6.5.8 Example H The equation 2yk V1– y? (6.111) Yk+1 = can be transformed into a simpler form by letting yk = sin ak; this gives sin xk+1 = sin 2xk (6.112) %3D the solution of which is Cek+1 = (-1)"2xk + nA, (6.113) %3D where n is an integer. There are two cases to consider. Let n = 2m be an even integer. Therefore, ( Xk+1 – 2xk = 2m . (6.114) and = c2k – 2mr. (6.115) Consequently, Yk = sin(c2* – 2mn) = sin(c2*). (6.116) %3D | Let n = 2m +1 be an odd integer. In this instance, we have k+1 +2xk = (2m + 1)7, ) Ck = c(-2)* + /3(2m + 1)7) (6.117) (6.118) %3D and Yk = sin[c(-2)k + 3(2m + 1)7]. (6.119)
6.5.8 Example H The equation 2yk V1– y? (6.111) Yk+1 = can be transformed into a simpler form by letting yk = sin ak; this gives sin xk+1 = sin 2xk (6.112) %3D the solution of which is Cek+1 = (-1)"2xk + nA, (6.113) %3D where n is an integer. There are two cases to consider. Let n = 2m be an even integer. Therefore, ( Xk+1 – 2xk = 2m . (6.114) and = c2k – 2mr. (6.115) Consequently, Yk = sin(c2* – 2mn) = sin(c2*). (6.116) %3D | Let n = 2m +1 be an odd integer. In this instance, we have k+1 +2xk = (2m + 1)7, ) Ck = c(-2)* + /3(2m + 1)7) (6.117) (6.118) %3D and Yk = sin[c(-2)k + 3(2m + 1)7]. (6.119)
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 purple
![6.5.8
Example H
The equation
Yk+1 =
- 2yk V1 – y%
(6.111)
can be transformed into a simpler form by letting yk
sin xk;
this gives
sin xk+1
sin 2xk
(6.112)
the solution of which is
Pk+1 = (-1)"2xk + na,)
(6.113)
where n is an integer. There are two cases to consider.
Let n = 2m be an even integer. Therefore,
( xk+1 - 20k
= 2m7
(6.114)
and
)
Xk =
c2k – 2mn.
(6.115)
Consequently,
Yk
sin(c2* – 2m7) = sin(c2*).
(6.116)
Let n = 2m +1 be an odd integer. In this instance, we have
C:
(6.117)
Tk+1 + 2^k — (2m + 1)п,
c(-2)* + /3(2m+ 1)T)
(6.118)
and
Yk = sin[c(-2)* + 1/3(2m + 1)7],
(6.119)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52e77568-6ab0-477a-b8a0-d1d660cfb901%2F53b77976-f8bc-4f53-87a2-ff22a9c7296d%2Fgw8ywxm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:6.5.8
Example H
The equation
Yk+1 =
- 2yk V1 – y%
(6.111)
can be transformed into a simpler form by letting yk
sin xk;
this gives
sin xk+1
sin 2xk
(6.112)
the solution of which is
Pk+1 = (-1)"2xk + na,)
(6.113)
where n is an integer. There are two cases to consider.
Let n = 2m be an even integer. Therefore,
( xk+1 - 20k
= 2m7
(6.114)
and
)
Xk =
c2k – 2mn.
(6.115)
Consequently,
Yk
sin(c2* – 2m7) = sin(c2*).
(6.116)
Let n = 2m +1 be an odd integer. In this instance, we have
C:
(6.117)
Tk+1 + 2^k — (2m + 1)п,
c(-2)* + /3(2m+ 1)T)
(6.118)
and
Yk = sin[c(-2)* + 1/3(2m + 1)7],
(6.119)
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