Part 1. Obtain a solution to the following differential equation valid near the origin: 1.] For the given differential equation, x = 0 is a/an: A. Regular point B. Ordinary point C. Regular singular point D. Irregular singular point 2. Which of the following is the correct form for the solution y? 100 A. y = 1 anx-r B. y = Two anxn+r 100 C. y=n=0 anx" y"+3xy' +7y=0 100 D.y = E=1 anx" 3. The indicial equation for the given differential equation is: A.r(r-4)= 0 B. r(2r + 1) = 0 C. r(r-3) = 0 D. None of these/Not applicable 4. The recurrence relation for n ≥ 2 is: A. n(n-1)an + (3n+ 1)an-2 = 0 B. n(n-1)an + (3n+7)an-2 = 0 C. n(n+1)an-2 + (3n+ 1)an = 0 D. n(n+1)an-2 + (3n+7)an = 0
Part 1. Obtain a solution to the following differential equation valid near the origin: 1.] For the given differential equation, x = 0 is a/an: A. Regular point B. Ordinary point C. Regular singular point D. Irregular singular point 2. Which of the following is the correct form for the solution y? 100 A. y = 1 anx-r B. y = Two anxn+r 100 C. y=n=0 anx" y"+3xy' +7y=0 100 D.y = E=1 anx" 3. The indicial equation for the given differential equation is: A.r(r-4)= 0 B. r(2r + 1) = 0 C. r(r-3) = 0 D. None of these/Not applicable 4. The recurrence relation for n ≥ 2 is: A. n(n-1)an + (3n+ 1)an-2 = 0 B. n(n-1)an + (3n+7)an-2 = 0 C. n(n+1)an-2 + (3n+ 1)an = 0 D. n(n+1)an-2 + (3n+7)an = 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Show complete solution for 4 and 5
![Part 1. Obtain a solution to the following differential equation valid near the
origin:
1.] For the given differential equation, x = 0 is a/an:
A. Regular point
B. Ordinary point
y"+3xy' +7y=0
C. Regular singular point
D. Irregular singular point
2. Which of the following is the correct form for the solution y?
100
A. y = E=1 anxn-r
B.y = no anx+r
100
C. y = Σn=0 anx
D.y = Σn=1 anx¹
3. The indicial equation for the given differential equation is:
A. r(r-4)= 0
B. r(2r + 1) = 0
C. r(r-3) = 0
D. None of these/Not applicable
4. The recurrence relation for n ≥ 2 is:
A. n(n-1)an + (3n+ 1)an-2 = 0
B. n(n-1)an + (3n+7)an-2 = 0
C. n(n+1)an-2 + (3n+ 1)an = 0
D. n(n+1)an-2 + (3n+ 7)an = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8125e78e-cb90-4f62-b2bd-c75f18dfa9c3%2F8f028e24-a9e7-44bc-90a9-4abf394b5cec%2Fd27z2i_processed.png&w=3840&q=75)
Transcribed Image Text:Part 1. Obtain a solution to the following differential equation valid near the
origin:
1.] For the given differential equation, x = 0 is a/an:
A. Regular point
B. Ordinary point
y"+3xy' +7y=0
C. Regular singular point
D. Irregular singular point
2. Which of the following is the correct form for the solution y?
100
A. y = E=1 anxn-r
B.y = no anx+r
100
C. y = Σn=0 anx
D.y = Σn=1 anx¹
3. The indicial equation for the given differential equation is:
A. r(r-4)= 0
B. r(2r + 1) = 0
C. r(r-3) = 0
D. None of these/Not applicable
4. The recurrence relation for n ≥ 2 is:
A. n(n-1)an + (3n+ 1)an-2 = 0
B. n(n-1)an + (3n+7)an-2 = 0
C. n(n+1)an-2 + (3n+ 1)an = 0
D. n(n+1)an-2 + (3n+ 7)an = 0
![5. Which of the following is the expression for a2k in terms of a?
A. azk=
(-1) [10-16-22---(6k+4)]
(2k+1)!
(-1)k[7-13-19 (6k+1)]
(2k)!
(−1)k[6-12-18---(6k)]
(2k+1)!
B. azk=
C. azk=
D. azk=
A. azk+1 =
(-1)k[8.14-20(6k+2)]
(2k)!
B. azk+1 =
C. a₂k+1 =
-ao
ao
6. Which of the following is the expression for a2k+1 in terms of a₂?
D. azk+1 =
·ao
ao
(2k)!
(-1) [6-12-18 (6k)]
(2k+1)!
(-1) [10-16-22-(6k+4)]
(2k+1)!
(-1) [7-13-19--(6k+1)]
-a₁
(-1) [8-14-20-(6k+2)]
(2k)!
a₁
a₁
a₁](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8125e78e-cb90-4f62-b2bd-c75f18dfa9c3%2F8f028e24-a9e7-44bc-90a9-4abf394b5cec%2F0hovwq_processed.png&w=3840&q=75)
Transcribed Image Text:5. Which of the following is the expression for a2k in terms of a?
A. azk=
(-1) [10-16-22---(6k+4)]
(2k+1)!
(-1)k[7-13-19 (6k+1)]
(2k)!
(−1)k[6-12-18---(6k)]
(2k+1)!
B. azk=
C. azk=
D. azk=
A. azk+1 =
(-1)k[8.14-20(6k+2)]
(2k)!
B. azk+1 =
C. a₂k+1 =
-ao
ao
6. Which of the following is the expression for a2k+1 in terms of a₂?
D. azk+1 =
·ao
ao
(2k)!
(-1) [6-12-18 (6k)]
(2k+1)!
(-1) [10-16-22-(6k+4)]
(2k+1)!
(-1) [7-13-19--(6k+1)]
-a₁
(-1) [8-14-20-(6k+2)]
(2k)!
a₁
a₁
a₁
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