Lecture Notes in Differential Equations - Bruce E. Shapiro

200 LESSON 23. METHOD OF ANNIHILATORS
Since
D1 = 0 =⇒ D annihilates 1 (23.6)
D 2 t = 0 =⇒ D 2 annihilates t (23.7)
D 3 t 2 = 0 =⇒ D 3 annihilates t 2 (23.8)
. (23.9)
D n t n−1 = 0 =⇒ D n annihilates t k−1 (23.10)
Theorem 23.3. (D − a) n annihilates t n−1 e at .
Proof. (induction). For n = 1, the theorem states that D − a annihilates
e at . To verify this observe that
(D − a)e at = De at − ae at = ae at − ae at = 0 (23.11)
hence the conjecture is true for n = 1.
Inductive step: Assume that (D − a) n annihilates t n−1 e at . Thus
Consider
(D − a) n t n−1 e at = 0 (23.12)
(D − a) n+1 t n e at = (D − a) n (D − a)t n e at (23.13)
where the last line follows from (23.12).
= (D − a) n (Dt n e at − at n e at ) (23.14)
= (D − a) n (nt n−1 e at + t n ae at − at n e at ) (23.15)
= (D − a) n nt n−1 e at (23.16)
= n(D − a) n t n−1 e at (23.17)
= 0 (23.18)
Theorem 23.4. (D 2 +a 2 ) annihilates any linear combination of cos ax and
sin ax
Proof.
(D 2 + a 2 )(A sin at + B cos at) = AD 2 sin at + Aa 2 sin at (23.19)
+ BD 2 cos at + a 2 B cos at (23.20)
= −Aa 2 sin at + Aa 2 sin at+ (23.21)
− Ba s cos at + a 2 B cos at (23.22)
= 0 (23.23)