Consider the following equation: f(E)yk = (E – 2)²yk = 2k. Che solution to the homogeneous equation is (H) : (c1 + c2k)2*. (4.232) Now f(E) has a double root and we must apply Theorem 4.9' again. [For this ase g(E) = 1 and m = 2.] Thus, 2k-212 k22k (P) = (E – 2)–2 . 2k (4.233) %3D (1)(?)
Consider the following equation: f(E)yk = (E – 2)²yk = 2k. Che solution to the homogeneous equation is (H) : (c1 + c2k)2*. (4.232) Now f(E) has a double root and we must apply Theorem 4.9' again. [For this ase g(E) = 1 and m = 2.] Thus, 2k-212 k22k (P) = (E – 2)–2 . 2k (4.233) %3D (1)(?)
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 me the steps of determine and Th is there
![4.6.4 Example D
Consider the following equation:
f(E)yk = (E – 2)²yk
2k.
The solution to the homogeneous equation is
(H)
Yk
: (c1 + c2k)2*.
(4.232)
Now f(E) has a double root and we must apply Theorem 4.9' again. [For this
case g(E) = 1 and m = 2.] Thus,
(P)
2k-2k2
k22k
= (E – 2)–2 . 2k
Yk
(4.233)
(1)(2!)
8.
and the general solution of equation (4.232) is
k22k
Yk = (C1 + c2k)2* +
8
(4.234)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd81067d-18d6-4c10-b87e-e427303cac90%2F00444d08-fcc4-42a1-8d15-d5f7f76332c7%2Fav3bjjf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4.6.4 Example D
Consider the following equation:
f(E)yk = (E – 2)²yk
2k.
The solution to the homogeneous equation is
(H)
Yk
: (c1 + c2k)2*.
(4.232)
Now f(E) has a double root and we must apply Theorem 4.9' again. [For this
case g(E) = 1 and m = 2.] Thus,
(P)
2k-2k2
k22k
= (E – 2)–2 . 2k
Yk
(4.233)
(1)(2!)
8.
and the general solution of equation (4.232) is
k22k
Yk = (C1 + c2k)2* +
8
(4.234)
Transcribed Image Text:Theorem 4.9'. Let f(a) = 0, where f(E) = (E – a)"g(E) and g(a) + 0.
Then
-m km
ak-
f(E)-'ak
(4.206)
g(a)m!
Proof. We assume that f(r) has a zero atr=a of multiplicity m. Therefore,
ak
f(E)-'a* = (E – a)-"g(E)¯'a* = (E – a)*
g(a)
-1k
-m
ak
- (aЕ — а)-
ak-m
1
g(a)
ak-m k(m)
-(E – 1) ™. 1
(4.207)
-m
g(a)
k-m
-m
1 =
g(a)
g(a) m!
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