Lecture Notes in Differential Equations - Bruce E. Shapiro

231
Example 27.2. Find the general solution of y (4) − 2y ′′ + y = 0
The characteristic equation is
0 = r 4 − 2r 2 + r = (r 2 − 1) 2 = (r − 1) 2 (r + 1) 2 (27.26)
The roots are 1 and -1, each with multiplicity 2. Thus the solution is
y = e t (C 1 + C 2 t) + e −t (C 3 + C 4 t) (27.27)
Example 27.3. Find the general solution of y (4) − y = 0
The characteristic equation is
0 = r 4 − 1 = (r 2 − 1)(r 2 + 1) = (r − 1)(r + 1)(r − i)(r + i) (27.28)
There are four distinct roots r = ±1, ±i. The real roots give solutions e ±t ;
the complex roots r = 0 ± i give terms sin t and cos t. Hence the general
solution is
y = C 1 e t + C 2 e −t + C 3 cos t + C 4 sin t (27.29)
Example 27.4. Find the general solution of y (4) + y = 0
The characteristic equation is r 4 + 1 = 0.
Therefore r = (−1) 1/4 . By Euler’s equation,
− 1 = e iπ = e i(π+2kπ) (27.30)
hence the four fourth-roots are
[ ]
i(π + 2kπ)
(−1) 1/4 = exp
(27.31)
4
( π
= cos
4 + kπ ) ( π
+ i sin
2
4 + kπ )
, k = 0, 1, 2, 3 (27.32)
2
Therefore
r i = ±
√
2
2 ± i √
2
√
2
2 = 2
(±1 ± i) =
1
√
2
(±1 ± i) (27.33)
and the general solution of the differential equation is
[
y = e t/√ 2
C 1 cos √ t + C 2 sin
t ]
√
2 2
+ e −t/√ 2
[
C 3 cos
t √
2
+ C 4 sin
√ t ]
2
(27.34)