The equation Yk+1Yk – 2yk = -2 can be transformed into the linear equation X'k+2 - 2xk+1 + 2xk = 0 by means of the substitution Yk = Xk+1/xk -
The equation Yk+1Yk – 2yk = -2 can be transformed into the linear equation X'k+2 - 2xk+1 + 2xk = 0 by means of the substitution Yk = Xk+1/xk -
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 yellow
![6.3.3 Example C
The equation
Yk+1Yk
2yk = -2
(6.38)
can be transformed into the linear equation
Xk+2 – 2xk+1+2xk = 0
(6.39)
by means of the substitution
Yk = Xk+1/xk -
(6.40)
The characteristic equation for equation (6.39) is
2r + 2 = 0,
(6.41)
and its two complex conjugate roots are
r1,2 = 1±i = V2e±ir/4
(6.42)
Therefore, the general solution of equation (6.39) is
/2
æk = 26 [D1 cos(Tk/4) + D2 sin(Tk/4)],
(6.43)
and
5D1 cos[T(k + 1)/4] + D2 sin[r(k +1)/4]
Di cos(Tk/4) + D2 sin(īk/4)
Yk =
(6.44)
200
Difference Equations
Now, define the constant a such that in the interval -T/2 < a < T/2,
- D2/D1.
(6.45)
tan a
With this result, equation (6.44) becomes
V2 cos(7/4)(k+1) – a]
cos(Tk/4 – a)
Yk
(6.46)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25a3cfb3-619d-471e-bb5a-8c85b1316cb2%2F6aa299f4-10dc-4d20-ae69-4a051d4eac0e%2Fal695n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:6.3.3 Example C
The equation
Yk+1Yk
2yk = -2
(6.38)
can be transformed into the linear equation
Xk+2 – 2xk+1+2xk = 0
(6.39)
by means of the substitution
Yk = Xk+1/xk -
(6.40)
The characteristic equation for equation (6.39) is
2r + 2 = 0,
(6.41)
and its two complex conjugate roots are
r1,2 = 1±i = V2e±ir/4
(6.42)
Therefore, the general solution of equation (6.39) is
/2
æk = 26 [D1 cos(Tk/4) + D2 sin(Tk/4)],
(6.43)
and
5D1 cos[T(k + 1)/4] + D2 sin[r(k +1)/4]
Di cos(Tk/4) + D2 sin(īk/4)
Yk =
(6.44)
200
Difference Equations
Now, define the constant a such that in the interval -T/2 < a < T/2,
- D2/D1.
(6.45)
tan a
With this result, equation (6.44) becomes
V2 cos(7/4)(k+1) – a]
cos(Tk/4 – a)
Yk
(6.46)
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