X6n-2 = r X6n-2 = r n-2 2p+h p+h (2P II i=0 h n-2 2p+h II p+h i=0 (fei+sp + foi+7h) (fo1+39 + foi+2k) f6i+49 +f6i+3k foi+7P + foi+6h, h 2p+h p+h n-2 - ( ² P + h) II ( i=0 (2p+h) h+P n-2 (1 Jei+7p+for+h) (f6i+49+J6i+3k\ f6i+39+f6i+2k n-2 II i=0 (fai+sp + foi+7h) (fot+39 + foi+2k) f6i+49 +f6i+3k\ f6i+7P + foi+ch feng+fen-1kh fon+19+fenk P f6i+8p+f6i+7h f6i+49+f6i+3k f6i+7p+f6i+6h 1) ( f6i+39+f6i+2k h1+ feng+fen-1k fon+19+fenk (f6i+8P + f6i+7h (f6i+49 +f6i+3k\
X6n-2 = r X6n-2 = r n-2 2p+h p+h (2P II i=0 h n-2 2p+h II p+h i=0 (fei+sp + foi+7h) (fo1+39 + foi+2k) f6i+49 +f6i+3k foi+7P + foi+6h, h 2p+h p+h n-2 - ( ² P + h) II ( i=0 (2p+h) h+P n-2 (1 Jei+7p+for+h) (f6i+49+J6i+3k\ f6i+39+f6i+2k n-2 II i=0 (fai+sp + foi+7h) (fot+39 + foi+2k) f6i+49 +f6i+3k\ f6i+7P + foi+ch feng+fen-1kh fon+19+fenk P f6i+8p+f6i+7h f6i+49+f6i+3k f6i+7p+f6i+6h 1) ( f6i+39+f6i+2k h1+ feng+fen-1k fon+19+fenk (f6i+8P + f6i+7h (f6i+49 +f6i+3k\
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
Show me the steps of determine blue and information is here step by step. It complete
![Brn-1n-2
YIn-1 + 8xn-4'
= 0, 1, ..,
In+1 = aIn-2+
(1)
The following special case of Eq.(1) has been studied
Xn-1In-2
In+1 = In-2 +
(8)
In-1 + In-4'
where the initial conditions r-4, x-3, x-2, x-1,and ro are arbitrary non zero real
numbers.
Theorem 4. Let {In}-4 be a solution of Eq.(8). Then for n = 0,1, 2, ..
п-1
feip + fei-ih
( fei+29 + foi+1k
hII
fo
X6n-4
i-1p+ foi-2h
fei+19 + feik
i=0
п-1
fei+4p+ fei+3h
feig + foi-ik
foi+3P + fei+2h) ( Fei-19 + fei-2k)
n-1 ( fei+2P + Joi+" a39 + foi+2k ,
X6n-3
%3D
i=0
fei+49+ fei+3k
X6n-2
fei+1P + feih
i=0
n-1
foi+24+ fei+1k
foi+19 + feik
foi+6p+ fei+5h
PII
foi+5p + fei+ah
X6n-1
%3D
i=0
n-1
П
(fei+4p+ foi+3h\ ( foi+69 + föi+5k
foi+3p + fei+2h)
X6n
%3D
foi+59 + foi+ak,
i=0
п-1
( föi+8P+ fei+7h
П
foi+7P + foi+sh
foi+49 + f6i+3k
fei+39 + f6i+2k)
2р + h
T6n+1
%3D
p+h
i=0
where x-4 = h, x-3 = k, x-2 = r, x-1 = p, xo = q, {fm}m=-1
Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption
holds for n – 2. That is;
100
{1,0, 1, 1, 2, 3, 5, 8, ...}).
