Example C The second-order equation Yk+2 – Yk+1 – k²yk = 0 (3.159) - can be written in the following factored form: (E + k)(E – k)yk 0. (3.160) Note that (E – k)(E + k)yk = Yk+2 + Yk+1 – k²yk, (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)! E + B (3.167) Yk i! i=1 Therefore, the general solution to equation (3.159) is k-1 Yk = +B| (k – 1)!. (3.168) i=1
Example C The second-order equation Yk+2 – Yk+1 – k²yk = 0 (3.159) - can be written in the following factored form: (E + k)(E – k)yk 0. (3.160) Note that (E – k)(E + k)yk = Yk+2 + Yk+1 – k²yk, (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)! E + B (3.167) Yk i! i=1 Therefore, the general solution to equation (3.159) is k-1 Yk = +B| (k – 1)!. (3.168) 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 determine
Transcribed Image Text:Example C
The second-order equation
Yk+2 – Yk+1 – k²yk = 0
(3.159)
can be written in the following factored form:
(E + k)(E – k)yk = 0.
(3.160)
Note that
(E – k)(E+ k)Yk
= Yk+2 + Yk+1 – k“yk,
(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)
%3|
or
Yk+1 – kyr = A(-1)* (k – 1)!.
(3.166)
The solution to this latter equation is
(k-1
A(-1)*(i – 1)!
(k – 1)! E
+ B
(3.167)
Yk
i!
i=1
Therefore, the general solution to equation (3.159) is
k-1
(-1)*
+ B (k – 1)!.
(3.168)
Yk =
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