17 minutes

TUTORIAL 3.5

1. Determine the range, variance and standard deviation of the data below:
a. { 35, 22, 25, 23, 28, 33, 30}

b. (34.3, 25.0, 30.4, 34.6, 29.6, 28.7, 33.4, 32.7, 29.0. 31.3}

37

2. For the data given, find the DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY
Class interval
6-8 Frequency
9-11 3
2
12-14 6
15-17 4
18-20 1

a) Variance
b) Standard deviation
c) Range

38

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

3. The frequency distribution given below refers to the heights in centimeters on 100 people.

Heights (cm) Frequency
150-156 5
157-163 18
164-170 20
171-177 27
178-184 22
185-191 8

Determine:
a) Variance

b) Standard deviation

c) Range

39

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

4. The gain of 90 similar transistors is measured and the results are as shown

Transistors Frequency
83.5-85.5 6
86.5-88.5 39
89.5-91.5 27
92.5-94.5 15
95.5-97.5 3

Determine:
a) Variance

b) Standard deviation

c) Range

40

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

5. The frequency distribution for the values of resistance in ohms of 48 resistors is as shown.

Resistors Frequency
20.5-20.9 3
21.0-21.4 11
21.5-21.9 9
22.0-22.4 10
22.5-22.9 13
23.0-23.4 2

Calculate:
a) Variance

b) Standard deviation

c) Range

41

UNIT DUM 30172 ENGINEERING MATHEMATICS 3
LEARNING UNIT 3: STATISTICS AND PROBABILITY

After completing the unit, students should be able to:
 Processing data by appropriate tools for data in graphical form
 Processing data by appropriate tools for measure of Central
Tendency
 Processing Data by appropriate tools for Measure of Dispersion.

3.7 PROCESSING DATA BY APPROPRIATE TOOLS

3.7.1 Using MS Excel for Graphical Form
3.7.1.1 Bar Chart : Steps how to construct bar chart

Click Insert

Click Bar
Chart

Highlight the Click for adding titles,
data axis and label

Click for the
type of bar

chart

42

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

Figure 1: Example of the bar chart
3.7.1.2 Pie Chart : Steps how to construct pie chart

Click Insert

Click Pie chart

Highlight the
data

43

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

Click for the
type of pie

chart

Click for adding titles,
axis and label

44

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

Figure 2: Example of the pie chart
3.7.1.3 Histogram : Steps how to construct histogram

Click Insert

Click Bar chart

Highlight the
data

45

Click chart DUM 30172 ENGINEERING MATHEMATICS 3
design UNIT 3: STATISTICS AND PROBABILITY

Click bars

Click layout 8 for histogram
(no gap between bars)

Click solid line to
add lines at bars

Figure 3: Example of the histogram
46

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

3.7.1.4 Ogive : Steps how to construct ogive

Add “upper limit” for the
class before first class
with frequency “0” .

Create a table that Formula for cumulative
consist upper limit frequency

and cumulative Select the “scatter (X,Y)
frequency or bubble chart

Click insert

Select the “scatter with
straight lines and
markers

47

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

Modify the horizontal axis:
 Navigate the Axis option.
 Change the minimum

bounds with the lowest
value in boundary (80).
 Change the maximum
bounds with the highest
value in boundary (100).

Modify the vertical axis:
 Navigate the Axis option.
 Change the minimum

bounds with the lowest
value in boundary (0).
 Change the maximum
bounds with the highest
value in boundary (90).

Select “add data labels”
Select “add chart element” to
modify the chart titles and axis.

