Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
14th Edition
ISBN: 9780134438986
Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher: PEARSON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 17.5, Problem 1E
To determine

To find: The general solution of the differential equation y+2y=0 by using power series.

Expert Solution & Answer
Check Mark

Answer to Problem 1E

The general solution of the differential equation y+2y=0 is y(x)=a+be2x_.

Explanation of Solution

Formula used:

Power Series Method:

The power series method for solving a second-order homogenous differential equation consists of finding the coefficients of a power series y(x)=n=0cnxn=c0+c1x+c2x2+

Calculation:

The given equation is y+2y=0.

Assume that the series solution takes the form of y=n=0cnxn.

Calculate the derivative of y=n=0cnxn.

y=n=1ncnxn1

Now calculate the second derivative of y=n=0cnxn.

y=n=2n(n1)cnxn11=n=2n(n1)cnxn2

Substitute the derivatives of y and y in y+2y=0 as follows.

n=2n(n1)cnxn2+2n=1ncnxn1=0

Equate the coefficient of each power of x to zero as shown in the following table.

Power of xCoefficient equation
x02(1)c2+2c1=0 or c2=c1
x13(2)c3+2(2)c2=0 or c3=23c2=23c1
x24(3)c4+2(3)c3=0 or c4=12c3=13c1
x35(4)c5+2(4)c4=0 or c5=25c4=215c1
x46(5)c6+2(5)c5=0 or c6=13c5=245c1
xn(n+2)(n+1)cn+2+2(n+1)cn+1=0 or cn+2=2n+2cn+1

The recursive relation cn+2=2n+2cn+1 can be written as cn=2ncn1.

Thus,

cn=(2n)(2n1cn2)=(2n)(2n1)(2n2cn3)=(2)n1n!c1(n2)

Hence,

y=c0+c1xc1x2+23c1x313c1x4+215c1x5245c1x6+=c0+c12c12+c1(2x)2c12(2x)22!+c12(2x)33!c12(2x)44!+c12(2x)55!c12(2x)66!+=(c0+c12)c12(1(2x)+(2x)22!(2x)33!+(2x)44!(2x)55!+(2x)66!)=c0+c12c12n=0(2x)nn!

On further simplification,

y=(c0+c12)c12e2x=a+be2x

Where a=c0+c12 and b=c12.

Therefore, the general solution of the differential equation y+2y=0 is y(x)=a+be2x_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
18. Using the method of variation of parameter, a particular solution to y′′ + 16y = 4 sec(4t) isyp(t) = u1(t) cos(4t) + u2(t) sin(4t). Then u2(t) is equal toA. 1 B. t C. ln | sin 4t| D. ln | cos 4t| E. sec(4t)
Question 4. Suppose you need to know an equation of the tangent plane to a surface S at the point P(2, 1, 3). You don't have an equation for S but you know that the curves r1(t) = (2 + 3t, 1 — t², 3 − 4t + t²) r2(u) = (1 + u², 2u³ − 1, 2u + 1) both lie on S. (a) Check that both r₁ and r2 pass through the point P. 1 (b) Give the expression of the 074 in two ways Ət ⚫ in terms of 32 and 33 using the chain rule მყ ⚫ in terms of t using the expression of z(t) in the curve r1 (c) Similarly, give the expression of the 22 in two ways Əz ди ⚫ in terms of oz and oz using the chain rule Əz მყ • in terms of u using the expression of z(u) in the curve r2 (d) Deduce the partial derivative 32 and 33 at the point P and the equation of მე მყ the tangent plane
Coast Guard Patrol Search Mission The pilot of a Coast Guard patrol aircraft on a search mission had just spotted a disabled fishing trawler and decided to go in for a closer look. Flying in a straight line at a constant altitude of 1000 ft and at a steady speed of 256 ft/s, the aircraft passed directly over the trawler. How fast (in ft/s) was the aircraft receding from the trawler when it was 1400 ft from the trawler? (Round your answer to one decimal places.) 1000 ft 180 × ft/s Need Help? Read It SUBMIT ANSWER

Chapter 17 Solutions

Thomas' Calculus (14th Edition)