п-2
foi+4p + fei+3h
fei+3p + fei+2h) ( foi-19 + fei-2k )
foig + föi-ik
X6n-9
%3D
i=0
п-2
fei+2P + fei+1h( fei+49 + foi+3k
I.
fei+1p + feih
)
X6n-8
%3D
fei+39 + fei+2k,
i=0
п-2
П
( foi+6P + fei+sh\ ( foi+29 + foi+ik
fsi+sP+ fei+ah,
X6n-7
%3D
fei+14 + foik
i=0
п-2
foi+4P + foi+3h
föi+3P + fei+2h ) \foi+59 + foi+ak )
foi+69 + foi+sk`
qII
X6n-6
%3D
i=0
n-2
föi+8P + foi+7h
foi+7P + fei+sh
föi+49 + fei+3k
) Foi+:39 + foi+2k,
2p + h
T6n-5
p+h
i=0
Now, it follows from Eq.(8) that
X6n-6X6n-7
X6n-4 = x6n-7 +
X6n-6 + X6n-9
11](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2F4e0d7a34-2475-411b-b2d5-e2af9eaa9959%2Fbgrpzqe_processed.png&w=3840&q=75)
Transcribed Image Text:Brn-1n-2
YIn-1 + 8xn-4'
= 0, 1, ..,
In+1 = aIn-2+
(1)
The following special case of Eq.(1) has been studied
Xn-1In-2
In+1 = In-2 +
(8)
In-1 + In-4'
where the initial conditions r-4, x-3, x-2, x-1,and ro are arbitrary non zero real
numbers.
Theorem 4. Let {In}-4 be a solution of Eq.(8). Then for n = 0,1, 2, ..
п-1
feip + fei-ih
( fei+29 + foi+1k
hII
fo
X6n-4
i-1p+ foi-2h
fei+19 + feik
i=0
п-1
fei+4p+ fei+3h
feig + foi-ik
foi+3P + fei+2h) ( Fei-19 + fei-2k)
n-1 ( fei+2P + Joi+" a39 + foi+2k ,
X6n-3
%3D
i=0
fei+49+ fei+3k
X6n-2
fei+1P + feih
i=0
n-1
foi+24+ fei+1k
foi+19 + feik
foi+6p+ fei+5h
PII
foi+5p + fei+ah
X6n-1
%3D
i=0
n-1
П
(fei+4p+ foi+3h\ ( foi+69 + föi+5k
foi+3p + fei+2h)
X6n
%3D
foi+59 + foi+ak,
i=0
п-1
( föi+8P+ fei+7h
П
foi+7P + foi+sh
foi+49 + f6i+3k
fei+39 + f6i+2k)
2р + h
T6n+1
%3D
p+h
i=0
where x-4 = h, x-3 = k, x-2 = r, x-1 = p, xo = q, {fm}m=-1
Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption
holds for n – 2. That is;
100
{1,0, 1, 1, 2, 3, 5, 8, ...}).
п-2
foi+4p + fei+3h
fei+3p + fei+2h) ( foi-19 + fei-2k )
foig + föi-ik
X6n-9
%3D
i=0
п-2
fei+2P + fei+1h( fei+49 + foi+3k
I.
fei+1p + feih
)
X6n-8
%3D
fei+39 + fei+2k,
i=0
п-2
П
( foi+6P + fei+sh\ ( foi+29 + foi+ik
fsi+sP+ fei+ah,
X6n-7
%3D
fei+14 + foik
i=0
п-2
foi+4P + foi+3h
föi+3P + fei+2h ) \foi+59 + foi+ak )
foi+69 + foi+sk`
qII
X6n-6
%3D
i=0
n-2
föi+8P + foi+7h
foi+7P + fei+sh
föi+49 + fei+3k
) Foi+:39 + foi+2k,
2p + h
T6n-5
p+h
i=0
Now, it follows from Eq.(8) that
X6n-6X6n-7
X6n-4 = x6n-7 +
X6n-6 + X6n-9
11
![п-2
2р + h
П
föi+49 + f6i+3k`
fsi+39 + föi+2k ,
foi+8P + fei+7h`
)
X6n-2
p+h
fei+7P
+ fei+6h,
i=0
п-2
-() II
2p+h
f6i+8p+f6i+7h
fei+7p+f6i+6h
i=0
föi+49+f6i+3k
f6i+39+f6i+2k
h r
p+h
+
fönq+ fên-1k
h
fon+19+fönk
h+ P
п-2
