application of this method considers the Clairaut difference equation As an Yk = kAyk + f(Ayk), (6.52)

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Explain the determine purple and the eqaution is here

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)
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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
6.4.1
Example
Let the function f be equal to (Ayk)²; therefore, equation (6.52) becomes
Yk = kxk + x,
(6.61)
where we have substituted x =
Ayk. Operating with A gives
(k +1)A¤k + 2xrAær + (Aæx)² =
= 0.
(6.62)
Thus, we conclude that either
Axk = 0 and
Yk = ck + c²,
(6.63)
%3D
or
Ark + 2xk + k +1= xk+1 + xk + k +1 = 0.
(6.64)
The solution to the last equation is
= c(-1)* – 1/½k – 1/4,
(6.65)
which gives for equation (6.61) the second solution
Yk = [c(-1)* – 1/4]² – 1¼k².
(6.66)
Transcribed Image Text:6.4.1 Example Let the function f be equal to (Ayk)²; therefore, equation (6.52) becomes Yk = kxk + x, (6.61) where we have substituted x = Ayk. Operating with A gives (k +1)A¤k + 2xrAær + (Aæx)² = = 0. (6.62) Thus, we conclude that either Axk = 0 and Yk = ck + c², (6.63) %3D or Ark + 2xk + k +1= xk+1 + xk + k +1 = 0. (6.64) The solution to the last equation is = c(-1)* – 1/½k – 1/4, (6.65) which gives for equation (6.61) the second solution Yk = [c(-1)* – 1/4]² – 1¼k². (6.66)
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