Ch. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - Prob. 13ECh. 17.1 - Prob. 14ECh. 17.1 - Prob. 15ECh. 17.1 - Prob. 16ECh. 17.1 - Prob. 17ECh. 17.1 - Prob. 18ECh. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - Prob. 20ECh. 17.1 - Prob. 21ECh. 17.1 - Prob. 22ECh. 17.1 - Prob. 23ECh. 17.1 - Prob. 24ECh. 17.1 - Prob. 25ECh. 17.1 - Prob. 26ECh. 17.1 - Prob. 27ECh. 17.1 - Prob. 28ECh. 17.1 - Prob. 29ECh. 17.1 - Prob. 30ECh. 17.1 - In Exercises 31 − 40, find the unique solution of...Ch. 17.1 - Prob. 32ECh. 17.1 - Prob. 33ECh. 17.1 - Prob. 34ECh. 17.1 - Prob. 35ECh. 17.1 - Prob. 36ECh. 17.1 - Prob. 37ECh. 17.1 - In Exercises 31 − 40, find the unique solution of...Ch. 17.1 - Prob. 39ECh. 17.1 - Prob. 40ECh. 17.1 - Prob. 41ECh. 17.1 - In Exercises 41 – 55, find the general...Ch. 17.1 - Prob. 43ECh. 17.1 - Prob. 44ECh. 17.1 - Prob. 45ECh. 17.1 - Prob. 46ECh. 17.1 - Prob. 47ECh. 17.1 - Prob. 48ECh. 17.1 - Prob. 49ECh. 17.1 - Prob. 50ECh. 17.1 - Prob. 51ECh. 17.1 - Prob. 52ECh. 17.1 - Prob. 53ECh. 17.1 - Prob. 54ECh. 17.1 - Prob. 55ECh. 17.1 - In Exercises 56−60, solve the initial value...Ch. 17.1 - Prob. 57ECh. 17.1 - Prob. 58ECh. 17.1 - Prob. 59ECh. 17.1 - Prob. 60ECh. 17.1 - Prob. 61ECh. 17.1 - Prob. 62ECh. 17.1 - Prob. 63ECh. 17.1 - Prob. 64ECh. 17.1 - Prob. 65ECh. 17.2 - Solve the equations in Exercises 1−16 by the...Ch. 17.2 - Prob. 2ECh. 17.2 - Prob. 3ECh. 17.2 - Prob. 4ECh. 17.2 - Prob. 5ECh. 17.2 - Prob. 6ECh. 17.2 - Prob. 7ECh. 17.2 - Prob. 8ECh. 17.2 - Prob. 9ECh. 17.2 - Prob. 10ECh. 17.2 - Prob. 11ECh. 17.2 - Prob. 12ECh. 17.2 - Prob. 13ECh. 17.2 - Prob. 14ECh. 17.2 - Prob. 15ECh. 17.2 - Prob. 16ECh. 17.2 - Prob. 17ECh. 17.2 - Prob. 18ECh. 17.2 - Prob. 19ECh. 17.2 - Prob. 20ECh. 17.2 - Prob. 21ECh. 17.2 - Prob. 22ECh. 17.2 - Prob. 23ECh. 17.2 - Prob. 24ECh. 17.2 - Prob. 25ECh. 17.2 - Prob. 26ECh. 17.2 - Prob. 27ECh. 17.2 - Prob. 28ECh. 17.2 - Prob. 29ECh. 17.2 - Prob. 30ECh. 17.2 - Prob. 31ECh. 17.2 - Prob. 32ECh. 17.2 - Prob. 33ECh. 17.2 - Prob. 34ECh. 17.2 - Prob. 35ECh. 17.2 - Prob. 36ECh. 17.2 - Prob. 37ECh. 17.2 - Prob. 38ECh. 17.2 - Prob. 39ECh. 17.2 - Prob. 40ECh. 17.2 - Prob. 41ECh. 17.2 - Prob. 42ECh. 17.2 - Prob. 43ECh. 17.2 - Prob. 44ECh. 17.2 - Prob. 45ECh. 17.2 - Prob. 46ECh. 17.2 - Prob. 47ECh. 17.2 - Prob. 48ECh. 17.2 - Prob. 49ECh. 17.2 - Prob. 50ECh. 17.2 - Prob. 51ECh. 17.2 - Prob. 52ECh. 17.2 - Prob. 53ECh. 17.2 - Prob. 54ECh. 17.2 - Prob. 55ECh. 17.2 - Prob. 56ECh. 17.2 - Prob. 57ECh. 17.2 - Prob. 58ECh. 17.2 - Prob. 59ECh. 17.2 - Prob. 60ECh. 17.3 - Prob. 1ECh. 17.3 - Prob. 2ECh. 17.3 - A 20-lb weight is hung on an 18-in. spring and...Ch. 17.3 - Prob. 4ECh. 17.3 - Prob. 5ECh. 17.3 - Prob. 6ECh. 17.3 - Prob. 7ECh. 17.3 - Prob. 8ECh. 17.3 - Prob. 9ECh. 17.3 - Prob. 10ECh. 17.3 - Prob. 