Therefore n-1 ( fsi+4P + föi+3h foi+3P + fei+2h ) (foi-19+ fei-2k , feiq + fei-1k Tên-3 = k| i=0 Also, from Eq.(8), we see that I6n-4T6n-5 X6n-2 = X6n-5 + I6n-4 + X6n-7 n-2 (2p+ h P+h fei+8p+ fei+7h fei+49+ fei+3k )Ip+ Jaseh) (Fursa + Saesk) foi+7p+ fei+6h i=0 n-1 n-2 2p+h) f6i+8P+S6i+7h Toi+19+f6ik imo i=0 n-2 n-1 h J6i+19+S6ik +p J6i+19+S6ik i=0 n-2 foi+8p+ fei+7h fei+7p + fei+sh, foi+49 + fei+3k \foi+39+ fei+2k, 2р + h П %3D p+h i=0 n-2 Sei+8P+S6i+7h)(!sit49+f6i+3k n-1 n-1 n-2 f6i-1p+S6i-2h fon+19+fônk) feipt fei-1h (f6i+6P+S6i+5h \ (Tên+19+S6nk) i=0 i=0 i=0 n-2 2p +h p+h foi+8P + foi+7h\ fei+7p + fei+sh)fi+39 + fei+2k fei+49 + foi+3k %3D i=0 n-2 (2p+h \ foi+sp+fei+zh P+h11(Toi+7p+f6i+sh) Tei+39+S6i+2k fei+49+Sai+3k h i=0 n-1 n-2 foi-1p+fei-ah feng+Sen-1k h+p(Fon+19+Sonk, i=0 n-1 I foi+sp+Sei+sh f6ip+ Sei-ih i=0 15 n-2 fei+8P+ fei+7h Sei+7P+ foi+sh, fei+19 + fei+3k` foi+39 + f6i+2k , 2p+ h I6n-2 %3D p+h i=0 n-2 foit49+Sei+ak Joi+39+ Sei+2k h Jei+7p+Sei+ah) i=0 fong+fén-ik h Jen+19+Senk p h+p n-2 2p+ h' p+h foi+8P+ fei+7h foi+7P+ foi+sh, foi+49 + fei+3k fei+39 + fei+2k, X6n-2 %3D i-0 n-2 (2p+h I(feitSp+fei+7h feit49+fei+3k Joi+39+Sei+2k fei+7p+fei+sh ) i=0 + hl1+ lông+ fen-1k Jon +19+fenk n-2 2р + h II fei+8P + fei+7h \fói+7p + f6i+6h, fei+49 + f6i+3k fôi+39 + f6i+2k X6n-2 %3D p+h i=0 n-2 (fei+49+f6i+3k` Jei+39+Sei+2k p+h Jei+7p+fei+sh i=0 fen +19+fenk+fênq+Sén–1k fon+19+Senk n-2 2p +h p+h fei+sP + fei+7h' fei+7P+ fei+sh, foi+49 + foi+3k foi+39 + foi+2k, i=0 n-2 Soi+7p+fei+sh ) i=0 feit49+ fei43k foi+39+ fei+2k pth fon+29+fon+1k Sen+19+fenk n-2 fei+49+ fei+3k fei+39 + fei+2k, fön+19 + fenk + (2p +h fei+8P + fei+7h` II fei+7P + fei+6h, = r p+h fen+29 + fen+1k i=0 n-2 fei+8P + fei+7h' fei+7P + fei+sh, fei+49+ f6i+3k fei+39 + fei+2k fon+29 + fen+1k + fen+19+ fenk] fen+29 + fon+1k 2р + h %3D p+h i=0 n-2 2p +h p+h fei+8P + fei+7h` fei+7P + fei+sh, fei+49 + fei+3k foi+39 + fei+2k, fen+39+ fen+2k] [ fön+29 + føn+1k] = r i=0
Therefore n-1 ( fsi+4P + föi+3h foi+3P + fei+2h ) (foi-19+ fei-2k , feiq + fei-1k Tên-3 = k| i=0 Also, from Eq.