Yk+1Yk – 2yk - -2 into the linear o quation ea

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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6.3.3 Еxample 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
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
= 2*/2[D1 cos(Tk/4)+ D2 sin(rk/4)],
(6.43)
Xk
and
Di cos[r(k+ 1)/4] + D2 sin[r(k +1)/4]
Dị cos(Tk/4) + D2 sin(tk/4)
Yk
V2-
(6.44)
200
Difference Equations
Now, define the constant a such that in the interval -T/2 < a <T/2,
tan a =
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(Tk/4 – a).
(6.47)
This is the general solution to equation (6.38).
Note that since tan(0+T) = 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(T/4 – a),
1- tan(T/2 – a),
Y1 =
(6.48)
Y2
Y3
1- tan(37/4 – a).
%3D
An easy calculation shows that yo = Y4.
Transcribed Image Text:6.3.3 Еxample 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 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 = 2*/2[D1 cos(Tk/4)+ D2 sin(rk/4)], (6.43) Xk and Di cos[r(k+ 1)/4] + D2 sin[r(k +1)/4] Dị cos(Tk/4) + D2 sin(tk/4) Yk V2- (6.44) 200 Difference Equations Now, define the constant a such that in the interval -T/2 < a <T/2, tan a = 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(Tk/4 – a). (6.47) This is the general solution to equation (6.38). Note that since tan(0+T) = 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(T/4 – a), 1- tan(T/2 – a), Y1 = (6.48) Y2 Y3 1- tan(37/4 – a). %3D An easy calculation shows that yo = Y4.
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