Show me the steps of determine purple and the eqaution is here 6.4.1 ExampleLet the function f be equal to (Ayık)²; therefore, equation (6.52) becomes= kxk + x%,(6.61)where we have substituted xk =Ayk. Operating with A gives(k + 1)A¤k + 2x1Axk + (Axk)² = 0.(6.62)Thus, we conclude that eitherandYkck + c²,(6.63)orAxk + 2xk + k +1The solution to the last equation isXk+1+ Xk + k + 1= 0.(6.64)Xj = c(-1)* – 1/2k – 1/4,(6.65)which gives for equation (6.61) the second solutionYk = [c(-1)* – 1/4]² – 1/¼k².(6.66) ull stc ksa3:03 PMC @ 1 40%As an application of this method considers the Clairaut difference equationYk = kAyk + f (Ayk),(6.52)CancelActual Size (434 KB)Choose

"Linear" equations are equations with just a plain old variable like "x", rather than something more…

6.4.1 Еxample Let the function f be equal to (Ayr)²; therefore, equation (6.52) becomes Yk = kx + x, where we have substituted x = Ayk. Operating with A gives (6.61) (k + 1)Aæk + 2xAxk + (Axk)² = 0. (6.62) Thus, we conclude that either

6.4.1 Еxample Let the function f be equal to (Ayr)²; therefore, equation (6.52) becomes Yk = kx + x, where we have substituted x = Ayk. Operating with A gives (6.61) (k + 1)Aæk + 2xAxk + (Axk)² = 0. (6.62) Thus, we conclude that either

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Chapter1: Introduction

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Question

Show me the steps of determine purple and the eqaution is here

6.4.1 Example
Let the function f be equal to (Ayık)²; therefore, equation (6.52) becomes
= kxk + x%,
(6.61)
where we have substituted xk =
Ayk. Operating with A gives
(k + 1)A¤k + 2x1Axk + (Axk)² = 0.
(6.62)
Thus, we conclude that either
and
Yk
ck + c²,
(6.63)
or
Axk + 2xk + k +1
The solution to the last equation is
Xk+1+ Xk + k + 1= 0.
(6.64)
Xj = c(-1)* – 1/2k – 1/4,
(6.65)
which gives for equation (6.61) the second solution
Yk = [c(-1)* – 1/4]² – 1/¼k².
(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

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