48

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

Figure 4: Example of the ogive
Note: Histograms, frequency polygons and ogive also can be constructed using the data analysis in MS
Excel.
3.7.2 Using MS Excel for Measure of Central Tendency

 Mean

Value of mean

49

 Median DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

Value of median

 Mode

Value of mode

50

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

3.7.3 Using MS Excel for Measure of Dispersion
 Variance

Value of variance

 Standard deviation

Value of standard
deviation

51

DUM 30172 ENGINEERING MATHEMATICS 3
UNIT 3: STATISTICS AND PROBABILITY

TUTORIAL 3.7

1. By using Microsoft Excel, present this data on a bar chart and pie chart

Volume (ml) Temperature (C)
100 20
150 35
200 15
250 30
300 28
350 13

2. The time taken in hours to the failure of 50 specimens of a metal subjected to fatigue failure tests are as
shown.

28 22 23 20 12 24 37 28 21 25
21 14 30 23 27 13 23 7 26 19
24 22 26 3 21 24 28 40 27 24
20 25 23 26 47 21 29 26 22 33
27 9 13 35 20 16 20 25 18 22

By using Microsoft Excel,
a. Construct a histogram depicting the data
b. Construct an ogive for the data
c. Find the mean, median and mode for the data
d. Find the variance and standard deviation.

3. The diameter in millimeters of a reel of wire is measured in 48 places and the results are as shown.
2.10 2.29 2.32 2.21 2.14 2.22 2.28 2.18 2.17 2.20

2.23 2.13 2.26 2.10 2.21 2.17 2.28 2.15 2.16 2.25

2.23 2.11 2.27 2.34 2.24 2.05 2.29 2.18 2.24 2.16

2.15 2.22 2.14 2.27 2.09 2.21 2.11 2.17 2.22 2.19

2.12 2.20 2.23 2.07 2.13 2.26 2.16 2.12

By using Microsoft Excel,
a. Construct a histogram depicting the data
b. Construct an ogive for the data
c. Find the mean, median and mode for the data
d. Find the variance and standard deviation.

52

UNIT 4 DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

ORDINARY DIFFERENTIAL

EQUATIONS

InINTRODUCTION

An equation containing the derivatives or differentials of one or more dependent variables,
with respect to one or more than one independent variables, is called a differential equation
(DE). Differential equations can be classified as ordinary differential equations (ODE) or partial
differential equations (PDE).

Differential equations arise in many engineering problems and other applications. Some
examples are given below:

water level Resistor, R

y
E Inductor, L

Water flowing out, dh = −k h Falling stone, y′′ = g = Current i in RL circuit,
dt
constant Li′ + Ri = E

CONTENT
4.1 Introduction to ODE
4.2 First Order Differential Equations
4.3 Second Order Differential Equations

1

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

4.1 INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS (ODE)

After completing the unit, students should be able to:
• Determine the degree and order of differential equations.

UNIT LEARNING
OUTCOME

WWhat is diferential equation?

A differential equation is relationship between an independent variable x,
a dependent variable y and one or more derivatives of y with respect to x.

Here are some examples of ODE:

y is dependent variable
a)

First derivative = 2

x is independent variable

b)+2 y is dependent variable

2 + 2 + First derivative
2
10 = sin 2

x is independent variable

Second derivative

Notes: (Differential Notations)
, 2 2 ,…………..
⇒Leibnitz’s notation

′, ̇ ,………………. ( ) ⇒ Prime notation
″, ̈ ,……………….. ( ) ⇒Newton’s dot notation
EXERCISE 4.1 (a)

For each of the following equations, state whether the equation is differential equation or not.

(a) = 3 + 1 (b) = 3 + 2 2

(c) � 2 2 �2 + = 3 (d) =


(e) ′( ) = + (f) 2 + 4 =
2

Ans: DE, not , DE, DE, DE, DE

2

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

What is ORDER?
The order of a differential equation is the highest derivatives in the differential equation.

What is DEGREE?

The degree of a differential equation is the power of its highest derivative.

Here are some examples of ODE with their order and degree:

a) Power = 2
= + 1
First derivative � �2 +

ODE with first order and second degree

b) Power = 1
� �3
Second derivative 2 + 2 − 15 = 0
2

ODE with second order and first degree

EXERCISE 4.1 (b)

For each of differential equations, state the order, degree, dependent variable and independent variable.

