p –1 = (r² + 1)(r2 – 1) = (r + i)(r – i)(r + 1)(r – 1), (= %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Show me the steps of determine red

4.2.9 Example I
The equation
kk+4 – Yk = 0
(4.71)
has the characteristic equation
( nt – 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(±iT/2)
±i, we
||
have the following fundamental set of solutions:
Yk
(1)
-ink/2
= e?Tk/2
= e
(4.73)
(3)
Y = (-1)*, y
= 1.
Note that y
,(1)
(2)
and
can be written in the equivalent forms
T = cos(Tk/2), = sin(Tk/2).
(4.74)
Therefore, the general solution to equation (4.71) is
Yk = C1 Cos(tk/2) + c2 sin(rk/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 ( nt – 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(±iT/2) ±i, we || have the following fundamental set of solutions: Yk (1) -ink/2 = e?Tk/2 = e (4.73) (3) Y = (-1)*, y = 1. Note that y ,(1) (2) and can be written in the equivalent forms T = cos(Tk/2), = sin(Tk/2). (4.74) Therefore, the general solution to equation (4.71) is Yk = C1 Cos(tk/2) + c2 sin(rk/2)+ c3(-1)* + c4, (4.75) where c1, c2, C3, and c4 are arbitrary constants.
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