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
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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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).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff127b632-c7db-43de-8096-c44c67b91c3f%2Fcaf3bc39-720d-42be-96a7-2e523693d73f%2Ftd3fbf_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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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