Lecture Notes in Differential Equations - Bruce E. Shapiro

236 LESSON 27. HIGHER ORDER EQUATIONS
is invertible, then a solution of (27.64) is
c = M −1 u 0 (27.66)
where c = (c 1 c 2 · · · c n ) T and u 0 = (u 0 u 1 · · · u n−1 ) T . But M is
invertible if and only if det M ≠ 0. We will prove this by contradiction.
Suppose that det M = 0. Then Mc = 0 for some non-zero vector c, i.e,
⎛
⎞ ⎛ ⎞
y 1 (t 0 ) y 2 (t 0 ) · · · y n (t 0 ) c 1
y 1(t ′ 0 ) y 2(t ′ 0 ) y n(t ′ 0 )
c 2 ⎜
.
⎝ .
.. ⎟ ⎜ ⎟
. ⎠ ⎝ . ⎠ = 0 (27.67)
y (n−1)
1 (t 0 ) y (n−1)
2 (t 0 ) y n (n−1) (t 0 ) c n
and line by line,
v (j) (t 0 ) =
n∑
i=1
Using the norm defined by (27.35),
c i y (j)
i (t 0 ) = 0, j = 0, 1, ..., n − 1 (27.68)
n−1
∑
‖v(t 0 )‖ 2 ∣
= ∣v (j) (t 0 ) ∣ 2 = 0 (27.69)
i=0
By the fundamental inequality, since v(t) is a solution,
‖v(t 0 )‖ e −K|t−t0| ≤ ‖v(t)‖ ≤ ‖v(t 0 )‖ e K|t−t0| (27.70)
Hence ‖v(t)‖ = 0, which means v(t) = 0 for all t. Since all of the y i (t) are
linearly independent, this means that all of the c i = 0, i.e., c = 0. But this
contradicts (27.67), so it must be true that det M ≠ 0.
Since det M ≠ 0, the solution given by (27.66) exists. Thus v(t), which exists
as a linear combination of the y i is a solution of the same initial value
problem as u(t). Thus v(t) and u(t) must be identical by uniqueness, and
our assumptions that u(t) was not a linear combination of the y i is contradicted.
This must mean that no such solution exists, and every solution of
L n y = 0 must be a linear combination of the y i .
The Particular Solution
The particular solution can be found by the method of undetermined coefficients
or annihilators, or by a generalization of the expression that gives