where αn, β nare to be determined. There are two unknowns and consequently two
conditions are needed to determine these quantities. As with ordinary differential
equations, assume for our first condition the relation
∆αnun+1 + ∆ βnvn+1 = 0. (10 .62)
The second condition is obtained by substituting the assumed solution, given by
equation (10.61), into the given difference equation. Starting with the assumed
solution given by equation (10.61) show
yn+1 =yn+ ∆ yn=αnun+βnvn+ ∆(αnun+βnvn)
yn+1 =αnun+βnvn+αn∆un+βn∆vn+ [(∆ αn)un+1 + (∆ βn)vn+1].
This equation simplifies since by assumption equation (10.62) must hold. One can
then show that yn+1 reduces to
yn+1 =αnun+1 +βnvn+1. (10 .63)
In equation (10.63) replace nby n+ 1 everywhere and establish the result
yn+2 =αn+1un+2 +βn+1vn+2. (10 .64)
By substituting equations (10.61), (10.63) and (10.64) into the equation (10.60),
a second condition for determining the unknown constants is found. This second
condition is that αn and βnmust satisfy the equation
αn+1un+2 +βn+1vn+2 +a 1 (n)(αnun+1 +βnvn+1) + a 2 (n)(αnun+βnvn) = fn.
Rearrange terms in this equation, and show it can be written in the form
(αn+1 −αn)un+2 + (βn+1 −βn)vn+2 +αnL(un) + βnL(vn) = fn. (10 .65)
By hypothesis L(un) = 0 and L(vn) = 0, thus simplifying the equation (10.65). The
equations (10.62) and (10.65) are produce the two conditions
∆αnun+1 + ∆ βnvn+1 = 0
∆αnun+2 + ∆ βnvn+2 =fn
for determining the constants αn and βn. This system of equations can be solve by
Cramer’s rule and written
∆αn=αn+1 −αn=−fnvn+1
Cn+1
, ∆βn=βn+1 −βn=fnun+1
Cn+1
, (10 .66)
