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home / study / math / applied mathematics / applied mathematics textbook solutions / numerical methods for engineers / 6th edition / chapter 25

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Numerical Methods for Engineers (6th Edition) View more editions

Solutions

Mohammad P. UCLA

Solutions for Chapter 25

Nick H.

861 step-by-step solutions

Numerical...

University of Washin…

4186

235

Solved by publishers, professors & experts Natalia G.

iOS, Android, & web

The University Of C…

GET SOLUTIONS

SAMPLE SOLUTION

Chapter:

FIND ME A TUTOR

CH25

Problem:

1P

75% (4 ratings)

Step 1 of 14 (a) Consider the following differential equation,

Find common factor on the right,

Now integrate both sides,

Calculate the value of the constant c from the initial condition

Put

, in the above equation

Put

in the solution of the differential equation

Then,

Hence, the solution of the differential equation is

Step 2 of 14 Use maple to draw graph of equation

Step 3 of 14 (b) Consider the solution of the differential equation at width

.

Consider the following differential equation,

The Euler formula is

Where, by the equation. The initial condition is and

.

So,

The next condition will be

Step 4 of 14 And,

Therefore,

So,

The next condition will be

And,

Therefore,

The next condition will be

And,

Therefore,

The next condition will be

And

Finally, From the above solution, the value of

and

are given in the below.

The Resolution of the differential equation at width ( h ) =0.25 The initial condition for the Euler method is And

then

The next condition will be

And,

The next condition will be

And, Further solve

Therefore,

The next condition will be

And,

Therefore,

The next condition will be

And,

The next condition is

And,

Therefore,

The next condition will be

And,

The next condition will be

And,

The next condition will be

And,

Step 5 of 14

Further,

Finally, From the above solution, the value of

and

are given in the below table.

Step 6 of 14 The result of the analytical solution and numerical solution are plotted below.

Step 7 of 14 (c) The formula of the Midpoint m e t h o d is , Where,

From the initial condition ,

and

The next condition will be

From the midpoint formula

Put

then,

Step 8 of 14

The next condition will be

Put

in the midpoint formula

Further solve

The next condition will be

Put

in the midpoint formula

Further solve,

The next condition will be

Put

in the midpoint formula

Finally, From the solution, the value of

1

-1.5

0.5

0.536133

-0.73718

1

0.346471

-0.17324

1.5

0.415176

0.778417

2

1.591802

10.34671

and

are given in the below table.

Step 9 of 14 Graph is shown below,

Step 10 of 14 (d) The fourth order Runge-Kutta method is

Where,

The initial condition is given as And

The further solve,

Further solve,

The further solve,

In the fourth-order Runge-Kutta method

Put

The next condition be,

Then,

Step 11 of 14

Step 12 of 14 In the fourth-order Runge-Kutta method. Put

,

Further solve,

The third condition will be

Then,

The further solve,

Now,

The further solve,

Put i=2, In the fourth-order Runge-Kutta method

The next condition will be

Step 13 of 14

Then,

The further solve,

Put

,

In the fourth-order Runge-Kutta method

The next condition will be

Then,

Further solve,

Further solve,

Further solve,

Finally, From the solution, the value of

0

1

-1.50000

-0.9277

and

-1.1401

are given below.

-0.5912

Step 14 of 14 0.5 0.4811

-0.6615

-0.3404

-0.4269

-0.1338

1

0.2869

-0.1435

0.1138

0.1429

0.6720

1.5

0.3738

0.7008

2.1186

3.4866

13.7607

2

2.5131

16.3350

65.2466

186.1883

13.504523

The result of the analytical solution and numerical solution are plotted below,

Back to top

Corresponding Textbook Numerical Methods for Engineers | 6th Edition ISBN-13:

9780077417109

ISBN:

0077417100

Authors:

Raymond Canale, Raymond P Canale, Steven C Chapra, Stephen Chapra, Steven Chapra

Rent | Buy

Solutions by Chapter Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Chapter 19

Chapter 20

Chapter 21

Chapter 22

Chapter 23

Chapter 24

Chapter 25

Chapter 26

Chapter 27

Chapter 28

Chapter 29

Chapter 30

Chapter 31

Chapter 32

Back to top Need an extra hand? Browse hundreds of Math tutors.

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home / study / math / applied mathematics / applied mathematics textbook solutions / numerical methods for engineers / 6th edition / chapter 25

Math Chegg tutors who can help right now

Numerical Methods for Engineers (6th Edition) View more editions

Solutions

Mohammad P. UCLA

Solutions for Chapter 25

Nick H.

861 step-by-step solutions

Numerical...

