Example B The equation z(k+2,€) = 4z(k, l+ 1) (5.36) can be written (E – 4E2)z(k, l) = 0. (5.37) %3D We will use two methods to solve this equation. First, write equation (5.37) in the form (E2 – 1¼E})z(k, l) = 0. (5.38) The solution to this equation is z(k, l) = (¼E†)' A(k) = 4¬€ A(k + 20). (5.39) %3D Second, we can write equation (5.37) in the factored form 1/2, (E1 – 2E,²)(E1 +2E;²)z(k, l) = 0. (5.40)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Example B
The equation
z(k + 2, e) = 42(k, l+ 1)
(5.36)
can be written
(E – 4E2)z(k, l) = 0.
(5.37)
%3D
We will use two methods to solve this equation.
First, write equation (5.37) in the form
(E2 – 1¼E†)z(k, l) = 0.
(5.38)
-
%3D
The solution to this equation is
z(k, l) = (¼E†)* A(k) = 4¬€ A(k + 20).
(5.39)
Second, we can write equation (5.37) in the factored form
(E1 – 2E;²)(E1 + 2E;²)z(k, e) = 0.
(5.40)
172
Difference Equations
Now,
(E1 ±2E,²)z(k, l) = 0
(5.41)
%3D
has the solution
z(k, l) = (#2E;"²)*+(e) = (#2)*¢±(e + ½k),
(5.42)
where ø4 and ø_ are arbitrary functions of l. Thus, the general solution to
equation (5.36) can also be expressed as
z(k, l) = 2* [ø_(e + ½k) + (-)*$+(l + 1/2k)].
(5.43)
Note that equation (5.36) is second order in k and first order in l. This
is the reason for obtaining two arbitrary functions in the solution given by
equation (5.43) and one in the solution of equation (5.39).
Transcribed Image Text:Example B The equation z(k + 2, e) = 42(k, l+ 1) (5.36) can be written (E – 4E2)z(k, l) = 0. (5.37) %3D We will use two methods to solve this equation. First, write equation (5.37) in the form (E2 – 1¼E†)z(k, l) = 0. (5.38) - %3D The solution to this equation is z(k, l) = (¼E†)* A(k) = 4¬€ A(k + 20). (5.39) Second, we can write equation (5.37) in the factored form (E1 – 2E;²)(E1 + 2E;²)z(k, e) = 0. (5.40) 172 Difference Equations Now, (E1 ±2E,²)z(k, l) = 0 (5.41) %3D has the solution z(k, l) = (#2E;"²)*+(e) = (#2)*¢±(e + ½k), (5.42) where ø4 and ø_ are arbitrary functions of l. Thus, the general solution to equation (5.36) can also be expressed as z(k, l) = 2* [ø_(e + ½k) + (-)*$+(l + 1/2k)]. (5.43) Note that equation (5.36) is second order in k and first order in l. This is the reason for obtaining two arbitrary functions in the solution given by equation (5.43) and one in the solution of equation (5.39).
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