Example K Consider the equation Yk+1 (2 cos o)yk + Yk-1 = 0, cos o + 0, (4.82) - where o is a constant. The characteristic equation is p2 – (2 cos ø)r +1 = 0, (4.83) and has the solutions r+ = 1/2(2 cos o± V4 cos? o – 4) (4.84) = cos o ± i sin o = e±iø Therefore, a fundamental set of solutions is (1) (2) cos(øk), y sin(øk), (4.85) and the general solution is Yk = c1 cos(øk) + c2 sin(øk), (4.86) where C1 and C2 are arbitrary constants.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Example K
Consider the equation
Yk+1 – (2 cos )Yk + Yk-1 = 0,
cos o + 0,
(4.82)
where o is a constant. The characteristic equation is
p2 – (2 cos o)r +1=0,
(4.83)
COS
and has the solutions
r+ = /2(2 cos ø± V4 cos?
$ – 4)
(4.84)
= cos o ± i sin o = e±iø
Therefore, a fundamental set of solutions is
(2)
9 = cos(øk), y sin(øk),
Yk
(4.85)
and the general solution is
Yk = C1 cos(øk) + c2 sin(øk),
(4.86)
where
C1
and
C2 are arbitrary constants.
Transcribed Image Text:Example K Consider the equation Yk+1 – (2 cos )Yk + Yk-1 = 0, cos o + 0, (4.82) where o is a constant. The characteristic equation is p2 – (2 cos o)r +1=0, (4.83) COS and has the solutions r+ = /2(2 cos ø± V4 cos? $ – 4) (4.84) = cos o ± i sin o = e±iø Therefore, a fundamental set of solutions is (2) 9 = cos(øk), y sin(øk), Yk (4.85) and the general solution is Yk = C1 cos(øk) + c2 sin(øk), (4.86) where C1 and C2 are arbitrary constants.
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