3.5.2 Example B (1) It is straightforward to show that y = k-1 is a solution to the homogeneous equation 2k +1 k Yk = 0. (3.151) Yk+2 -Yk+1+ k k - 1 We will now use this to determine the general solution to the inhomogeneous equation 2k +1 k Yk+1+ k Yk = 1 = k(k +1). (3.152) Yk+2 k With the identification k (1) Uk, Yk = Yk Rx = k(k + 1), (3.153) Ik k – 1 equation (3.120) becomes (k + 1)Aur+1 – kAux = k(k + 1). (3.154) %3D This equation has the solution Auk = A/k+ /3(k² – 1), (3.155) where A is an arbitrary constant. If we define k-1 1 $(k) (3.156) i=1 104 Difference Equations then equation (3.155) has the solution (Uk = Ap(k) + B+½sk(k +1)(2k – 5), (3.157) where B is a second arbitrary constant. Therefore, the general solution of (3.152) is (1) Yk = Y Uk (k – 1)uk (3.158) = A¢(k)(k – 1) + B(k – 1) + 1/18(k – 1)k(k +1)(2k – 5). IWI

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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3.5.2
Example B
(1)
It is straightforward to show that y
equation
= k-1 is a solution to the homogeneous
2k +1
k
Yk+1+
Yk
k
- 1
Yk+2
= 0.
(3.151)
k
We will now use this to determine the general solution to the inhomogeneous
equation
2k +1
Yk+1+
k
Yk = k(k +1).
1
(3.152)
Yk+2
k
With the identification
k
(1)
Uk,
Yk = Yk
Ik =
k
Rk = k(k + 1),
1'
(3.153)
equation (3.120) becomes
(k + 1)Aur+1 – kAuk = k(k +1).
(3.154)
This equation has the solution
Auk = A/k + 1/3(k² – 1),
(3.155)
where A is an arbitrary constant. If we define
k-1
1
$(k)
(3.156)
104
Difference Equations
then equation (3.155) has the solution
( uk = Ap(k) +B+ ½18k(k+1)(2k – 5),
(3.157)
where B is a second arbitrary constant. Therefore, the general solution of
(3.152) is
(1)
Yk = Y Uk =
(k – 1)uk
(3.158)
= A¢(k)(k – 1) + B(k – 1) + 1/18(k – 1)k(k + 1)(2k – 5).
Transcribed Image Text:3.5.2 Example B (1) It is straightforward to show that y equation = k-1 is a solution to the homogeneous 2k +1 k Yk+1+ Yk k - 1 Yk+2 = 0. (3.151) k We will now use this to determine the general solution to the inhomogeneous equation 2k +1 Yk+1+ k Yk = k(k +1). 1 (3.152) Yk+2 k With the identification k (1) Uk, Yk = Yk Ik = k Rk = k(k + 1), 1' (3.153) equation (3.120) becomes (k + 1)Aur+1 – kAuk = k(k +1). (3.154) This equation has the solution Auk = A/k + 1/3(k² – 1), (3.155) where A is an arbitrary constant. If we define k-1 1 $(k) (3.156) 104 Difference Equations then equation (3.155) has the solution ( uk = Ap(k) +B+ ½18k(k+1)(2k – 5), (3.157) where B is a second arbitrary constant. Therefore, the general solution of (3.152) is (1) Yk = Y Uk = (k – 1)uk (3.158) = A¢(k)(k – 1) + B(k – 1) + 1/18(k – 1)k(k + 1)(2k – 5).
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