The equation Yk+1Yk – 2yk = -2 (6.38) can be transformed into the linear equation Xk+2 - 2xk+1+ 2xk = 0 (6.39) %3D 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±ir/4. (6.42) Therefore, the general solution of equation (6.39) is æk = 2*/2[D1 cos(Tk/4) + D2 sin(tk/4)], (6.43)
The equation Yk+1Yk – 2yk = -2 (6.38) can be transformed into the linear equation Xk+2 - 2xk+1+ 2xk = 0 (6.39) %3D 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±ir/4. (6.42) Therefore, the general solution of equation (6.39) is æk = 2*/2[D1 cos(Tk/4) + D2 sin(tk/4)], (6.43)
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Chapter2: Second-order Linear Odes
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Explain the determine yellow
![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±in/4
(6.42)
Therefore, the general solution of equation (6.39) is
X = 2*/2[D1 cos(rk/4)+ D2 sin(k/4)],
(6.43)
and
Di cos(T(k+1)/4] + D2 sin[r(k +1)/4]
Di cos(Tk/4) + D2 sin(Tk/4)
Yk =
(6.44)
200
Difference Equations
Now, define the constant a such that in the interval -7/2 < 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(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(T/4 – a),
tan(T/2 – a),
Y1 =
(6.48)
Y2
= -
Y3
1- tan(37/4 – 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%2Ff881a6e8-0e99-4f44-a53a-8cbadf3de3d0%2Fn20sbi9_processed.jpeg&w=3840&q=75)
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±in/4
(6.42)
Therefore, the general solution of equation (6.39) is
X = 2*/2[D1 cos(rk/4)+ D2 sin(k/4)],
(6.43)
and
Di cos(T(k+1)/4] + D2 sin[r(k +1)/4]
Di cos(Tk/4) + D2 sin(Tk/4)
Yk =
(6.44)
200
Difference Equations
Now, define the constant a such that in the interval -7/2 < 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(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(T/4 – a),
tan(T/2 – a),
Y1 =
(6.48)
Y2
= -
Y3
1- tan(37/4 – a).
An easy calculation shows that yo = Y4.
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