8.7.1 Example A Consider the following equation (D – a)(A – b)y(æ) = 0, (8.268) where (a, b) are constants. Written out, we have dy(x) A: dx aAy(x) – - dy(x) + aby(x) = 0, dx (8.269) or dy(x + 1) dy(x) dx - a[y(x+1) – y(x)] + b- dy(x) + aby(x) = 0, (8.270) dx dx and finally, dy(x + 1) + (b – 1) dy(x) + a(1+b)y(x) – ay(x + 1) = 0. dx (8.271) dx To construct the solution, inspection of equation (8.268) shows that y(x) = y1(x) + y2(x), (8.272) where y1 (x) and y2(x) are, respectively, solutions of (D – a)y1 (x) = 0, (A- b)y2(x) = 0. (8.273) Therefore, Y1 (x) = Aea", A = constant, (8.274) Y2(x) = B(x)(1+ b)ª, B(x+1) = B(x), (8.275) and it is important to point out that while A is an arbitrary constant, B(x) is an arbitrary function of period 1. We conclude that the general solution to equation (8.268) or (8.271) is y(x) = Ae + B(x)(1+b)*, B(x+1) = B(x). (8.276) %3D

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8.7.1
Example A
Consider the following equation
(D – a)(A – b)y(æ) = 0,
(8.268)
where (a, b) are constants. Written out, we have
dy(x)
- aAy(x) – -
dy(x)
+ aby(x) = 0,
dx
(8.269)
dx
or
dy(x + 1)
dy(x)
– aly(x + 1) – y(x)] +6dy(x)
+ aby(x) = 0,
dx
(8.270)
dx
dx
and finally,
dy(x + 1)
dy(x)
+ (b – 1).
dx
+ a(1+ b)y(x) – ay(x + 1) = 0.
(8.271)
dx
To construct the solution, inspection of equation (8.268) shows that
y(x) = y1 (x) + Y2 (x),
(8.272)
where y1 (x) and y2(x) are, respectively, solutions of
(D – a)yı (æ) = 0, (A- b)y2(x) = 0.
(8.273)
Therefore,
1 (х) — Ае",
A = constant,
(8.274)
Y2(x) = B(x)(1+ b)ª, B(x+1) = B(x),
(8.275)
and it is important to point out that while A is an arbitrary constant, B(æ)
is an arbitrary function of period 1.
We conclude that the general solution to equation (8.268) or (8.271) is
y(x) = Ae* + B(x)(1+b)", B(x + 1) = B(x).
(8.276)
Transcribed Image Text:8.7.1 Example A Consider the following equation (D – a)(A – b)y(æ) = 0, (8.268) where (a, b) are constants. Written out, we have dy(x) - aAy(x) – - dy(x) + aby(x) = 0, dx (8.269) dx or dy(x + 1) dy(x) – aly(x + 1) – y(x)] +6dy(x) + aby(x) = 0, dx (8.270) dx dx and finally, dy(x + 1) dy(x) + (b – 1). dx + a(1+ b)y(x) – ay(x + 1) = 0. (8.271) dx To construct the solution, inspection of equation (8.268) shows that y(x) = y1 (x) + Y2 (x), (8.272) where y1 (x) and y2(x) are, respectively, solutions of (D – a)yı (æ) = 0, (A- b)y2(x) = 0. (8.273) Therefore, 1 (х) — Ае", A = constant, (8.274) Y2(x) = B(x)(1+ b)ª, B(x+1) = B(x), (8.275) and it is important to point out that while A is an arbitrary constant, B(æ) is an arbitrary function of period 1. We conclude that the general solution to equation (8.268) or (8.271) is y(x) = Ae* + B(x)(1+b)", B(x + 1) = B(x). (8.276)
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