The equation kk+4 – Yk = (4.71) has the characteristic equation pd – 1 = (r2 + 1)(r² – 1) = (r + i)(r – i)(r + 1)(r – 1), (4.72) and roots r1 = -i, r2 = +i, r3 = -1, r4 = +1. Since exp(±in/2) = ±i, we have the following fundamental set of solutions: .(1) = e e-ink/2. (2) Yk = eink/2 (4.73) (3) (4) y = (-1)*, = 1.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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4.2.9
Example I
The equation
kk+4 -
- Yk = 0
(4.71)
has the characteristic equation
pd – 1 = (r2 + 1)(r² – 1) = (r + i)(r – i)(r + 1)(r – 1),
(4.72)
and roots rị
-i, r2 = +i, r3 = -1, r4 = +1. Since exp(±in/2) = ±i, we
have the following fundamental set of solutions:
.(1)
e-ink/2
(2)
= e?nk/2
(4.73)
(3)
(4)
= (-1)*, y = 1.
(1)
Note that Y
can be written in the equivalent forms
Yk
(2)
and
(1)
= cos(Tk/2), T = sin(Tk/2).
(4.74)
Therefore, the general solution to equation (4.71) is
k
Yk = C1 cos(rk/2)+ c2 sin(Tk/2) + c3(-1)* + c4,
(4.75)
where c1, c2, C3, and c4 are arbitrary constants.
Transcribed Image Text:4.2.9 Example I The equation kk+4 - - Yk = 0 (4.71) has the characteristic equation pd – 1 = (r2 + 1)(r² – 1) = (r + i)(r – i)(r + 1)(r – 1), (4.72) and roots rị -i, r2 = +i, r3 = -1, r4 = +1. Since exp(±in/2) = ±i, we have the following fundamental set of solutions: .(1) e-ink/2 (2) = e?nk/2 (4.73) (3) (4) = (-1)*, y = 1. (1) Note that Y can be written in the equivalent forms Yk (2) and (1) = cos(Tk/2), T = sin(Tk/2). (4.74) Therefore, the general solution to equation (4.71) is k Yk = C1 cos(rk/2)+ c2 sin(Tk/2) + c3(-1)* + c4, (4.75) where c1, c2, C3, and c4 are arbitrary constants.
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