4. Use power series to find two linearly independent solutions of this differential equation (x - 1)y" + 3y = 0. a. Find the recurrence relation needed for the solution. b. Find the first three terms of the two linearly independent solutions.
4. Use power series to find two linearly independent solutions of this differential equation (x - 1)y" + 3y = 0. a. Find the recurrence relation needed for the solution. b. Find the first three terms of the two linearly independent solutions.
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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Please solve #4 doing similar work to question #1 in the first pic
Transcribed Image Text:4. Use power series to find two linearly independent solutions of this differential equation
(x-1)y" + 3y = 0.
a. Find the recurrence relation needed for the solution.
b. Find the first three terms of the two linearly independent solutions. 14 points
![1. Find the recurrence relation needed to find a power series solution for the differential equation
y" + 3y = 0. You do not need to find the actual solution to the equation, just the recurrence
relation.
∞
y=
Σε
Cnx, y = Encnxn-1, y" = En(n-1cnxn-2
n=0
n=1
n=2
y" + 3y = 0
∞
Στη
1)Cnxn-2 + 3 Σ
Cnxn = 0
η=2
n=0
00
∞
Σ« + 2
+ 1)ck+2** +
3ckx* = 0
k=0
k=0
00
Στα + 2)(k + 1)ck+2** + 3cx*] = 0
k=0
0
Στα + 2 + 1)ck+2 + 3ck]x* = 0
k=0
(k + 2)(k + 1)ck+2 + 3ck = 0 → (k + 2)(k + 1)ck+2 = −3ck → Ck+2 =
-
-
Ck+2
=3ck
for k = 0, 1, 2, ...
(k+2)(k+1)
η(η –
–3ck
(k + 2)(k + 1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd936c6c0-8e3b-4a92-9646-2c460fec57ae%2Fa21d0160-6b11-49e1-a4f7-cd10f0d6d8a9%2Ftnqkmu4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. Find the recurrence relation needed to find a power series solution for the differential equation
y" + 3y = 0. You do not need to find the actual solution to the equation, just the recurrence
relation.
∞
y=
Σε
Cnx, y = Encnxn-1, y" = En(n-1cnxn-2
n=0
n=1
n=2
y" + 3y = 0
∞
Στη
1)Cnxn-2 + 3 Σ
Cnxn = 0
η=2
n=0
00
∞
Σ« + 2
+ 1)ck+2** +
3ckx* = 0
k=0
k=0
00
Στα + 2)(k + 1)ck+2** + 3cx*] = 0
k=0
0
Στα + 2 + 1)ck+2 + 3ck]x* = 0
k=0
(k + 2)(k + 1)ck+2 + 3ck = 0 → (k + 2)(k + 1)ck+2 = −3ck → Ck+2 =
-
-
Ck+2
=3ck
for k = 0, 1, 2, ...
(k+2)(k+1)
η(η –
–3ck
(k + 2)(k + 1)
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