U8n-6 U8n-10 U8n-4 = W8n-6 U8n-8 - U8n-8-U8n-10

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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U8n-6 U8n-10
U8n-4 = U8n-6 – U8n-8
U8n-8-U8n–10
Transcribed Image Text:U8n-6 U8n-10 U8n-4 = U8n-6 – U8n-8 U8n-8-U8n–10
11. EXACT SOLUTION OF EQ. (2) WHEN a =B =y= 6 = 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 =
a, u-4 = b, u_3 = c, u_2 = d, u-1 = e, uo = f. Then, for n = 0,1,2, ..., the
solutions of Eq. (18) are given by the following formulas:
e2n 2n-1
IT (ie + c)(ic +a)'
f2n 2n-1
IT (if + d)(id + b)
c2n e2n
2n-1
U8n-5 =
2n-1
U8n-4 =
2n-1
U8n-3 = =
IT (ic +a) IIT (ie + c)
d2n f2n
IT (id + b) IT, (if + d)'
U8n-2 =
e2n+1 2n
U8n-1 =
IT", (ie + c)(ic + a)
fan+l2n
IT, (if + d)(id + b)
U8n =
2n
c2n+1e2n+1
U8n+1 =
2n+1
IT (ic + a) I (ie + c)'
d?n+1 f2n+1
U8n+2 =
T2n+1
IIT" (id + b) IT (if + d)
Proof. It can be easily seen that the solutions are true for n = 0. We suppose
that n >0 and assume that our assumption holds for n – 1. That is,
e2n-2,2n-3
U8n-13 =
2n-3
IE1"(ie + c)(ic+ a)
f2n-2d2n-3
U8n-12 =
T2n-3
IT (if + d)(id + b)
c2n-22n-2
U8n-11 =
T2n-2
2n-3
II (ic + a) II,", (ie + c)
d?n-2 f2n-2
U8n-10 =
12л -2,
II (id + b) [I (if + d)
e2n-1,2n-2
2n-3
U8n-9 =
II (ie + c)(ic +a)'
f2n-12n-2
T2n-2
II", (if + d)(id + b)
c2n-1e2n-1
U8n-8 =
U8n-7 =
2n-
IT", (ic + a) IT",(ie +c)
dan-1 f2n-1
IT (id + b) IIif + d)
U8n-6 =
i=1
Moreover, Eq. (18) gives us that
U8n-6U8n-10
U8n-4 = U8n-6
U8n-8 + U8n–10
d2n-1 f2n-1
d2n-1f2n-1
ךלק
d2n-2 f2n-2
T2n-3
IT (id+b) I?",?(if+d) [T?,?(id+b) [T?"(if+d)
f2n-1 d2n-2
IT",?(iS+d)(id+b)
d²n-2 f2n–2
I (id + b) [I(if+d)
dn-1 f2n-1
IT (id + b) T",(if + d)
-1
+
II?(id+b) 1?",³(if+d)
d2n-1 2n-2
T (id+b) I",(if+d) IT(if+d)(
d2n-1 f2n-2
dn-1 f2n-1
II,"(id + b) II(if +d)
2n-2
=D1
2п-1
li=1
II (id + b) II(if +d) (en-7+a +
d?n-1 f2n-1((2n - 2)f + d)
IT (id + b) II" (if + d) ((2n – 1) f +d)
(2n – 2)f + d`
(2n – 1)f + d
2n-2
den-1 f2n-1
an-1(id+b) [T (if+d)
2n-1 f2n–1
2n-2
2n-1
2n-2
i=1
i=1
i=1
%D
1-
I (id + b) IT,if + d)
d2n-1 f2n
I, (id + b) II{if + d)((2n – 1)f + d)
f2n ď²n-1
2n-'(if + d)(id +b)
Transcribed Image Text:11. EXACT SOLUTION OF EQ. (2) WHEN a =B =y= 6 = 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 = a, u-4 = b, u_3 = c, u_2 = d, u-1 = e, uo = f. Then, for n = 0,1,2, ..., the solutions of Eq. (18) are given by the following formulas: e2n 2n-1 IT (ie + c)(ic +a)' f2n 2n-1 IT (if + d)(id + b) c2n e2n 2n-1 U8n-5 = 2n-1 U8n-4 = 2n-1 U8n-3 = = IT (ic +a) IIT (ie + c) d2n f2n IT (id + b) IT, (if + d)' U8n-2 = e2n+1 2n U8n-1 = IT", (ie + c)(ic + a) fan+l2n IT, (if + d)(id + b) U8n = 2n c2n+1e2n+1 U8n+1 = 2n+1 IT (ic + a) I (ie + c)' d?n+1 f2n+1 U8n+2 = T2n+1 IIT" (id + b) IT (if + d) Proof. It can be easily seen that the solutions are true for n = 0. We suppose that n >0 and assume that our assumption holds for n – 1. That is, e2n-2,2n-3 U8n-13 = 2n-3 IE1"(ie + c)(ic+ a) f2n-2d2n-3 U8n-12 = T2n-3 IT (if + d)(id + b) c2n-22n-2 U8n-11 = T2n-2 2n-3 II (ic + a) II,", (ie + c) d?n-2 f2n-2 U8n-10 = 12л -2, II (id + b) [I (if + d) e2n-1,2n-2 2n-3 U8n-9 = II (ie + c)(ic +a)' f2n-12n-2 T2n-2 II", (if + d)(id + b) c2n-1e2n-1 U8n-8 = U8n-7 = 2n- IT", (ic + a) IT",(ie +c) dan-1 f2n-1 IT (id + b) IIif + d) U8n-6 = i=1 Moreover, Eq. (18) gives us that U8n-6U8n-10 U8n-4 = U8n-6 U8n-8 + U8n–10 d2n-1 f2n-1 d2n-1f2n-1 ךלק d2n-2 f2n-2 T2n-3 IT (id+b) I?",?(if+d) [T?,?(id+b) [T?"(if+d) f2n-1 d2n-2 IT",?(iS+d)(id+b) d²n-2 f2n–2 I (id + b) [I(if+d) dn-1 f2n-1 IT (id + b) T",(if + d) -1 + II?(id+b) 1?",³(if+d) d2n-1 2n-2 T (id+b) I",(if+d) IT(if+d)( d2n-1 f2n-2 dn-1 f2n-1 II,"(id + b) II(if +d) 2n-2 =D1 2п-1 li=1 II (id + b) II(if +d) (en-7+a + d?n-1 f2n-1((2n - 2)f + d) IT (id + b) II" (if + d) ((2n – 1) f +d) (2n – 2)f + d` (2n – 1)f + d 2n-2 den-1 f2n-1 an-1(id+b) [T (if+d) 2n-1 f2n–1 2n-2 2n-1 2n-2 i=1 i=1 i=1 %D 1- I (id + b) IT,if + d) d2n-1 f2n I, (id + b) II{if + d)((2n – 1)f + d) f2n ď²n-1 2n-'(if + d)(id +b)
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