Lecture Notes in Differential Equations - Bruce E. Shapiro

228 LESSON 27. HIGHER ORDER EQUATIONS
initial value problem is
L n y = f(t)
where y 0 , ..., y n ∈ R are constants.
y(t 0 ) = y 0
y ′ (t 0 ) = y 1
.
⎫⎪ ⎬
⎪ ⎭
(27.7)
y (n) (t 0 ) = y n
While the roots of P n (r) = 0 may be complex we will restrict the coefficients
and initial conditions to be real. The linear operator has the following
properties:
L n (y) = P n (D)y (27.8)
L n (e rt ) = e rt P n (r) (27.9)
L n (e rt y) = e rt P n (D + r)y (27.10)
Equations (27.8) and (27.9) are straightforward; (27.10) can be proven by
induction.
Proof of (27.10) by induction (inductive step only). ∗ Assume L n (e rt y) =
e rt P n (D + a). Then
where
L n+1 (e rt y) = a n+1 (D − r 1 ) · · · (D − r n+1 )e rt y (27.11)
= (D − r 1 )z (27.12)
z = a n+1 (D − r 2 )(D − r 3 ) · · · (D − r n+1 )(e rt y) (27.13)
Since (27.12) is an nth order equation, the inductive hypothesis holds for
it, namely, that
z = e rt a n+1 (D + r − r 2 )(D + r − r 3 ) · · · (D + r − r n+1 )y = e rt u (27.14)
where
Substituting,
u = a n+1 (D + r − r 2 ) · · · (D + r − r n+1 )y (27.15)
L n+1 (e rt y) = (D − r 1 )z = (D − r 1 )e rt u
= re rt u + e rt u ′ − r 1 e rt u
= e rt (D + r − r 1 )u
(27.16)
Substituting back for u from (27.15) and applying the definition of P n+1 (x)