Lecture Notes in Differential Equations - Bruce E. Shapiro

230 LESSON 27. HIGHER ORDER EQUATIONS
The Homogeneous Equation
The solutions we found for second order equations generalize to higher order
equations. In fact, the solutions to the homogeneous equation are the
functions e rt , te rt ,..., t k−1 e rt where each r is a root of P n (r) with multiplicity
k, and these are the only solutions to the homogeneous equation, as
we prove in the following two theorems.
Theorem 27.3. Let r be a root of P n (r) with multiplicity k. Then
e rt , te rt , ..., t k−1 e rt are solutions to the homogeneous equation L n y = 0.
Proof. Since r is a root of P n (r), then P n (r) = 0. Hence
L n e rt = e rt P n (r) = 0 (27.20)
Suppose that r has multiplicity k. Renumber the roots r 1 , r 2 , ..., r n as
r 1 , r 2 , ..., r n−k , r, ..., r. Then
} {{ }
k times
P n (x) = a n (x − r 1 )(x − r 2 ) · · · (x − r n−k )(x − r) k (27.21)
Let s ∈ {0, 1, 2, ..., k−1} so that s < k is an integer. Then by the polynomial
shift property (27.10) and equation (27.21)
L n (e rt t s ) = e rt P n (D + r)t s (27.22)
because D k (t s ) = 0 for any integer s < k.
Theorem 27.4. Suppose that the roots of P n (r) = a n r n + a n−1 r n−1 +
· · · + a 0 are r 1 , ..., r k with multiplicities m 1 , m 2 , ..., m k (where m 1 + m 2 +
· · · + m k = n). Then the general solution of L n y = 0 is
y H =
k∑
i=1
m∑
i−1
e rit
j=0
C ij t j (27.23)
Before we prove theorem 27.3 will consider several examples
Example 27.1. Find the general solution of y ′′′ − 3y ′′ + 3y ′ − y = 0 The
characteristic equation is
r 3 − 3r 2 + 3r − 1 = (r − 1) 3 = 0 (27.24)
which has a single root r = 1 with multiplicity 3. The general solution is
therefore
y = (C 1 + C 2 t + C 3 t 3 )e t (27.25)