Lecture Notes in Differential Equations - Bruce E. Shapiro

252 LESSON 27. HIGHER ORDER EQUATIONS
Variation of Parameters
Theorem 27.12 (Variation of Parameters.). Suppose that y 1 , ..., y n are a
fundamental set of solutions to
L n y = a n (t)y (n) + a n−1 (t)y (n−1) + · · · + a 0 (t)y = 0 (27.194)
Then a particular solution to
is given by
L n y = a n (t)y (n) + a n−1 (t)y (n−1) + · · · + a 0 (t)y = f(t) (27.195)
∫
∫
W 1 (s)f(s)
y P = y 1
t W (s)a n (s) ds + y W 2 (s)f(s)
2
t W (s)a n (s) ds
∫
W n (s)f(s)
+ · · · + y n ds (27.196)
W (s)a n (s)
t
t
where W j (t) is the determinant of W[y 1 , ..., y n ](t) with the jth column
replaced by a vector with all zeroes except for a 1 in the last row. In
particular, for n = 2, a particular solution to a(t)y ′′ + b(t)y ′ + c(t)y = f(t)
is
∫
∫
y 2 (s)f(s)
y p = −y 1 (t)
W (s)a(s) ds + y y 1 (s)f(s)
2(t)
ds (27.197)
W (s)a(s)
Proof for general case. Look for a solution of the form
t
y = u 1 y 1 + · · · + u n y n (27.198)
This is under-determined so we can make additional assumptions; in particular,
we are free to assume that
Then
u ′ 1y 1 + · · · + u ′ ny n = 0
u ′ 1y ′ 1 + · · · + u ′ ny ′ n = 0
.
u ′ 1y (n−2)
1 + · · · + u ′ ny n (n−2) = 0
(27.199)
y ′ = u 1 y 1 ′ + · · · + u n y n
′
y ′′ = u 1 y 1 ′′ + · · · + u n y n
′′
.
y (n−1) = u 1 y (n−1)
1 + · · · + u n y n
(n−1)
y (n) = u 1 y (n)
1 + · · · + u n y n
(n)
+ u ′ 1y (n−1)
1 + · · · + u ′ ny n
(n−1)
(27.200)