Differential Equation Order Degree Dependent Independent
Variable Variable

a) = 2


b) = cos + ( )5


c) + 2 = 3 2


) 2 + 2 = 5
2

) � 2 2 �3 − = 3


) 3 − � �4 + = 0
3

) ( ′′′)4 − 2( ′′)5 + 4 =

ℎ) � 4 4 �2 − 1 = 3 � �3

Ans: a) 1,1, , b) 1,1, , c) 1,1, , d) 2,1, , e) 2,3, , f) 3,1, , g) 3,4, , h) 4,2, ,

3

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

TUTORIAL 4.1

Determine the order and degree for each equation. ORDER DEGREE
EQUATION

1. 2 ′ + ( ′′)3 + =

2. ′′′ − ′ − 5 =

3. 2 + = 0
2

4. 4 (5) − ′′ + 6 = 0

5. � 4 4 �2 + 2 = 0

6. ʺ + ʹ =

7. � ʺ�2 + ‴ − = 1

8. � �2 = 3 −
3

9. � 4 4 �3 + 2 + � �5 = 0
2

10. 2 + − = 0
2

11. 3 + 4 � �2 = 2 +
3 2

12. 2 2 + ( ) − =
2

13. � 2 2 �3 + =


14. + = ; , , are constants


Ans: 1) 2,3, , 2) 3,1, , 3) 2,1, , 4) 5, 1, , 5) 4,2, , 6) 2, 1, , 7) 3,1, , 4
8) 3,1, , 9) 4,3, , 10) 2,1, , 11) 3,1, , 12) 2,1, , 13) 2,3, , 14) 1,1, ,

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

4.2 FIRST ORDER DIFFERENTIAL EQUATIONS

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
• Solve first order differential equations with separable variables.
• Solve first order differential equations by using integrating factor.

How do we solve first order differential equations?
There are two methods which can be used to solve first order differential equations.

1. Separation of Variables
2. Integrating Factor

SEPARATION OF VARIABLES

These are differential equations that can be written in the form

= ( ) ( )


with ( ) is a function of only while ( ) is a function of only.

Solution is obtained by separating the variables and performing direct integration

� = � ( )
( )

INTEGRATING FACTOR

Step 1 + ( ) = ( ) − − − − − − − − − −①

General form:

Step 2

Find the integrating factor: = ( ) = e∫ ( )

Step 3

Then, multiply the to the equation ①

e∫ ( ) . + e∫ ( ) . ( ) = e∫ ( ) . ( )


�e∫ ( ) � = e∫ ( ) . ( ) [LHS: Product Rule)

�e∫ ( ) � = ∫ e∫ ( ) . ( )

Therefore, the general solution is

. = � . ( )

5

EXAMPLE Separation of Variables Method DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

1. Solve the following differential equations by separating the variables.

a) Notes: INTEGRATION FORMULA

=

Solution: � = +

= � = +1 +
+ 1
� = �
1
∴The general solution, � = − +

2 = 2 + � = 1 +
2 2

is the constant of integration. � = 1 +


b) � 2 = 1 +
2 − 3
= 2 + 1
� = 1 +


Solution: � 1 = +


� 2 + 1� = (2 − 3) � 1 = | + | +
�� 2 + 1� = �(2 − 3) +
∴ 3 + = 2 − 3 +
� 1 = 1 | + | +
3 +

�( + ) = ( + ) +1 +
( + 1)

c)

= +

Solution :

=

1
=

� − = �

∴ − − = +

6

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

d)

= +

Solution:

= (1 + )

� = � 1 +


� = � �1 + 1�

∴ = + +

2. Determine the particular solution of = −4 , given that = 0 when = 1 .

Solution:

= −4


� = � −4

= −4 2 +
2

= 0 when = 1:

0 = −4(1)2 +
2

= 2

∴The particular solution is = −2 2 + 2

3. Given a differential equation.

= 2

a) Find the general solutions of the equation.
b) Determine the particular solution given � 4 � = 2.
Solution:

= 2

� 1 = � 2

1
∴ General solutions : = − 2 cos 2 +

Given � 4 � = 2: 1 � 4 �
2
2 = − cos 2 +

= 2 + 1
2
1 1
∴ Particular solution : = − 2 cos 2 + 2 + 2

7

EXERCISE 4.2 (a) DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

1. Solve the following differential equations by separating the variables.

a) 4 b)
Solution: = Solution: =

2 Ans:
2
= 2 2 + Ans:
= +

c) −3 d) 2
Solution: = ( + 5) = 2

Solution:

