Σnn - 1cnx" + 2 nn - 1)x"-2 – 6 Σεx" = 0 n=2 n=2 n=0. 00 Σκα - 1cx*+ Σ« + 2k + 1)c+2x* – 6 Σax* = 0 k=2 A=0 A=0 2c₂ - 6c +(6c3-6c₁ )x+ Σ [(k² − k − 6)c₂ + (k+ 2)(k+1)C₁+2] x* = 0 k=2 Thus, 2cy – 6co = 0,6c3 – 6ci = 0 and (k - 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0. Simplifying the above equations gives c2 = 3co and c3 = cı and ck+2 = - =- (k−3)ck for k = 2, 3, 4, ... (k+1) Substituting k = 2. 3. 4. ... into the last formula gives.
Σnn - 1cnx" + 2 nn - 1)x"-2 – 6 Σεx" = 0 n=2 n=2 n=0. 00 Σκα - 1cx*+ Σ« + 2k + 1)c+2x* – 6 Σax* = 0 k=2 A=0 A=0 2c₂ - 6c +(6c3-6c₁ )x+ Σ [(k² − k − 6)c₂ + (k+ 2)(k+1)C₁+2] x* = 0 k=2 Thus, 2cy – 6co = 0,6c3 – 6ci = 0 and (k - 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0. Simplifying the above equations gives c2 = 3co and c3 = cı and ck+2 = - =- (k−3)ck for k = 2, 3, 4, ... (k+1) Substituting k = 2. 3. 4. ... into the last formula gives.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
I'm not sure how they get this line from the line prior (marked in image). When I try it, I get zero c_0 and zero c_1 but here they have nonzero coefficients.
https://www.bartleby.com/solution-answer/chapter-62-problem-16e-a-first-course-in-differential-equations-with-modeling-applications-mindtap-course-list-11th-edition/9781305965720/aea0e5fe-7b16-4fe7-afe5-5e2756118845
![Given:
The given differential equation is (x² + 1)y" - 6y = 0 and x = 0 is the ordinary point.
Definition used:
If x = xo is an ordinary point of the differential equation then there exist two linearly independent solution in
the form of power series centered at xo.
y = =Σ Cn(x-xo)"
n=0
Calculation:
Given differential equation is (x² + 1)y" - 6y=0 so consider power series y =
n=2
Substitute y = nx", and the second derivative y" = Σcn(n-1)x"-² in the differential equation as,
00
00
Σn(n-1)cnx" + n(n − 1)c₁x²-² – 6
k=2
n=0
Σk(k-1)cx +
C4 =
00
n=2
00
00
Σ cnx" = 0
n=0
(k+ 2)(k + 1)ck+2x² − 6
k=0
00
n=2
2₁₁x² = 0
k=0
00
2c₂ - 6€ + (6c3 - 6c₁ )x+ Σ [(k² − k − 6)ck + (k+2)(k+ 1)Ck+2] x² = 0
k=2
Thus, 2c₂6c0 = 0, 6c36c₁ = 0 and (k - 3)(k+ 2)ck + (k+2)(k+ 1)ck+2 = 0.
Simplifying the above equations gives c₂= 3co and c3 = C₁ and C+2 =
Substituting k = 2, 3, 4, ... into the last formula gives,
= Co
C5 = 0
C6 = 300
Similarly find other coefficients using the above formula.
ΣC.X².
n=0
Substituting all the coefficients in the power series y = cx" gives,
ΣΩ
n=0
(k-3) for k= 2, 3, 4, ...
(k+1)
y = co + c₁x + c₂x² + C3x²³ + C₁x² + c6x² + ...
y = co + c₁x +3cox² + c₁x²³ + cox² = {{coxº +...
y = co(1 + 3x² + x4 − ½ x6 + ...) + c₁(x + x³)
The above solution is of the form y = co(vi) + ci(v2).
Thus, the two power series solutions are y₁ = 1 + 3x² + x² - ²x² + ... and y₂ = x + x³](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8693a44-c76f-4d3e-8b43-577ed05feba6%2F9229eb99-6d0b-4b9b-9556-9b4f568f1ded%2Fdp7qgds_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given:
The given differential equation is (x² + 1)y" - 6y = 0 and x = 0 is the ordinary point.
Definition used:
If x = xo is an ordinary point of the differential equation then there exist two linearly independent solution in
the form of power series centered at xo.
y = =Σ Cn(x-xo)"
n=0
Calculation:
Given differential equation is (x² + 1)y" - 6y=0 so consider power series y =
n=2
Substitute y = nx", and the second derivative y" = Σcn(n-1)x"-² in the differential equation as,
00
00
Σn(n-1)cnx" + n(n − 1)c₁x²-² – 6
k=2
n=0
Σk(k-1)cx +
C4 =
00
n=2
00
00
Σ cnx" = 0
n=0
(k+ 2)(k + 1)ck+2x² − 6
k=0
00
n=2
2₁₁x² = 0
k=0
00
2c₂ - 6€ + (6c3 - 6c₁ )x+ Σ [(k² − k − 6)ck + (k+2)(k+ 1)Ck+2] x² = 0
k=2
Thus, 2c₂6c0 = 0, 6c36c₁ = 0 and (k - 3)(k+ 2)ck + (k+2)(k+ 1)ck+2 = 0.
Simplifying the above equations gives c₂= 3co and c3 = C₁ and C+2 =
Substituting k = 2, 3, 4, ... into the last formula gives,
= Co
C5 = 0
C6 = 300
Similarly find other coefficients using the above formula.
ΣC.X².
n=0
Substituting all the coefficients in the power series y = cx" gives,
ΣΩ
n=0
(k-3) for k= 2, 3, 4, ...
(k+1)
y = co + c₁x + c₂x² + C3x²³ + C₁x² + c6x² + ...
y = co + c₁x +3cox² + c₁x²³ + cox² = {{coxº +...
y = co(1 + 3x² + x4 − ½ x6 + ...) + c₁(x + x³)
The above solution is of the form y = co(vi) + ci(v2).
Thus, the two power series solutions are y₁ = 1 + 3x² + x² - ²x² + ... and y₂ = x + x³
Expert Solution
Step 1
I am going to solve the problem by using some simple calculus to get the required result of the given problem.
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning