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k
vk( + 1 ) = ∑ Bj() + C (2.37)
j=0 lj( + 1 )

where C is a constant.

Example 2.5
Find the general solution of the following first-order difference equation:

2
y(k + 1) – k y(k) = 0
with y(1) = 1.

Solution:

2
2
2
yk ( + 1 ) = k y k ( ) = k k ( − 1 ) 2 y k ( − 1 ) = k k ( − 1 ) 2 k ( − 2 ) 2 yk ( − 2 )
= kk ( − 1 ) 2 k ( − 2 ) 2 k ( − 3 ) 2 y k ( − 3 ) = …
2
= kk ( − 1 ) 2 k ( − 2 ) 2 k ( − 3 ) … 2 2 2 y() =1 k ( !) 2 2
2
2
( ) () 1
Example 2.6
Find the general solution of the following first-order difference equation:

(k + 1)y(k + 1) – ky(k) = k 2

with y(1) = 1.

Solution: Reducing this equation to the standard form, we have:

Ak () =− k and Bk = k 2
()
k + 1 k + 1
The homogeneous solution is given by:

lk( + 1 ) = k! = 1
k ( + 1 )! k ( + 1 )

The particular solution is given by:

k
∑ j 2 ∑ 2 k ( + 12 k + 1 k )
k
)(
vk( + 1 ) = j ( + 1 ) + C = j + C = + C
j=1 j ( + 1 ) j=1 6
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