Ans: Ans:

2 + 5 = − 3 2 + − −1 = 1 +
2 2 2

e) = f) =
Solution: Solution:

Ans: 2 Ans:
= + 2
= − cos +

8

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

2. Solve the following differential equations by separating the variables.

a) = 2 b) = 4
Solution: Solution:

Ans: Ans:
= 2 + = 2 2 +

c) = 1 d) = 2 − 4 3
Solution: 2
Solution:

Ans: 4 Ans:
3
3 = − − + = 2 − 3 +
3

e) 2 3 − 1 f)
Solution: = 3 − 2
= ( + 3)( + 1)

Solution:

4 Ans: Ans:
2
3 − 2 = − + | + 3| = 2 + +
2

9

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

3. Solve the following differential equations by separating the variables.

a) b)

= 3 + 2 ( − ) = 4

Solution: Solution:

3 Ans: Ans:
2
= 2 + 2 + 1 − 1 = +
4 4

c) 2 + 2 d) = −2
Solution: = 2 − 2 Solution:

Ans: 1 Ans:
+ −1 = − −1 + + 2
− − = − −2 +

10

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

4. Determine the particular solution of = −3 − 4 , given that = 0 when = 1.


Solution:

Notes:

General Solution
A solution of a differential
equation from which all
solutions can be derived by
substituting values for
arbitrary constants is called
the general solution of the
equation.

Particular Solution
A condition that specifies the
value of the unknown function
y(x) at the some point x=xo is

called an initial condition.

Ans:
= −3 − 2 2 + 5

5. Find the particular solution of the following equation.

= 2 , (0) = 4


Solution:

1 1 Ans:
2 2
ln = + ln 4

11

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

6. Find the particular solution of the following equation.

= 5 , (0) = 0


Solution:

2 Ans:
2
− 1 −5 = − 1
5 5

7. Solve the following differential equation subject to the given initial condition.

cos ( ) = 2
= ,

Solution:

2 Ans:
2
= sin + 2

12

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

EXAMPLE Integrating Factor Method

1. Solve the following differential equations by finding the integrating factor.

a) Notes:
Integrating Factor Method
+ 3 = e

Solution:

General form,

+ = + ( ) = ( )
+ ( ) =
( )

( ) = Integrating factor,
= ( ) = e∫ ( )
Find the integrating factor,

= ( ) = e∫ =

∴The general solution, General solution is
. = � . � � . = � . ( )

e3 = � e +3

e3 = � e4

∴ e3 = 1 e4 + c
4

b)

− = 3 + 3 2 − 2

Solution:

Rewrite, − = + − Let eln −1 =
This is in eln −1 =
ln −1 ln =
the form + ( ) = ( ) ln −1(1) =
ln −1 =
Find the integrating factor, ∴ −1 =
= ( ) = e∫ − = e−ln = eln −1 = [ln = ]
−1 =

∴The general solution,
� �
. = �� + − �.

= � � + 3 − 2 �

2
∴ = 2 + 3 − 2 +

13

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

2. Solve the differential equation

+ = 3

Given that = 3 when = 2.

Solution:

+ =
+
( ) ( ) = ( )

Let eln =
= eln =
ln =
Find the integrating factor, = ( ) = e∫ = eln = ln (1) =
∴ =
The general solution,
= � 3.

= � 4

= 5 +
5
Given = 3 when = 2,
(2)5
3(2) = 5 +

6 = 32 +
5

∴ = − 2
5

∴The particular solution is
= 5 − 2
55

14

EXERCISE 4.2 (b) DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

1. Find the general solutions of the following differential equations.

a) + 3 = 3 b) − 2 = 10
Solution: Solution:

Ans: 3 = 3 + Ans: −2 = −5 −2 +

c) + = 2 d) − = 3 + 5
Solution: Solution:

3 Ans: Ans:
3
= + −1 = 2 + 5 +
2

15

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

e) + = 2 f) − 3 =
Solution: Solution:

