then 7(1 - а)М? + 6(1 - а) Мт %3D ВM?, 7(1 - а)т? + 6(1 — а)Мт %3D Вт?, Subtracting we obtain Y(1 – a)(M² – m²) = 3(M² – m²), (1- a) + ß. Thus М — т.
then 7(1 - а)М? + 6(1 - а) Мт %3D ВM?, 7(1 - а)т? + 6(1 — а)Мт %3D Вт?, Subtracting we obtain Y(1 – a)(M² – m²) = 3(M² – m²), (1- a) + ß. Thus М — т.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Show me the steps of determine red and information is here step by step
![or,
BM2
Вт?
M(1 – a) =
т(1 — а)
YM + Sm'
Ym + 8M'
then
7(1 - а)М? + 6(1 — а) Мт %—D ВМ?, (1- а)т? + ӧ(1 — а)Мт %— Вт?,
Subtracting we obtain
- a)(M² – m²) = B(M² – m²), (1– a) + ß.
Thus
CN=m.
It follows by Theorem B that ī is a global attractor of Eq.(1) and then the proof is
complete.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2F0436d8be-c9f1-44fe-b03e-5e9311bc0294%2Fd2t7gmq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:or,
BM2
Вт?
M(1 – a) =
т(1 — а)
YM + Sm'
Ym + 8M'
then
7(1 - а)М? + 6(1 — а) Мт %—D ВМ?, (1- а)т? + ӧ(1 — а)Мт %— Вт?,
Subtracting we obtain
- a)(M² – m²) = B(M² – m²), (1– a) + ß.
Thus
CN=m.
It follows by Theorem B that ī is a global attractor of Eq.(1) and then the proof is
complete.
![Bxn-10n-2
Xn+1 = axn-2+
п 3D 0, 1, ...,
(1)
YXn-1 + dxn-4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2F0436d8be-c9f1-44fe-b03e-5e9311bc0294%2Fbic17d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Bxn-10n-2
Xn+1 = axn-2+
п 3D 0, 1, ...,
(1)
YXn-1 + dxn-4
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