L(yk) := a0yk+d + a1yk+d-1 + ... + adyk = gk
See Rosen 6/e section 7.2, esp. Theorems 4 and 6, or Introduction to Difference Equations p. 146.
Characteristic equation (auxiliary equation)
a0 r2 + a1r + a2 = 0 if order/degree d = 2.
Roots of the characteristic equation
General solution of the homogeneous equation
Particular solution of the nonhomogeneous equation: See below for trial solutions.
General solution of the nonhomogeneous equation
Solve the initial-condition (or boundary-condition) equations for the constants in the general solution.
Solution of the given nonhomogeneous equation satisfying the given initial conditions
Check: Compare values of yk for the recurrence plus initial conditions with the values of your solution for k = 0, 1, 2, ....
Trial solutions for the nonhomogeneous equation L(yk)=gk
|Right-hand term gk||Trial solution yk*|
|Geometric sequence, a rk||C rk|
or polynomial in k of degree n
| Polynomial in k of degree n:
A0 + A1k + A2k2 + ... + Ankn
or geometric times polynomial
| Geometric times polynomial:
rk( A0 + A1k + A2k2 + ... + Ankn )
| sin(bk) or cos(bk)
or linear combination thereof
|C sin(bk) + D cos(bk)|
When the trial solution has one or more terms in common with the general solution of the homogeneous equation, we must modify the trial solution by multiplying it by kp where p is a just-high-enough integer power to make sure that the trial solution does not have any terms in common with the homogeneous solution. See Goldberg p. 146 for an example.