University of Washin…

4186

235

Solved by publishers, professors & experts Natalia G.

iOS, Android, & web

The University Of C…

GET SOLUTIONS

SAMPLE SOLUTION

Chapter:

FIND ME A TUTOR

CH25

Problem:

1P

75% (4 ratings)

Step 1 of 14 (a) Consider the following differential equation,

Find common factor on the right,

Now integrate both sides,

Calculate the value of the constant c from the initial condition

Put

, in the above equation

Put

in the solution of the differential equation

Then,

Hence, the solution of the differential equation is

Step 2 of 14 Use maple to draw graph of equation

Step 3 of 14 (b) Consider the solution of the differential equation at width

.

Consider the following differential equation,

The Euler formula is

Where, by the equation. The initial condition is and

.

So,

The next condition will be

Step 4 of 14 And,

Therefore,

So,

The next condition will be

And,

Therefore,

The next condition will be

And,

Therefore,

The next condition will be

And

Finally, From the above solution, the value of

and

are given in the below.

The Resolution of the differential equation at width ( h ) =0.25 The initial condition for the Euler method is And

then

The next condition will be

And,

The next condition will be

And, Further solve

Therefore,

The next condition will be

And,

Therefore,

The next condition will be

And,

The next condition is

And,

Therefore,

The next condition will be

And,

The next condition will be

And,

The next condition will be

And,

Step 5 of 14

Further,

Finally, From the above solution, the value of

and

are given in the below table.

Step 6 of 14 The result of the analytical solution and numerical solution are plotted below.

Step 7 of 14 (c) The formula of the Midpoint m e t h o d is , Where,

From the initial condition ,

and

The next condition will be

From the midpoint formula

Put

then,

Step 8 of 14

The next condition will be

Put

in the midpoint formula

Further solve

The next condition will be

Put

in the midpoint formula

Further solve,

The next condition will be

Put

in the midpoint formula

Finally, From the solution, the value of

1

-1.5

0.5

0.536133

-0.73718

1

0.346471

-0.17324

1.5

0.415176

0.778417

2

1.591802

10.34671

and

are given in the below table.

Step 9 of 14 Graph is shown below,

Step 10 of 14 (d) The fourth order Runge-Kutta method is

Where,

The initial condition is given as And

The further solve,

Further solve,

The further solve,

In the fourth-order Runge-Kutta method

Put

The next condition be,

Then,

Step 11 of 14

Step 12 of 14 In the fourth-order Runge-Kutta method. Put

,

Further solve,

The third condition will be

Then,

The further solve,

Now,

The further solve,

Put i=2, In the fourth-order Runge-Kutta method

The next condition will be

Step 13 of 14

Then,

The further solve,

Put

,

In the fourth-order Runge-Kutta method

The next condition will be

Then,

Further solve,

Further solve,

Further solve,

Finally, From the solution, the value of

0

1

-1.50000

-0.9277

and

-1.1401

are given below.

-0.5912

Step 14 of 14 0.5 0.4811

-0.6615

-0.3404

-0.4269

-0.1338

1

0.2869

-0.1435

0.1138

0.1429

0.6720

1.5

0.3738

0.7008

2.1186

3.4866

13.7607

2

2.5131

16.3350

65.2466

186.1883

13.504523

The result of the analytical solution and numerical solution are plotted below,

Back to top

Corresponding Textbook Numerical Methods for Engineers | 6th Edition ISBN-13:

9780077417109

ISBN:

0077417100

Authors:

Raymond Canale, Raymond P Canale, Steven C Chapra, Stephen Chapra, Steven Chapra

Rent | Buy

Solutions by Chapter Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Chapter 19

Chapter 20

Chapter 21

Chapter 22

Chapter 23

Chapter 24

Chapter 25

Chapter 26

Chapter 27

Chapter 28

Chapter 29

Chapter 30

Chapter 31

Chapter 32

Back to top Need an extra hand? Browse hundreds of Math tutors.

ABOUT CHEGG Media Center

RESOURCES Site Map

TEXTBOOK LINKS Return Your Books

STUDENT SERVICES Chegg Play

COMPANY Jobs

LEARNING SERVICES Online Tutoring

College Marketing

Mobile

Textbook Rental

Chegg Coupon

Customer Service

Chegg Study Help

Privacy Policy

Publishers

eTextbooks

Scholarships

Give Us Feedback

Solutions Manual

Your CA Privacy Rights

Join Our Affiliate

Used Textbooks

Career Search

Chegg For Good

Tutors by City

Terms of Use

Program

Cheap Textbooks

Internships

Become a Tutor

GPA Calculator

General Policies

Advertising Choices

College Textbooks

College Search

Sell Textbooks

College Majors

Intellectual Property Rights Investor Relations

Test Prep

Scholarship Redemption

Enrollment Services

Over 6 million trees planted © 2003-2016 Chegg Inc. All rights reserved.

17

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