Ans: Ans:

e = 1 e3 + e−3 = − 1 e−2 +
3 2

g) 2 + 2 = 1 h) + 2 = −4
Solution: Solution:

Ans: 1 Ans:
2 = + 2
2 = − −2 +

16

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

2. Solve the following differential equation subject to the given initial condition.

+ 2 = 5, (0) = 2

Solution:

5 Ans:
2 1
2 = 2 − 2

3. Find the particular solution of the differential equation

+ 6 = 3

which satisfies the condition = 0 when = 0

Solution:

1 Ans:
9 1
6 = 9 − 9

17

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

4. Given a differential equation.

2 + = 1

a) Find the general solutions of the equation.
b) Determine the particular solution when (1) = 2

Solution:

Ans:
a) = +
b) = + 2

5. Given differential equation

+ = 4
2

a) Determine the integrating factor.
b) Solve the differential equation, given that (1) = 2.
Solution:

a) = 2 Ans:

b) 2 = 1 6 + 5
3 3

18

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

TUTORIAL 4.2

1. Solve the following differential equations by separating the variables.

a) b)
2 4 3
= 2 = ( + 2)

Solution: Solution:

3 Ans: Ans:
3
= 2 + 2 + 2 = 4 +
2

c) d)
Solution:
= 4 = 3 2

Solution:

Ans: Ans:

= 1 4 + 1 = 1 sin 2 +
4 3 2

19

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

e) 2 2 f) 3 + 1
Solution: = 2 + 2 Solution: = 3 2 2 + 2

3 Ans: 2 Ans:
3 2
−2 −1 + = + 3 + = − −1 +

g) 2 − = 1 h) = 4 +
Solution: Solution:

Ans: Ans:

|1 + | = 1 + − − = 1 4 +
2 4

20

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

2. Solve the following differential equations by by finding the integrating factor.

a) b)

+ 3 = 8 − 5 = 20

Solution: Solution:

8 Ans:
3
3 = 3 + Ans:
−5 = −4 −5 +

c) + 3 = d) − = 2 − 6
Solution: Solution:

5 Ans:
5
3 = + Ans:
−1 = 2 − 6 +

21

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

e) + 3 = 4 f) − 5 = 3
Solution: Solution:

1 Ans: 1 Ans:
7 2
3 = 7 + −5 = − −2 +

g) + = h) + 2 = 5 2 + 4
Solution: Solution:

1 Ans: 4 Ans:
2 3
= 2 + 2 = 5 + 3 +

22

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

3. Given a differential equation. 4. Given a differential equation

= 3 = 4 3

a) Find the general solutions of the equation. a) Find the general solutions of the equation.
b) Determine the particular solution given b) Determine the particular solution given

(0) = 2. � 2 � = 3.
Solution:
Solution:

3 Ans:
2
a) = 2 + Ans:
a) 4 = sin +
b) = 3 2 + 2 b) 4 = sin + 80
2
23

5. Given a differential equation. DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)
+ 3 = 2
6. Given a differential equation
a) Determine the integrating factor. 2
+ = 3
b) Solve the differential equation given that
(0) = 4 a) Determine the integrating factor.
b) Solve the differential equation given that
Solution:
(2) = 3
Solution:

a) = 3 Ans: Ans:
a) = 2
b) 3 = 1 5 + 19 b) 2 = 3 + 4
5 5
24

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

4.3 SECOND ORDER DIFFERENTIAL EQUATIONS

After completing the unit, students should be able to:
• Solve second order differential equations.

UNIT LEARNING
OUTCOME

What is second-order linear differential equation?

A second-order linear differential equation with a constant coefficient has the general form:

2 + + = ( ), a ≠ 0.
2

• If the function ( ) = , then the differential equation is a homogeneous equation.
Solution of the homogenous give the complementary function (CF) denoted as

• If the function ( ) ≠ , then the differential equation is a non-homogeneous equation
and the solution consists of two parts the complementary function (CF) and the
particular integral (PI) denoted as
Hence,

2 + + = ( ), a ≠ 0 and ( ) ≠ 0
2

gives the solution : = + =CF +PI

How to solve second-order linear differential equation?