foi+8P + fói+7h
П
fei+7P + fei+6h,
( föi+49 + f6i+3k
foi+39 + fei+2k,
2р + h
X6n-2
p+h
i=0
п-2
2p+h
p+h
П
f6i+8p+f6i+7h
fei+7p+f6i+6h
f6i+49+ƒ6i+3k
fei+39+f6i+2k
h r
i=0
fönq+ fén-1k
fön+19+fénk
h|1+
п-2
foi+8P + foi+7h`
П
foi+7P + fói+6h ,
föi+49 + f6i+3k`
föi+39 + f6i+2k ,
2р + h
X6n-2
p+h
i=0
п-2
2p+h
p+h
П
f6i+8p+f6i+7h
f6i+7p+ f6i+6h
foi+49+f6i+3k
fei+39+f6i+2k
i=0
+
fon+19+f6nk+fënq+fên–1k
fon+19+ fönk
п-2
foitsP + fói+7h
foi+7P + f6i+6h
foi+49 + f6i+3k
fei+39 + fei+2k,
2р + h
p+h
i=0
n-2
2p+h
p+h
П
f6i+8p+f6i+7h
f6i+7p+f6i+6h
fei+49+f6i+3k
foi+39+f6i+2k /
i=0
fön+29+fön+1k
fön+19+f6nk
п-2
2р + h
П
foi+sP + föi+7h
foi+rP + fei+sh ) Foi+39 + foi+2k ,
(foi+49 + fói+3k
fön+19 + fönk
1+
p+h
fon+29 + fön+1k ]
i=0
п-2
2р + h
П
foi+8P + f6i+7h`
foi+7P+ foi+ch ) \ foi+39 + föi+2k /
(foi+49 + fói+3k \ [ fön+29 + fön+1k + fon+19 + fönk
=
r
p+h
fön+29 + fön+1k
i=0
п-2
föi+8P + f6i+7h`
П
Jöi+7P + fei+6h ) Cein2g i fn) Tôn+39 + fön+2k]
(2р + h
foi+49 + föi+3k`
p+h
[ fön+29 + fön+1k ]
i=0
16](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2F4e0d7a34-2475-411b-b2d5-e2af9eaa9959%2Fsh1sh3m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:п-2
2р + h
П
föi+49 + f6i+3k`
fsi+39 + föi+2k ,
foi+8P + fei+7h`
)
X6n-2
p+h
fei+7P
+ fei+6h,
i=0
п-2
-() II
2p+h
f6i+8p+f6i+7h
fei+7p+f6i+6h
i=0
föi+49+f6i+3k
f6i+39+f6i+2k
h r
p+h
+
fönq+ fên-1k
h
fon+19+fönk
h+ P
п-2
foi+8P + fói+7h
П
fei+7P + fei+6h,
( föi+49 + f6i+3k
foi+39 + fei+2k,
2р + h
X6n-2
p+h
i=0
п-2
2p+h
p+h
П
f6i+8p+f6i+7h
fei+7p+f6i+6h
f6i+49+ƒ6i+3k
fei+39+f6i+2k
h r
i=0
fönq+ fén-1k
fön+19+fénk
h|1+
п-2
foi+8P + foi+7h`
П
foi+7P + fói+6h ,
föi+49 + f6i+3k`
föi+39 + f6i+2k ,
2р + h
X6n-2
p+h
i=0
п-2
2p+h
p+h
П
f6i+8p+f6i+7h
f6i+7p+ f6i+6h
foi+49+f6i+3k
fei+39+f6i+2k
i=0
+
fon+19+f6nk+fënq+fên–1k
fon+19+ fönk
п-2
foitsP + fói+7h
foi+7P + f6i+6h
foi+49 + f6i+3k
fei+39 + fei+2k,
2р + h
p+h
i=0
n-2
2p+h
p+h
П
f6i+8p+f6i+7h
f6i+7p+f6i+6h
fei+49+f6i+3k
foi+39+f6i+2k /
i=0
fön+29+fön+1k
fön+19+f6nk
п-2
2р + h
П
foi+sP + föi+7h
foi+rP + fei+sh ) Foi+39 + foi+2k ,
(foi+49 + fói+3k
fön+19 + fönk
1+
p+h
fon+29 + fön+1k ]
i=0
п-2
2р + h
П
foi+8P + f6i+7h`
foi+7P+ foi+ch ) \ foi+39 + föi+2k /
(foi+49 + fói+3k \ [ fön+29 + fön+1k + fon+19 + fönk
=
r
p+h
fön+29 + fön+1k
i=0
п-2
föi+8P + f6i+7h`
П
Jöi+7P + fei+6h ) Cein2g i fn) Tôn+39 + fön+2k]
(2р + h
foi+49 + föi+3k`
p+h
[ fön+29 + fön+1k ]
i=0
16
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