Theorem 9 Let {um}=-5 be a solution to Eq. (18) and suppose that u-s = a, u-4 = b, u_3 = c, u_2 = d, u-1 = e, uo = f. Then, for n = solutions of Eq. (18) are given by the following formulas: 0, 1, 2, ..., the e2n 2n-1 (ie + c)(ic+a)' f2n2n-1 IT(if + d)(id + b)' U8n-5 = 2n-1 U8n-4 U8n-3 = 2n-1 IT, (ic + a) IT (ie + c)' dan f2n n-1 U8n-2 IT, (id + b) IT'(if +d)' e2n+12n U8n-1 ITE, (ie + c)(ic + a) f2n+!d2n IT, (if + d)(id + b)' U8n = c2n+1@2n+1 U8n+1 2n IT (ic+ a) (ie + c)' d?n+1 f2n+1 IT*(id + b) [T": (if + d) U8n+2 = i=1
Theorem 9 Let {um}=-5 be a solution to Eq. (18) and suppose that u-s = a, u-4 = b, u_3 = c, u_2 = d, u-1 = e, uo = f. Then, for n = solutions of Eq. (18) are given by the following formulas: 0, 1, 2, ..., the e2n 2n-1 (ie + c)(ic+a)' f2n2n-1 IT(if + d)(id + b)' U8n-5 = 2n-1 U8n-4 U8n-3 = 2n-1 IT, (ic + a) IT (ie + c)' dan f2n n-1 U8n-2 IT, (id + b) IT'(if +d)' e2n+12n U8n-1 ITE, (ie + c)(ic + a) f2n+!d2n IT, (if + d)(id + b)' U8n = c2n+1@2n+1 U8n+1 2n IT (ic+ a) (ie + c)' d?n+1 f2n+1 IT*(id + b) [T": (if + d) U8n+2 = i=1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![11. EXACT SOLUTION OF EQ. (2) WHEN a = B = y = 8 = 1
This section shows the exact solutions of the following equation:
Un-1un-5
Un+1 = Un-1
n = 0, 1, ...,
(18)
Un-3 + un-5
where the initial conditions are selected to be positive real numbers.
Theorem 9 Let {un}5 be a solution to Eq. (18) and suppose that u-5 =
а, и-4 3D 6, и-з 3D с, и-2%3D d, и-1 — е, ио 3D f. Then, for n 3 0, 1, 2, .., the
solutions of Eq. (18) are given by the following formulas:
e2n 2n-1
II (ie + c)(ic+a)*
f2n d2n-1
IT (if + d)(id + b)'
c2n e2n
U8n-5 =
2n-
U8n-4 =
2n-1
U8n-3 =
2n-1
IT (ic + a) IT (ie + c)
d2n f2n
U8n-2 =
IT, (id + b) II if + d)
=D1
e2n+1c2n
U8n-1 =
IT, (ie + c)(ic+a)
f2n+1d2n
II (if + d)(id + b)'
U8n =
c2n+1e2n+1
U8n+1 =
2n+1
ITT (ic+ a) II, (ie + c)
d2n+1 f2n+1
U8n+2 =
2n+1
IIT (id + b) [I, (if + d)'
Proof.
that n >0 and assume that our assumption holds for n- 1. That is,
It can be easily seen that the solutions are true for n = 0. We suppose
e2n-2,2n-3
U8n-13 =
2n-3
IT (ie + c)(ic + a)
f2n-2?n-3
IT(if + d)(id + b)
2n-2 2n-2
U8n-12 =
2n-3
U8n-11 =
2n-2
li=1
2n-3
I (ic+a) I (ie + c)
d2n-2 f2n-2
U8n-10 =
T2n-2
IT(id + b) IT (if + d)
li=1
e2n-12n-2
U8n-9 =
2n-2
ITE (ie + c)(ic +
f2n-12n-2
n-2
U8n-8 =
II(if + d)(id + b)
c2n-1e2n-1
U8n-7 =
2n-2
(ic+ a) II(ie +c)
d2n-1 f2n-1
IT (id + b) II (if + d)
U8n-6 =
In addition, Eq. (18) leads to
U8n-5U8n-9
U8n-3 = U8n-5
U8n-7 + ug8n-9
2n 2n-1
2n-12n-2
e2n c2n–1
IT (ie+c)(ic+a) IT¿²(ie+c)(ic+a)
li=1
c2n-1 e2n-1
IT"'(ic+a) [I?",?(ie+c)
2n-1(ie+c)(ic +a)
e2n –1 c2n–2
T (ie+c)(ic+a)
c2n-2e2n
2n-2
+
e2n c2n-1
'ie+c)(ic+a) I'(ie + c)(ic+a) I (ic+ a)(Ttieta) + eI!n-
-1
li=1 (ic+a)
e2n 2n-1
c2n-2e2n
%3D
II"(ie + c)(ic + a) II(ie +c)(ic + a) (en-ijeta + )
2n-1
i3D1
(2n–1)c+a
c2n-1e2n ((2n – 1)c+a)
I (ie + c)(ic +a) II (ie + c)(ic+ a) ((2n)c + a)
(2n – 1)c+ a)
(2n)c+ a
e2n _2n-1
2n-1
2n-1
=1
i=1
e2n c2n-1
%3D
1-
2n-1
II (ie + c)(ic+a)
i=1
e2n 2n
c2n e2n
%3D
2п-1
II"T (ie + c)(ic + a)((2n)c + a) II", (ic + a) II" (ie + c)
2n-1
Similarly, one can prove other solutions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc212f4a0-6bcd-4a11-b8da-e45b9d5abb23%2F71de453f-792a-4911-831b-f47c0b8d9bf6%2F9gkket7_processed.png&w=3840&q=75)
Transcribed Image Text:11. EXACT SOLUTION OF EQ. (2) WHEN a = B = y = 8 = 1
This section shows the exact solutions of the following equation:
Un-1un-5
Un+1 = Un-1
n = 0, 1, ...,
(18)
Un-3 + un-5
where the initial conditions are selected to be positive real numbers.
