Difference equations 27
3) If D c^2 - 4ab < 0, equation (46) possesses two complex-conjugate
roots
c+JIDTi.
q 1 =
2
b = p(cost.p + i sin <p) = pe''P,
c-JIDTi..
q 2 =
2
b = p(cos<p-i sin <p) = pe-''P,
where
p=~, <p = arctg c
The appropriate functions
or
are just particular solutions to equation ( 45). In this case both pairs of so-
lutions are linearly independent in connection with the linear independence
of the functions sin k<p and cos kt.p: l:!.k, k+i -:f 0. Then the general solution
acquires the form
Yk = /' ( C 1 cos kt.p + C 2 sin k<p).
In order to illustrate om· approach a little better, it seems worthwhile giving
several simple examples.
Example 1 We are interested in giving the general solution to the equation
Yk+l - 2pyk + Yk-t = 0, p > 0.
Still using the above framework with D = 4 (p^2 - 1), the following three
cases need investigation.
a) Let p < 1. Setting p = cos CY, CY -:f 0, and Yk = qk we obtain the quadratic
equation for q:
q^2 - 2 COS CY q + 1 = 0