Explain the detemaine 2.8.5Example EThe graph of the nonlinear equationy+1 = Yk(2.203)is shown in Figure 2.11. There are two fixed points, yk = 0 and yk = 1.The slopes of the curve at these fixed points are, respectively, o and 1/2;consequently, yk = 0 is an unstable fixed point and yk = 1 is a stable fixedpoint.Note that the lower branch of the curve does not enter into considerationsince it does not correspond to real values of yk under continued iteration.The transformationUk = log yk(2.204)will allow us to obtain the exact solution to equation (2.203). We find10220k+1 = Vk,(2.205)the solution of which isUk+1 =A(!/½)*,(2.206)where A is an arbitrary constant. Therefore,Yk = exp[A(!/½)*],(2.207)which has the asymptotic expansionYk = 1+ A(1/½)* + 1½A²(1/½)²k + 1/6A³ (!/2)³k + · . . .(2.208)Rewrite equation (2.203) as follows:Yk+1 = V(2.209)and letYk = 1+ uk,(2.210)where uk is assumed to be small. Substitution of equation (2.210) into equation(2.209) givesUk+1 = /2uk,(2.211)the solution of which is- A(/½)*,(2.212)Uk =where A is an arbitrary constant. Therefore, the first approximation to yk is[Yk = 1+ A(1/½)*.(2.213)Now letA(/2)*, z(t) = Yk-(2.214)t =FIRST-ORDER DIFFERENCE EQUATIONS79Consequently, equation (2.203) becomesC:(/2t)² = z(t).(2.215)Assuming a solution of the formC z(t) = 1+t + Azť² + A3t³ + ...(2.216)and substituting this result into equation (2.215) gives2(/2t)? = 1+t+(/½A2 + /¼)t² + (/¼A3 + ¼A2)t³ + · ...(2.217)Comparison of equations (2.216) and (2.217) gives1/2A2 + 1/4 = A2, ¼A3 + 1/4A2 = A3(2.218)andA2 = 1/2, A3 = !/.(2.219)Therefore,z(t) = 1+t+ 1/2t² + 1/6t³ + ·..,which, on using the definition of t in equation (2.214), gives the same asymp-totic expansion as equation (2.208).

Name: Explain the detemaine 2.8.5Example EThe graph of the nonlinear equatio
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Description: Explain the detemaine 2.8.5Example EThe graph of the nonlinear equationy+1 = Yk(2.203)is shown in Figure 2.11. There are two fixed points, yk = 0 and yk = 1.The slopes of the curve at these fixed points are, respectively, o and 1/2;consequently, yk =

Given difference equation is yk+12=yk (i) It is a second order non-linear…

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the solution of which is Uk+1 = A(/½)*, where A is an arbitrary constant. Therefore, (2.206) - exp[A(t/½)*], (2.207) Yk = which has the asymptotic expansion Yk = 1+ A(/2)* + 1/2A² ('/½)²k + \/6A°(1/½)³* + - ... (2.208) Rewrite equation (2.203) as follows: Yk+1 = VYk;, (2.209) and let Yk = 1+ uk; (2.210) where u is assumed to be small. Substitution of equation (2.210) into equation (2.209) gives Uk+1 = 1/2uk, (2.211) the solution of which is u = A(½)*, (2.212) where A is an arbitrary constant. Therefore, the first approximation to yk is Yk = 1+ A(/2)*. (2.213) Now let t = A(1/½)*, z(t) = Yk- (2.214) FIRST-ORDER DIFFERENCE EQUATIONS 79 Consequently, equation (2.203) becomes C <(/½t)? = z(t). (2.215) Assuming a solution of the form C 2(t) = 1+t + A2t2 + A3t3 +... (2.216) and substituting this result into equation (2.215) gives (/2t)? = 1+t+ (½A2 + /4)t² + (/¼A3 + ¼A2)t³ + · ... (2.217) Comparison of equations (2.216) and (2.217) gives 1/2A2 + 4 = A2, 14A3 +4A2 = A3 (2.218) %3D %3D and A2 = 2, A3 = 1/6. (2.219) Therefore, z(t) = 1+t+/2t² + 1/6t³ + • · · , which, on using the definition of t in equation (2.214), gives the same asymp- totic expansion as equation (2.208).