Theorem 9 Let {un}5 be a solution to Eq. (18) and suppose that u-5 =
а, и-4 3D 6, и-з 3D с, и-2%3D d, и-1 — е, ио 3D f. Then, for n 3 0, 1, 2, .., the
solutions of Eq. (18) are given by the following formulas:
e2n 2n-1
II (ie + c)(ic+a)*
f2n d2n-1
IT (if + d)(id + b)'
c2n e2n
U8n-5 =
2n-
U8n-4 =
2n-1
U8n-3 =
2n-1
IT (ic + a) IT (ie + c)
d2n f2n
U8n-2 =
IT, (id + b) II if + d)
=D1
e2n+1c2n
U8n-1 =
IT, (ie + c)(ic+a)
f2n+1d2n
II (if + d)(id + b)'
U8n =
c2n+1e2n+1
U8n+1 =
2n+1
ITT (ic+ a) II, (ie + c)
d2n+1 f2n+1
U8n+2 =
2n+1
IIT (id + b) [I, (if + d)'
Proof.
that n >0 and assume that our assumption holds for n- 1. That is,
It can be easily seen that the solutions are true for n = 0. We suppose
e2n-2,2n-3
U8n-13 =
2n-3
IT (ie + c)(ic + a)
f2n-2?n-3
IT(if + d)(id + b)
2n-2 2n-2
U8n-12 =
2n-3
U8n-11 =
2n-2
li=1
2n-3
I (ic+a) I (ie + c)
d2n-2 f2n-2
U8n-10 =
T2n-2
IT(id + b) IT (if + d)
li=1
e2n-12n-2
U8n-9 =
2n-2
ITE (ie + c)(ic +
f2n-12n-2
n-2
U8n-8 =
II(if + d)(id + b)
c2n-1e2n-1
U8n-7 =
2n-2
(ic+ a) II(ie +c)
d2n-1 f2n-1
IT (id + b) II (if + d)
U8n-6 =
In addition, Eq. (18) leads to
U8n-5U8n-9
U8n-3 = U8n-5
U8n-7 + ug8n-9
2n 2n-1
2n-12n-2
e2n c2n–1
IT (ie+c)(ic+a) IT¿²(ie+c)(ic+a)
li=1
c2n-1 e2n-1
IT"'(ic+a) [I?",?(ie+c)
2n-1(ie+c)(ic +a)
e2n –1 c2n–2
T (ie+c)(ic+a)
c2n-2e2n
2n-2
+
e2n c2n-1
'ie+c)(ic+a) I'(ie + c)(ic+a) I (ic+ a)(Ttieta) + eI!n-
-1
li=1 (ic+a)
e2n 2n-1
c2n-2e2n
%3D
II"(ie + c)(ic + a) II(ie +c)(ic + a) (en-ijeta + )
2n-1
i3D1
(2n–1)c+a
c2n-1e2n ((2n – 1)c+a)
I (ie + c)(ic +a) II (ie + c)(ic+ a) ((2n)c + a)
(2n – 1)c+ a)
(2n)c+ a
e2n _2n-1
2n-1
2n-1
=1
i=1
e2n c2n-1
%3D
1-
2n-1
II (ie + c)(ic+a)
i=1
e2n 2n
c2n e2n
%3D
2п-1
II"T (ie + c)(ic + a)((2n)c + a) II", (ic + a) II" (ie + c)
2n-1
Similarly, one can prove other solutions.
![U8n-5 U8n-9
U8n-3
U8n-5 – U8n-7 –
U8n-7-U8n-9](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc212f4a0-6bcd-4a11-b8da-e45b9d5abb23%2F71de453f-792a-4911-831b-f47c0b8d9bf6%2F7jiqm5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:U8n-5 U8n-9
U8n-3
U8n-5 – U8n-7 –
U8n-7-U8n-9
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