Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Explain the determine purple and the equation is here
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)
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![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)](https://content.bartleby.com/qna-images/question/57a26049-260e-45aa-910f-dd852150673b/322d6936-ff8a-4046-acf2-fe5d5532502d/7msan3_processed.jpeg)
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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