Example C The following two functions are linearly independent: Yk (2) = e rọk, 0 = constant. (4.99) To determine the second-order difference equation that has these two functions as solutions, form the determinant Yk 1 1 sin(20) Yk+1+ Yk sin ø eiø (-2i sin ø) ( Yk+2 Yk+1 eiø Yk+2 e2iø e-2iø |(4.100) (-2i sin ø)[yk+2 – (2 cos o)yk+1+ yk]. | Setting the determinant equal to zero gives Yk+2 (2 cos o)yk+1 + Yk = 0, (4.101) which is the desired equation.
Example C The following two functions are linearly independent: Yk (2) = e rọk, 0 = constant. (4.99) To determine the second-order difference equation that has these two functions as solutions, form the determinant Yk 1 1 sin(20) Yk+1+ Yk sin ø eiø (-2i sin ø) ( Yk+2 Yk+1 eiø Yk+2 e2iø e-2iø |(4.100) (-2i sin ø)[yk+2 – (2 cos o)yk+1+ yk]. | Setting the determinant equal to zero gives Yk+2 (2 cos o)yk+1 + Yk = 0, (4.101) which is the desired equation.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain the determine
![Example C
The following two functions are linearly independent:
(1)
Yk
eiøk
(2)
-iøk
= e-1ok
constant.
(4.99)
To determine the second-order difference equation that has these two functions
as solutions, form the determinant
Yk
1
1
sin(20)
Yk+1+Yk
sin o
piø
-2i sin ø) ( Yk+2
Yk+1
eiø
e2iø
|(4.100)
Yk+2
e-2iø
-2i sin ø) [yk+2 - (2 cos )yk+1 + yk].
Setting the determinant equal to zero gives
Yk+2 – (2 cos ¢)Yk+1 + Yk = 0,
(4.101)
which is the desired equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F284e70ee-eed6-4dfe-aa18-32502476453c%2Fad275424-61c2-43a3-b04c-d0238471e8dc%2Fy2l8ndi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example C
The following two functions are linearly independent:
(1)
Yk
eiøk
(2)
-iøk
= e-1ok
constant.
(4.99)
To determine the second-order difference equation that has these two functions
as solutions, form the determinant
Yk
1
1
sin(20)
Yk+1+Yk
sin o
piø
-2i sin ø) ( Yk+2
Yk+1
eiø
e2iø
|(4.100)
Yk+2
e-2iø
-2i sin ø) [yk+2 - (2 cos )yk+1 + yk].
Setting the determinant equal to zero gives
Yk+2 – (2 cos ¢)Yk+1 + Yk = 0,
(4.101)
which is the desired equation.
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