Explain the determain and the Q is complete Adobe Scan .l2:24 AMT @ 9 93% 4Done2.8.3 Example CThe equationYk+1 = Yk + 1/yk(2.183)does not cross the line yk+1 = Yk; see Figure 2.9. Therefore, we expect ykto become unbounded as k0o.we cannot use any of the above-discussed techniques to determine its asymptotic behavior.Note that we cannot use the fact that Yk > 1/yk in equation (2.183)because the solution of the difference equation yk+1 = Yk is a constant; thisviolates the condition that yk - x∞ as k → .98To proceed, we square both sides of equation (2.183) and obtain= y% + 2+1/y.(2.184)For this equation, it is valid to neglect y? in comparison with 2 as k → o.The resulting asymptotic difference equation(2.185)k+1can be solved by making the transformation vk = y. Its solution is%3DVk = 2k, k → 00,(2.186)andYk = V2k,k → 00.(2.187)This is the dominant asymptotic behavior of yk as k → 0o.The next-order correction to the result of equation (2.187) is obtained bylettingYk = V2k + €k,(2.188)where we assume that e « 2k for k → o. Substituting equation (2.188) intoequation (2.184) and neglecting €k in comparison with 2k in the denominatorof the fraction gives99Ek+1 – Ek = 1/2k, k → o,(2.189)the solution of which is/½ log k,(2.190)Ek =k → 0.From the resultV2k + 4n = /2k (1+ )" = v2k (1 + +)1/2= V2k (1+= V2k (1+2k.1 log k4 V2k+..(2.191)V2k+..we conclude that the asymptotic behavior of equation (2.183) is1 log k+...(2.192)Yk = V2k +4 V2k

The equation yk+1=yk+1yk To Procced we square both sides of the equation and obtain…

The equation Yk+1 = Yk + 1/yk (2.183) %3D does not cross the line yk+1 = Yk; see Figure 2.9. Therefore, we expect yk to become unbounded as k → o. However, we cannot use any of the above- discussed techniques to determine its asymptotic behavior. Note that we cannot use the fact that yk > 1/yk in equation (2.183)

The equation Yk+1 = Yk + 1/yk (2.183) %3D does not cross the line yk+1 = Yk; see Figure 2.9. Therefore, we expect yk to become unbounded as k → o. However, we cannot use any of the above- discussed techniques to determine its asymptotic behavior. Note that we cannot use the fact that yk > 1/yk in equation (2.183)

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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Question

Explain the determain and the Q is complete

Adobe Scan .l
2:24 AM
T @ 9 93% 4
Done
2.8.3 Example C
The equation
Yk+1 = Yk + 1/yk
(2.183)
does not cross the line yk+1 = Yk; see Figure 2.9. Therefore, we expect yk
to become unbounded as k
0o.
we cannot use any of the above-
discussed techniques to determine its asymptotic behavior.
Note that we cannot use the fact that Yk > 1/yk in equation (2.183)
because the solution of the difference equation yk+1 = Yk is a constant; this
violates the condition that yk - x∞ as k → .
98
To proceed, we square both sides of equation (2.183) and obtain
= y% + 2+1/y.
(2.184)
For this equation, it is valid to neglect y? in comparison with 2 as k → o.
The resulting asymptotic difference equation
(2.185)
k+1
can be solved by making the transformation vk = y. Its solution is
%3D
Vk = 2k, k → 00,
(2.186)
and
Yk = V2k,
k → 00.
(2.187)
This is the dominant asymptotic behavior of yk as k → 0o.
The next-order correction to the result of equation (2.187) is obtained by
letting
Yk = V2k + €k,
(2.188)
where we assume that e « 2k for k → o. Substituting equation (2.188) into
equation (2.184) and neglecting €k in comparison with 2k in the denominator
of the fraction gives
99
Ek+1 – Ek = 1/2k, k → o,
(2.189)
the solution of which is
/½ log k,
(2.190)
Ek =
k → 0.
From the result
V2k + 4n = /2k (1+ )" = v2k (1 + +)
1/2
= V2k (1+
= V2k (1+
2k.
1 log k
4 V2k
+..
(2.191)
V2k
+..
we conclude that the asymptotic behavior of equation (2.183) is
1 log k
+...
(2.192)
Yk = V2k +
4 V2k

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