11ECh. 17.3 - Prob. 12ECh. 17.3 - Prob. 13ECh. 17.3 - Prob. 14ECh. 17.3 - Prob. 15ECh. 17.3 - Prob. 16ECh. 17.3 - Prob. 17ECh. 17.3 - Prob. 18ECh. 17.3 - Prob. 19ECh. 17.3 - Prob. 20ECh. 17.3 - Prob. 21ECh. 17.3 - Prob. 22ECh. 17.3 - Prob. 23ECh. 17.3 - Prob. 24ECh. 17.3 - Prob. 25ECh. 17.3 - Prob. 26ECh. 17.4 - Prob. 1ECh. 17.4 - Prob. 2ECh. 17.4 - Prob. 3ECh. 17.4 - Prob. 4ECh. 17.4 - Prob. 5ECh. 17.4 - Prob. 6ECh. 17.4 - Prob. 7ECh. 17.4 - Prob. 8ECh. 17.4 - Prob. 9ECh. 17.4 - Prob. 10ECh. 17.4 - Prob. 11ECh. 17.4 - Prob. 12ECh. 17.4 - Prob. 13ECh. 17.4 - Prob. 14ECh. 17.4 - Prob. 15ECh. 17.4 - Prob. 16ECh. 17.4 - Prob. 17ECh. 17.4 - Prob. 18ECh. 17.4 - Prob. 19ECh. 17.4 - Prob. 20ECh. 17.4 - Prob. 21ECh. 17.4 - Prob. 22ECh. 17.4 - Prob. 23ECh. 17.4 - Prob. 24ECh. 17.4 - Prob. 25ECh. 17.4 - Prob. 26ECh. 17.4 - Prob. 27ECh. 17.4 - Prob. 28ECh. 17.4 - Prob. 29ECh. 17.4 - Prob. 30ECh. 17.5 - Prob. 1ECh. 17.5 - Prob. 2ECh. 17.5 - Prob. 3ECh. 17.5 - Prob. 4ECh. 17.5 - Prob. 5ECh. 17.5 - Prob. 6ECh. 17.5 - Prob. 7ECh. 17.5 - Prob. 8ECh. 17.5 - Prob. 9ECh. 17.5 - Prob. 10ECh. 17.5 - Prob. 11ECh. 17.5 - Prob. 12ECh. 17.5 - Prob. 13ECh. 17.5 - Prob. 14ECh. 17.5 - In Exercises 1–18, use power series to find the...Ch. 17.5 - Prob. 16ECh. 17.5 - Prob. 17ECh. 17.5 - Prob. 18E
Knowledge Booster
Background pattern image
Calculus
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Calculus: Early Transcendentals
Calculus
ISBN:9781285741550
Author:James Stewart
Publisher:Cengage Learning
Text book image
Thomas' Calculus (14th Edition)
Calculus
ISBN:9780134438986
Author:Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:PEARSON
Text book image
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:9780134763644
Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:PEARSON
Text book image
Calculus: Early Transcendentals
Calculus
ISBN:9781319050740
Author:Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:W. H. Freeman
Text book image
Precalculus
Calculus
ISBN:9780135189405
Author:Michael Sullivan
Publisher:PEARSON
Text book image
Calculus: Early Transcendental Functions
Calculus
ISBN:9781337552516
Author:Ron Larson, Bruce H. Edwards
Publisher:Cengage Learning
01 - What Is A Differential Equation in Calculus? Learn to Solve Ordinary Differential Equations.; Author: Math and Science;https://www.youtube.com/watch?v=K80YEHQpx9g;License: Standard YouTube License, CC-BY
Higher Order Differential Equation with constant coefficient (GATE) (Part 1) l GATE 2018; Author: GATE Lectures by Dishank;https://www.youtube.com/watch?v=ODxP7BbqAjA;License: Standard YouTube License, CC-BY
Solution of Differential Equations and Initial Value Problems; Author: Jefril Amboy;https://www.youtube.com/watch?v=Q68sk7XS-dc;License: Standard YouTube License, CC-BY