the solution of which is Uk+1 = A(/½), where A is an arbitrary constant. Therefore, (2.206) - exp[A(t/½)], (2.207) Yk = which has the asymptotic expansion Yk = 1+ A(/2)* + 1/2A² ('/½)²k + \/6A°(1/½)³* + - ... (2.208) Rewrite equation (2.203) as follows: Yk+1 = VYk;, (2.209) and let Yk = 1+ uk; (2.210) where u is assumed to be small. Substitution of equation (2.210) into equation (2.209) gives Uk+1 = 1/2uk, (2.211) the solution of which is u = A(½), (2.212) where A is an arbitrary constant. Therefore, the first approximation to yk is Yk = 1+ A(/2). (2.213) Now let t = A(1/½)*, z(t) = Yk- (2.214) FIRST-ORDER DIFFERENCE EQUATIONS 79 Consequently, equation (2.203) becomes C <(/½t)? = z(t). (2.215) Assuming a solution of the form C 2(t) = 1+t + A2t2 + A3t3 +... (2.216) and substituting this result into equation (2.215) gives (/2t)? = 1+t+ (½A2 + /4)t² + (/¼A3 + ¼A2)t³ + · ... (2.217) Comparison of equations (2.216) and (2.217) gives 1/2A2 + 4 = A2, 14A3 +4A2 = A3 (2.218) %3D %3D and A2 = 2, A3 = 1/6. (2.219) Therefore, z(t) = 1+t+/2t² + 1/6t³ + • · · , which, on using the definition of t in equation (2.214), gives the same asymp- totic expansion as equation (2.208).

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 detemaine

2.8.5
Example E
The graph of the nonlinear equation
y+1 = Yk
(2.203)
is shown in Figure 2.11. There are two fixed points, yk = 0 and yk = 1.
The slopes of the curve at these fixed points are, respectively, o and 1/2;
consequently, yk = 0 is an unstable fixed point and yk = 1 is a stable fixed
point.
Note that the lower branch of the curve does not enter into consideration
since it does not correspond to real values of yk under continued iteration.
The transformation
Uk = log yk
(2.204)
will allow us to obtain the exact solution to equation (2.203). We find
102
20k+1 = Vk,
(2.205)
the solution of which is
Uk+1 =
A(!/½)*,
(2.206)
where A is an arbitrary constant. Therefore,
Yk = exp[A(!/½)*],
(2.207)
which has the asymptotic expansion
Yk = 1+ A(1/½)* + 1½A²(1/½)²k + 1/6A³ (!/2)³k + · . . .
(2.208)
Rewrite equation (2.203) as follows:
Yk+1 = V
(2.209)
and let
Yk = 1+ uk,
(2.210)
where uk is assumed to be small. Substitution of equation (2.210) into equation
(2.209) gives
Uk+1 = /2uk,
(2.211)
the solution of which is
- A(/½)*,
(2.212)
Uk =
where A is an arbitrary constant. Therefore, the first approximation to yk is
[
Yk = 1+ A(1/½)*.
(2.213)
Now let
A(/2)*, z(t) = Yk-
(2.214)
t =
FIRST-ORDER DIFFERENCE EQUATIONS
79
Consequently, equation (2.203) becomes
C:(/2t)² = z(t).
(2.215)
Assuming a solution of the form
C z(t) = 1+t + Azť² + A3t³ + ...
(2.216)
and substituting this result into equation (2.215) gives
2(/2t)? = 1+t+(/½A2 + /¼)t² + (/¼A3 + ¼A2)t³ + · ...
(2.217)
Comparison of equations (2.216) and (2.217) gives
1/2A2 + 1/4 = A2, ¼A3 + 1/4A2 = A3
(2.218)
and
A2 = 1/2, A3 = !/.
(2.219)
Therefore,
z(t) = 1+t+ 1/2t² + 1/6t³ + ·..,
which, on using the definition of t in equation (2.214), gives the same asymp-
totic expansion as equation (2.208).

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