1 1 1 1 1 (z + 1)(z + 3) z +4 (z + 1)(z + 3) 1 1 2 z + 1 2 z + 3 3 1 1 1 2 z + 1 2 z +3' 1 1 1 1 + 10 z +3* 1 (z – 2)(z+1)(2 +3) 15 z - 2 6 z +1

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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Show me the steps of determine red

Example F
Taking the z-transform of each term of the second-order equation
Yk+2 +a1Yk+1 + a2Yk
R
(4.313)
gives
[2°F(2) – 2²y0 – zyı] + a1[zF(2) – zyo) + azF(z) = G(2),
(4.314)
or
G(2) + Yoz² + (a1Y0 + Y1)z
F(2)
(4.315)
z2 + a1% + a2
where G(z) = Z(Rr).
Let a1 = 0, a2 = 4, and Rk
equation
= 0. This corresponds to the difference
Yk+2 –
4yk
= 0.
(4.316)
The z-transform is, from equation (4.315), given by the expression
F(z)
2y0 – Y1
2Yo + Y1
(z + 2)(z – 2)
1
2y0 + Y1
1
(4.317)
4
z + 2
4
2 - 2
From Table 4.1, we can obtain the inverse transform of F(z); doing this gives
Yk = ¼(2yo – Y1)(-2)* + '¼(2yo + Y1)2*.
(4.318)
Since yo and Y1 are arbitrary, this gives the general solution of equation
6).
The equation
Yk+2 + 4Yk+1 + 3yk
2k
(4.319)
%3D
corresponds to a1 = 4, a2
= 3, and Rk = 2k. Using the fact that
G(2) = Z(2*)
(4.320)
2
we can rewrite equation (4.315), for this case, as
F(2)
Yo(z + 4)
(z + 1)(z + 3)
Y1
1
(4.321)
(z + 1)(z + 3)
(z – 2)(z + 1)(z +3)'
The three terms on the right-hand side of this equation have the following
partial-fraction expansions:
1
1
1
1 1
(4.322)
(z + 1)(2 + 3)
z + 4
(z +1)(x + 3)
2 z + 1
2 z + 3
3
1
1
(4.323)
2 z + 1
2 z + 3
1
1
1
1
1
1
1
(4.324)
(z – 2)(z + 1)(z + 3)
15 z – 2
6 z + 1
10 z + 3
Transcribed Image Text:Example F Taking the z-transform of each term of the second-order equation Yk+2 +a1Yk+1 + a2Yk R (4.313) gives [2°F(2) – 2²y0 – zyı] + a1[zF(2) – zyo) + azF(z) = G(2), (4.314) or G(2) + Yoz² + (a1Y0 + Y1)z F(2) (4.315) z2 + a1% + a2 where G(z) = Z(Rr). Let a1 = 0, a2 = 4, and Rk equation = 0. This corresponds to the difference Yk+2 – 4yk = 0. (4.316) The z-transform is, from equation (4.315), given by the expression F(z) 2y0 – Y1 2Yo + Y1 (z + 2)(z – 2) 1 2y0 + Y1 1 (4.317) 4 z + 2 4 2 - 2 From Table 4.1, we can obtain the inverse transform of F(z); doing this gives Yk = ¼(2yo – Y1)(-2)* + '¼(2yo + Y1)2*. (4.318) Since yo and Y1 are arbitrary, this gives the general solution of equation 6). The equation Yk+2 + 4Yk+1 + 3yk 2k (4.319) %3D corresponds to a1 = 4, a2 = 3, and Rk = 2k. Using the fact that G(2) = Z(2*) (4.320) 2 we can rewrite equation (4.315), for this case, as F(2) Yo(z + 4) (z + 1)(z + 3) Y1 1 (4.321) (z + 1)(z + 3) (z – 2)(z + 1)(z +3)' The three terms on the right-hand side of this equation have the following partial-fraction expansions: 1 1 1 1 1 (4.322) (z + 1)(2 + 3) z + 4 (z +1)(x + 3) 2 z + 1 2 z + 3 3 1 1 (4.323) 2 z + 1 2 z + 3 1 1 1 1 1 1 1 (4.324) (z – 2)(z + 1)(z + 3) 15 z – 2 6 z + 1 10 z + 3
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