Explain the determaine 3.5.3 Example CThe second-order equationYk+2 – Yk+1 – k´yk(3.159)can be written in the following factored form:(E+k)(E – k)yk == 0.(3.160)Note that(E – k)(E + k)yYkYk+2 + Yk+1 – kʻyk,(3.161)which shows that the order of the factors is important.To solve equation (3.160), let21(k) = (E – k)Yk;(3.162)therefore, equation (3.160) becomes(E + k)z1(k) = 0,(3.163)which has the solution21(k) = A(-1)*(k – 1)!,(3.164)where A is an arbitrary function. We now have(E – k)yk = A(-1)* (k – 1)!,(3.165)orYk+1 – kyk = A(-1)*(k – 1)!.(3.166)The solution to this latter equation isk-1A(-1)*(i – 1)!(k – 1)! >+ B(3.167)Yk =i=1Therefore, the general solution to equation (3.159) isk-1(-1)*+ B (k – 1)!.(3.168)Yk

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Description: Explain the determaine 3.5.3 Example CThe second-order equationYk+2 – Yk+1 – k´yk(3.159)can be written in the following factored form:(E+k)(E – k)yk == 0.(3.160)Note that(E – k)(E + k)yYkYk+2 + Yk+1 – kʻyk,(3.161)which shows that the order of the fac

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which has the solution 21(k) = A(-1)*(k – 1)!, where A is an arbitrary function. We now have (E – k)yk = A(-1)*(k – 1)!, | or Yk+1 – kyk = A(-1)* (k – 1)!. The solution to this latter equation is k = (k – 1)! (A(-1)*(i – 1)! i! +B Therefore, the general solution to equation (3.159) is k-1 Yk = | A£ (-1)* +B (k – 1)!. i i=1

which has the solution 21(k) = A(-1)(k – 1)!, where A is an arbitrary function. We now have (E – k)yk = A(-1)(k – 1)!, | or Yk+1 – kyk = A(-1)* (k – 1)!. The solution to this latter equation is k = (k – 1)! (A(-1)(i – 1)! i! +B Therefore, the general solution to equation (3.159) is k-1 Yk = | A£ (-1) +B (k – 1)!. i i=1

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Explain the determaine

3.5.3 Example C
The second-order equation
Yk+2 – Yk+1 – k´yk
(3.159)
can be written in the following factored form:
(E+k)(E – k)yk =
= 0.
(3.160)
Note that
(E – k)(E + k)yYk
Yk+2 + Yk+1 – kʻyk,
(3.161)
which shows that the order of the factors is important.
To solve equation (3.160), let
21(k) = (E – k)Yk;
(3.162)
therefore, equation (3.160) becomes
(E + k)z1(k) = 0,
(3.163)
which has the solution
21(k) = A(-1)*(k – 1)!,
(3.164)
where A is an arbitrary function. We now have
(E – k)yk = A(-1)* (k – 1)!,
(3.165)
or
Yk+1 – kyk = A(-1)*(k – 1)!.
(3.166)
The solution to this latter equation is
k-1
A(-1)*(i – 1)!
(k – 1)! >
+ B
(3.167)
Yk =
i=1
Therefore, the general solution to equation (3.159) is
k-1
(-1)*
+ B (k – 1)!.
(3.168)
Yk

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