Lecture Notes in Differential Equations - Bruce E. Shapiro

202 LESSON 23. METHOD OF ANNIHILATORS
But by the inductive hypothesis these are all annihilated by (D 2 − 2aD +
(a 2 + b 2 )) n . Hence (D 2 − 2aD + (a 2 + b 2 )) n+1 annihilates t n e at cos bt. A
similar argument applies to the sin bt functions, completing the proof by
induction.
Example 23.1. To solve the differential equation
y ′′ − 6y ′ + 8y = t (23.40)
We first solve the homogeneous equation. Its characteristic equation is
which has roots at 2 and 4, so
r 2 − 6y + 8 = 0 (23.41)
y H = C 1 e 2 t + C 2 e 4 t (23.42)
Then we observe that D 2 is an annihilator of t. We rewrite the differential
equation as
The characteristic equation is
(D 2 − 6D + 8)y = t (23.43)
D 2 (D 2 − 6D − 8)y = D 2 = 0 (23.44)
(23.45)
r 2 (r − 4)(r − 2) = 0 (23.46)
so the roots are 4,2,0, and 0, giving us additional particular solutions of
The general solution is
To find A and B we differentiate,
y P = At + B (23.47)
y = C 1 e 2 t + C 2 e 4 t + At + B (23.48)
Substituting into the original differential equation,
y P ′ = A (23.49)
y P ′′ = 0 (23.50)
0 − 6A + 8(At + B) = t (23.51)
Equating coefficients gives A = 1/8 and 8B = 6A = 3/4 =⇒ B = 3/32.
Hence
y = C 1 e 2 t + C 2 e 4 t + 1 8 t + 3 32
(23.52)
The constants C 1 and C 2 depend on initial conditions.