Given the homogenous equation : + + =


Let = be a solution

Then, ′ = , ″ = 2

Substituting into the homogenous equation,
Thus, 2 + + = 0

( 2 + + ) = 0

If ≠ 0 then + + = known as the auxiliary equation.

Solving the quadratic equation we will get either one of the following roots:

Real and distinct, Real and both equal, Complex conjugates,
≠ = = = ±

then the solution is then the solution is then the solution is

= + = + = ( ) + ( )

where A and B are constant where A and B are constant where A and B are constant coefficients. 25
coefficients. coefficients.

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

EXAMPLE

1. Find the complementary functions of the following differential equations.

a) 2 b) 2
2 2
+ + = 0 − + = 0

Solution: Solution:

The auxiliary equation: The auxiliary equation:
2 + + = 0 2 − + = 0
⇒ ( + 2)( + 3) = 0 ⇒ ( − 1)(2 − 3) = 0
∴ = − or = −
Real and distinct, ∴ = or = Real and distinct,
Solution is ≠
= − + − Solution is ≠
then the solution is
then the solution is
= +
= + = +

c) 2 d) 2
2 2
+ + = 0 + + = 0

Solution: Solution:

The auxiliary equation: The auxiliary equation:
2 + + = 0 2 + + = 0
⇒ ( + 3)( + 3) = 0 ⇒ (2 + 3)(2 + 3) = 0
= − twice
Real and both equal, = − twice Real and both equal,
Solution is = = = =
= − + −
then the solution is then the solution is

= + Solution is = +
= − + −

e) 2 f) 2
2 2
+ + = 0 − + = 0


Solution: Solution:

The auxiliary equation: Complex conjugates, The auxiliary equation: Complex conjugates,
2 + + = 0 = ± 2 − + = 0 = ±
⇒ = − ±
then the solution is then the solution is

⇒ = ± .
= ( )
+ ( )
= ( )
+ ( )

Solution is Solution is
= − ( ) + − ( ) = ( . ) + ( . )

26

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

EXERCISE 4.3

1. Find the general solution of

a) 2 + − 2 = 0 b) 2 − 6 + 9 = 0
Solution: 2 Solution: 2

Ans: Ans:
= + − = 3 + 3

c) 2 d) 2
2 Solution: 2
6 − 7 + 2 = 0 − 2 + 5 = 0

Solution:

Ans: Ans:
= 23 + 12 = (2 ) + (2 )

e) 6 2 − − 2 = 0 f) 4 2 + 4 + = 0
Solution: 2 Solution: 2

Ans: Ans:
= 23 + −21 = −21 + −12

27

DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

g) 2 + 2 + 5 = 0 h) 6 2 + 5 − 6 = 0
Solution: 2 Solution: 2

Ans: Ans:
= − (2 ) + − (2 ) = 32 + −21

i) 4 2 − 5 + = 0 j) 2 − 7 + 12 = 0
Solution: 2 Solution: 2

Ans: Ans:
= + 41 = 4 + 3

k) l)
Solution:
2 + 6 + 13 = 0 2 + − 9 = 0
2 2

Ans: Ans:
= −3 (2 ) + −3 (2 ) = 2.54 + −3.54

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DUM 30172 MATHEMATICS FOR MECHANICAL ENGINEERING
UNIT 4 ORDINARY DIFFERENTIAL EQUATIONS (ODE)

TUTORIAL 4.3

1. Determine the complementary functions of the given differential equations.

a) 2 + 3 − 10 = 0 b) 2 + 8 + 16 = 0
Solution: 2 Solution: 2

Ans: Ans:
= 2 + −5 = −4 + −4

c) 2 + 2 − 3 = 0 d) 2 − 12 + 36 = 0
Solution: 2 Solution: 2

Ans: Ans:
= + −3 = 6 + 6

e) 2 − 2 + 10 = 0 f) 2 + 7 = 0
Solution: 2 Solution: 2

Ans: Ans:
= (3 ) + (3 ) = (2.65 ) + (2.65 )

29