Now consider the case where one or more of the roots of the characteristic equation are repeated. Suppose the root r1 has multiplicity m1s the root r2 has multiplicity m2, and the root rt has multiplicity mt such that
m1 + m2 +——- -me = n. (1.8)
The characteristic equation can be written as
(r – r1 )m1 (r – r2 )m2… (r – rl )mi = 0. (1.9)
Corresponding to a repeated root of the characteristic polynomial (1.9) of multiplicity m, the solution is
ук = (A1 + A2k + A3k ————— Amk>n ^ r, (1.10)
where A1, A2,…, Am are arbitrary constants.
example. Consider the general solution of the equation
Ук+2 – 6Ук+1 + 9Ук = 0.
The characteristic equation is
r2 – 6r + 9 = 0 or (r – 3)2 = 0.
Thus, there is a repeated root r = 3, 3. The general solution is
Ук = (A + Bk) 3к.