Lecture Notes in Differential Equations - Bruce E. Shapiro

253
So that
f(t) = a n (t)y[
(n) + a n−1 (t)y (n−1) + · · · + a 0 (t)y
]
= a n (t) u 1 y (n)
1 + · · · + u n y n
(n) + u ′ 1y (n−1)
1 + · · · + u ′ ny n
(n−1) +
[
]
a n−1 (t) u 1 y (n−1)
1 + · · · + u n y n
(n−1) + · · · +
a 1 (t) [ [u 1 y 1 ′ + · · · + u n y n] ′ + a 0 (t) ] [u 1 y 1 + · · · + u n y n ]
= a n (t) u ′ 1y (n−1)
1 + · · · + u ′ ny n
(n−1)
(27.201)
Combining (27.201) and (27.199) in matrix form,
⎛
⎞ ⎛ ⎞ ⎛
⎞
y 1 y 2 · · · y n u ′
y 1 ′ y 2 ′ y n
′ 1
0
⎟ ⎜ u ′ 2 ⎟ ⎜ 0
⎟
⎜
⎝
.
.
y (n−2)
1 y (n−2)
2 · · · y n
(n−2)
y (n−1)
1 y (n−1)
2 · · · y n
(n−1)
⎟ ⎜
⎠ ⎝
.
u ′ n−1
u ′ n
=
⎟ ⎜
⎠ ⎝
.
0
f(t)/a n (t)
(27.202)
The matrix on the left is the fundamental matrix of the differential equation,
and hence invertible, so that
⎛
⎜
⎝
u ′ 1
u ′ 2
.
u ′ n−1
u ′ n
⎞
⎛
=
⎟ ⎜
⎠ ⎝
y 1 · · · y n
y 1
′ y n
′
⎞ ⎛
.
.
y (n−2)
1 · · · y n
(n−2) ⎟ ⎜
⎠ ⎝
y (n−1)
1 · · · y n
(n−1)
0
0
.
0
f(t)/a n (t)
⎞
⎟
⎠
= f(t)
a n (t)
⎟
⎠
⎛ ⎞
[W −1 ] 1n
[W −1 ] 2n
⎜ .
⎟
⎝ ⎠
[W −1 ] nn
(27.203)
where W is the fundamental matrix and [W −1 ] ij denotes the ijth element
of W −1 .
Therefore
[W −1 ] jn = cof[W ] ni
det W
= W i(t)
W (t)
(27.204)
du i
dt = f(t)W i(t)
a n (t)W (t)
∫
u i (t) =
t
(27.205)
f(s)W i (s)
ds (27.206)
a n (s)W (s)
Substitution of equation (27.206) into equation (27.198) yields equation
(27.196).