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 (6.39) = 0 by means of the substitution Yk = Xk+1/k. (6.40) The characteristic equation for equation (6.39) is p2 – 2r + 2 = 0, (6.41) and its two complex conjugate roots are r1,2 = 1+i = V2e±i#/4 (6.42) Therefore, the general solution of equation (6.39) is Xk = 2*/2[D1 cos(Tk/4)+ D2 sin(Tk/4)], (6.43)
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 (6.39) = 0 by means of the substitution Yk = Xk+1/k. (6.40) The characteristic equation for equation (6.39) is p2 – 2r + 2 = 0, (6.41) and its two complex conjugate roots are r1,2 = 1+i = V2e±i#/4 (6.42) Therefore, the general solution of equation (6.39) is Xk = 2*/2[D1 cos(Tk/4)+ D2 sin(Tk/4)], (6.43)
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
p? – 2r + 2 = 0,
(6.41)
and its two complex conjugate roots are
r1,2 = 1±i = v
VZetin/4.
(6.42)
Therefore, the general solution of equation (6.39) is
æ = 24/2 [D1 cos(īk/4) + D2 sin(Tk/4)],
(6.43)
and
vD1 cos(T(k+ 1)/4] + D2 sin[r(k +1)/4]
D1 cos(Tk/4) + D2 sin(Tk/4)
Yk
(6.44)
200
Difference Equations
Now, define the constant a such that in the interval -T/2 < a <T/2,
tan α
D2/D1.
(6.45)
With this result, equation (6.44) becomes
V2 cos[(T/4)(k + 1) – a]
Yk
cos(Tk/4 – a)
(6.46)
or
Yk
1 – tan(rk/4 – a).
(6.47)
This is the general solution to equation (6.38).
Note that since tan(0+7) = tan 0, the solution has period 4; i.e., for given
constant a, equation (6.47) takes on only four values; they are
Yo
1
tan(-a),
1- tan(7/4 – a),
- tan(T/2 – a),
Y1 =
(6.48)
Y2
Y3 = 1 – tan(3T/4 – a).
An easy calculation shows that yo = Y4.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25a3cfb3-619d-471e-bb5a-8c85b1316cb2%2F22dd25e2-013f-44af-8246-5ba3dbc0bbce%2Fbqetrb_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
p? – 2r + 2 = 0,
(6.41)
and its two complex conjugate roots are
r1,2 = 1±i = v
VZetin/4.
(6.42)
Therefore, the general solution of equation (6.39) is
æ = 24/2 [D1 cos(īk/4) + D2 sin(Tk/4)],
(6.43)
and
vD1 cos(T(k+ 1)/4] + D2 sin[r(k +1)/4]
D1 cos(Tk/4) + D2 sin(Tk/4)
Yk
(6.44)
200
Difference Equations
Now, define the constant a such that in the interval -T/2 < a <T/2,
tan α
D2/D1.
(6.45)
With this result, equation (6.44) becomes
V2 cos[(T/4)(k + 1) – a]
Yk
cos(Tk/4 – a)
(6.46)
or
Yk
1 – tan(rk/4 – a).
(6.47)
This is the general solution to equation (6.38).
Note that since tan(0+7) = tan 0, the solution has period 4; i.e., for given
constant a, equation (6.47) takes on only four values; they are
Yo
1
tan(-a),
1- tan(7/4 – a),
- tan(T/2 – a),
Y1 =
(6.48)
Y2
Y3 = 1 – tan(3T/4 – a).
An easy calculation shows that yo = Y4.
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