Complex Roots

Since the coefficients of the characteristic polynomial are real, complex roots must appear as complex conjugate pairs. Suppose r and r* (* = complex conjugate) are roots of the characteristic equation; then, corresponding to these roots the solutions may be written as

y^^, У2}=(г* )к.

If a real solution is desired, these solutions can be recasted into a real form. Let r = Rei0, then an alternative set of fundamental solutions is

у{1 = R cos (кв), у(к> = Як sin (кв).

If r and r* are repeated roots of multiplicity m, then the set of fundamental solutions corresponding to these roots is

ук) = Як cos (кв) yf+У = R sin (кв)

yk) = кЯк cos (кв) у{к"+2) = кЯк sin (кв)

к . к. (1.11)

у(т) = кт-1^ cos (кв ) ykm) = к“-1^к sin (кв ) .

example. Find the general solution of

yk+2 – 4yk+1 + 8yk = 0

The characteristic equation is

r2 – 4r + 8 = 0.

The roots are r = 2 ± 2i = 2V2 e±l(n /4). Therefore, the general solution (can be verified by direct substitution) is

yk = A(2V2)k cos kj + B(2V2)k srn(|kj,

where A and B are arbitrary constants.

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