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+1+ k Yk = 0. - 1 (3.151) Yk+2 k We will now use this to determine the general solution to the inhomogeneous equation 2k +1 k Yk+2 Yk+1+ Yk = k(k + 1). (3.152) k - 1

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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 = 0.
- 1
(3.151)
Yk+2
Yk+1+
We will now use this to determine the general solution to the inhomogeneous
equation
2k + 1
k
Yk+1+
Yk
k
1
k(k + 1).
Yk+2
(3.152)
k
With the identification
k
Yk
Yk
Uk,
R: = k(k + 1),
(3.153)
k
equation (3.120) becomes
(k + 1)Auk+1 – kAuz = k(k + 1).
(3.154)
This equation has the solution
Auk = A/k + /3(k² – 1),
(3.155)
where A is an arbitrary constant. If we define
k-1
$(k)
(3.156)
IWI
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 = 0. - 1 (3.151) Yk+2 Yk+1+ We will now use this to determine the general solution to the inhomogeneous equation 2k + 1 k Yk+1+ Yk k 1 k(k + 1). Yk+2 (3.152) k With the identification k Yk Yk Uk, R: = k(k + 1), (3.153) k equation (3.120) becomes (k + 1)Auk+1 – kAuz = k(k + 1). (3.154) This equation has the solution Auk = A/k + /3(k² – 1), (3.155) where A is an arbitrary constant. If we define k-1 $(k) (3.156) IWI
(1)
(1)
Y+2(uk+2 – Uk+1) – qkY" (Uk+1 – Uk) = Rk,
(3.119)
or
(1)
.(1)
Yk+2Auk+1 – qkY'Auk
R.
(3.120)
Lot r.
Au: thoroforo
Transcribed Image Text:(1) (1) Y+2(uk+2 – Uk+1) – qkY" (Uk+1 – Uk) = Rk, (3.119) or (1) .(1) Yk+2Auk+1 – qkY'Auk R. (3.120) Lot r. Au: thoroforo
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