Lecture Notes in Differential Equations - Bruce E. Shapiro

201
Theorem 23.5. (D 2 − 2aD + (a 2 + b 2 )) n annihilates t n−1 e at cos bt and
t n−1 e at sin bt.
Proof. For n=1.
(D 2 − 2aD + (a 2 + b 2 )) 1 t 1−1 e at sin bt
= (D 2 − 2aD + a 2 + b 2 )(e at sin bt) (23.24)
= ((D − a) 2 + b 2 )(e at sin bt) (23.25)
= (D − a)(D − a)(e at sin bt) + b 2 e at sin bt (23.26)
= (D − a)(ae at sin bt + be at cos bt − ae at sin bt) + b 2 e at sin bt (23.27)
= (D − a)(be at cos bt) + b 2 e at sin bt (23.28)
= abe at cos bt − b 2 e at sin bt − abe at cos bt + b 2 e at sin bt (23.29)
= 0 (23.30)
For general n, assume that (D 2 − 2aD + (a 2 + b 2 )) n t n−1 e at cos bt = 0 and
similarly for sin bt. Consider first
(D 2 − 2aD + (a 2 + b 2 ))t n e at cos bt (23.31)
= [(D − a) 2 + b 2 ](t n e at cos bt) (23.32)
= (D − a) 2 (t n e at cos bt) + b 2 (t n e at cos bt) (23.33)
= (D − a)[nt n−1 e at cos bt + at n e at cos bt − bt n e at sin bt]
+ b 2 (t n e at cos bt) (23.34)
= n(n − 1)t n−2 e at cos bt + nat n−1 e at cos bt − nbt n−1 e at sin bt
+ nat n−1 e at cos bt + a 2 t n e at cos bt − abt n e at sin bt
− bnt n−1 e at sin bt − abt n e at sin bt − b 2 t n e at cos bt
− ant n−1 e at cos bt − a 2 t n e at cos bt + abt n e at sin bt
+ b 2 t n e at cos bt (23.35)
= n(n − 1)t n−2 e at cos bt + nat n−1 e at cos bt − nbt n−1 e at sin bt
− abt n e at sin bt − bnt n−1 e at sin bt (23.36)
The last line only contains terms such as
t n−1 e at cos bt (23.37)
t n−1 e at sin bt (23.38)
t n−1 e at cos bt (23.39)