6.5.6 Example F Consider the equation Yk+1YkYk-1 = A? (yk+1 + Yk + Yk-1), (6.98) where A is a constant. If we make use of the relationship tan 01 + tan 02 tan(01 + 02) : (6.99) 1 - tan 01 tan 02 and set Yk = A tan xk, (6.100) then equation (6.98) becomes tan(ek+1 + Xk + *k-1) = 0. (6.101) This last equation has the complete solution Ck+1 +xk + xk-1 = nT, (6.102) where n is an integer. The two roots to the characteristic equation corresponding to equation (6.102) are $ = 27/3. (6.103) ri = e' r2 = e Therefore, Xk = c1 cos(2kT/3) + c2 sin(2kT/3) + nT/3, (6.104) and Yk A tan[cı cos(2kT /3) + c2(2kT/3) +nn/3]. (6.105) where m is an integer, equation (6.105) shows Note that up to factors of that there are three classes of solutions, corresponding to the "phase," na/3, being equal to 0, T/3, or 27/3. MT,
6.5.6 Example F Consider the equation Yk+1YkYk-1 = A? (yk+1 + Yk + Yk-1), (6.98) where A is a constant. If we make use of the relationship tan 01 + tan 02 tan(01 + 02) : (6.99) 1 - tan 01 tan 02 and set Yk = A tan xk, (6.100) then equation (6.98) becomes tan(ek+1 + Xk + *k-1) = 0. (6.101) This last equation has the complete solution Ck+1 +xk + xk-1 = nT, (6.102) where n is an integer. The two roots to the characteristic equation corresponding to equation (6.102) are $ = 27/3. (6.103) ri = e' r2 = e Therefore, Xk = c1 cos(2kT/3) + c2 sin(2kT/3) + nT/3, (6.104) and Yk A tan[cı cos(2kT /3) + c2(2kT/3) +nn/3]. (6.105) where m is an integer, equation (6.105) shows Note that up to factors of that there are three classes of solutions, corresponding to the "phase," na/3, being equal to 0, T/3, or 27/3. MT,
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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Explain the determine blue
![6.5.6 Example F
Consider the equation
Yk+1YkYk–1 =
A (yk+1 + Yk + Yk-1),
(6.98)
where A is a constant. If we make use of the relationship
tan 01 + tan02
tan(01 + 02)
(6.99)
tan 01 tan 02
and set
Yk
A tan xk,
(6.100)
then equation (6.98) becomes
tan(xk+1 + Xk + xk-1) = 0.
(6.101)
This last equation has the complete solution
Ck+1 + xk +xk-1 = T,
(6.102)
where n is an integer.
The two roots to the characteristic equation corresponding to equation
(6.102) are
ri = e
r2 = e
-io
0 = 27/3.
(6.103)
Therefore,
Th — C1 сos(2kт/3) + с2 sin(2kт /3) + пп/3,
(6.104)
and
(ук3D A tan(c cos(2kr/3) + сә(2kт/3) + пп/3].
(6.105)
Yk =
Note that up to factors of mT, where m is an integer, equation (6.105) shows
that there are three classes of solutions, corresponding to the "phase," na/3,
being equal to 0, 1/3, or 27/3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52e77568-6ab0-477a-b8a0-d1d660cfb901%2F19eb6da0-7eec-4934-a589-2854493a8619%2Frzj7fro_processed.jpeg&w=3840&q=75)
Transcribed Image Text:6.5.6 Example F
Consider the equation
Yk+1YkYk–1 =
A (yk+1 + Yk + Yk-1),
(6.98)
where A is a constant. If we make use of the relationship
tan 01 + tan02
tan(01 + 02)
(6.99)
tan 01 tan 02
and set
Yk
A tan xk,
(6.100)
then equation (6.98) becomes
tan(xk+1 + Xk + xk-1) = 0.
(6.101)
This last equation has the complete solution
Ck+1 + xk +xk-1 = T,
(6.102)
where n is an integer.
The two roots to the characteristic equation corresponding to equation
(6.102) are
ri = e
r2 = e
-io
0 = 27/3.
(6.103)
Therefore,
Th — C1 сos(2kт/3) + с2 sin(2kт /3) + пп/3,
(6.104)
and
(ук3D A tan(c cos(2kr/3) + сә(2kт/3) + пп/3].
(6.105)
Yk =
Note that up to factors of mT, where m is an integer, equation (6.105) shows
that there are three classes of solutions, corresponding to the "phase," na/3,
being equal to 0, 1/3, or 27/3.
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