6.4.1 Example Let the function f be equal to (Ayk)²; therefore, equation (6.52) becomes Yk = kar + x?, (6.61) where we have substituted xk Ayk. Operating with A gives | (k + 1)Aak + 2xAxk + (Axk)² = 0. (6.62) %3D Thus, we conclude that either and Yk = ck +c², (6.63) or Axk + 2xk + k +1= xk+1 +Xk + k +1 = 0. (6.64) The solution to the last equation is (Ek = c(-1)* – 1/2k – 14, (6.65) which gives for equation (6.61) the second solution Yk = [c(-1)* – 1/4]² – 1¼k?. (6.66)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Explain the determine purple and the eqaution is here

6.4.1
Example
Let the function f be equal to (Ayk)²; therefore, equation (6.52) becomes
= kæk + xk,
(6.61)
where we have substituted xk
Ay. Operating with A gives
(k + 1)Axk + 2x ,Axx + (Axk)² = 0.
(6.62)
Thus, we conclude that either
and
Yk = ck + c?,
(6.63)
or
Axk + 2xk + k +1
Xk+1 + Xk + k +1= 0.
(6.64)
The solution to the last equation is
ak = c(-1)* – 1/2k – 1/4,
(6.65)
which gives for equation (6.61) the second solution
Yk =
[c(-1)* – 1/¼]² – 1/¾k?.
(6.66)
Transcribed Image Text:6.4.1 Example Let the function f be equal to (Ayk)²; therefore, equation (6.52) becomes = kæk + xk, (6.61) where we have substituted xk Ay. Operating with A gives (k + 1)Axk + 2x ,Axx + (Axk)² = 0. (6.62) Thus, we conclude that either and Yk = ck + c?, (6.63) or Axk + 2xk + k +1 Xk+1 + Xk + k +1= 0. (6.64) The solution to the last equation is ak = c(-1)* – 1/2k – 1/4, (6.65) which gives for equation (6.61) the second solution Yk = [c(-1)* – 1/¼]² – 1/¾k?. (6.66)
ull stc ksa
3:03 PM
C @ 1 40%
As an application of this method considers the Clairaut difference equation
Yk = kAyk + f (Ayk),
(6.52)
Cancel
Actual Size (434 KB)
Choose
Transcribed Image Text:ull stc ksa 3:03 PM C @ 1 40% As an application of this method considers the Clairaut difference equation Yk = kAyk + f (Ayk), (6.52) Cancel Actual Size (434 KB) Choose
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