(8), we see that I6n-4T6n-5 X6n-2 = X6n-5 + I6n-4 + X6n-7 n-2 (2p+ h P+h fei+8p+ fei+7h fei+49+ fei+3k )Ip+ Jaseh) (Fursa + Saesk) foi+7p+ fei+6h i=0 n-1 n-2 2p+h) f6i+8P+S6i+7h Toi+19+f6ik imo i=0 n-2 n-1 h J6i+19+S6ik +p J6i+19+S6ik i=0 n-2 foi+8p+ fei+7h fei+7p + fei+sh, foi+49 + fei+3k \foi+39+ fei+2k, 2р + h П %3D p+h i=0 n-2 Sei+8P+S6i+7h)(!sit49+f6i+3k n-1 n-1 n-2 f6i-1p+S6i-2h fon+19+fônk) feipt fei-1h (f6i+6P+S6i+5h \ (Tên+19+S6nk) i=0 i=0 i=0 n-2 2p +h p+h foi+8P + foi+7h\ fei+7p + fei+sh)fi+39 + fei+2k fei+49 + foi+3k %3D i=0 n-2 (2p+h \ foi+sp+fei+zh P+h11(Toi+7p+f6i+sh) Tei+39+S6i+2k fei+49+Sai+3k h i=0 n-1 n-2 foi-1p+fei-ah feng+Sen-1k h+p(Fon+19+Sonk, i=0 n-1 I foi+sp+Sei+sh f6ip+ Sei-ih i=0 15 n-2 fei+8P+ fei+7h Sei+7P+ foi+sh, fei+19 + fei+3k` foi+39 + f6i+2k , 2p+ h I6n-2 %3D p+h i=0 n-2 foit49+Sei+ak Joi+39+ Sei+2k h Jei+7p+Sei+ah) i=0 fong+fén-ik h Jen+19+Senk p h+p n-2 2p+ h' p+h foi+8P+ fei+7h foi+7P+ foi+sh, foi+49 + fei+3k fei+39 + fei+2k, X6n-2 %3D i-0 n-2 (2p+h I(feitSp+fei+7h feit49+fei+3k Joi+39+Sei+2k fei+7p+fei+sh ) i=0 + hl1+ lông+ fen-1k Jon +19+fenk n-2 2р + h II fei+8P + fei+7h \fói+7p + f6i+6h, fei+49 + f6i+3k fôi+39 + f6i+2k X6n-2 %3D p+h i=0 n-2 (fei+49+f6i+3k` Jei+39+Sei+2k p+h Jei+7p+fei+sh i=0 fen +19+fenk+fênq+Sén–1k fon+19+Senk n-2 2p +h p+h fei+sP + fei+7h' fei+7P+ fei+sh, foi+49 + foi+3k foi+39 + foi+2k, i=0 n-2 Soi+7p+fei+sh ) i=0 feit49+ fei43k foi+39+ fei+2k pth fon+29+fon+1k Sen+19+fenk n-2 fei+49+ fei+3k fei+39 + fei+2k, fön+19 + fenk + (2p +h fei+8P + fei+7h` II fei+7P + fei+6h, = r p+h fen+29 + fen+1k i=0 n-2 fei+8P + fei+7h' fei+7P + fei+sh, fei+49+ f6i+3k fei+39 + fei+2k fon+29 + fen+1k + fen+19+ fenk] fen+29 + fon+1k 2р + h %3D p+h i=0 n-2 2p +h p+h fei+8P + fei+7h` fei+7P + fei+sh, fei+49 + fei+3k foi+39 + fei+2k, fen+39+ fen+2k] [ fön+29 + føn+1k] = r i=0
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 purple and information is here step by step
![(Jên+19+S6nk )II 76ip+S6i-1h
Therefore
п-1
foi+aP + fei+3h'
I6n-3 = k11( fei+3P + fei+2h )
föiq + fei-1k
fei-19+ fei-2k
i=0
Also, from Eq.(8), we see that
X6n-4X6n-5
X6n-2 = X6n-5 +
X6n-4 + x6n-7
n-2
fei+8p+ f6i+7h'
fei+7P + fei+ch
fei+49+ fei+3k
fei+39 + fei+2k,
2р + h
p+h
i=0
n-1
n-2
( 2pth)
pth
f6i+8P+S6i+7h)((si+49+/6i+3k
i=0
i=0
+
n-2
n-
I e 6i+29+16i+1k)+pT(6it+6P+S6i+5h\(f6i+29+S6i+1*
Tôi+19+S6ik )
S6iP+S6i-1h
T6i+19+S6ik
Ssi+5P+S6i+4h ,
i=0
i=0
foi+8P + fei+7h
Jõi+7p + feineh )( 6i+49 + foi+3k
п-2
2р + h
= r
p+h
foi+39 + f6i+2k /
i=0
n-2
2p+h
S6i+8P+SBi+7h
S6i+49+S6i+3k
i=0
n-1
n-1
n-2
f6n9+Sên-1k
( S6i-1P+f6i-2hY
fên+19+f6nk)
f6ip+f6i-1h
+p
( f6i+6P+S6i+5hY
i=0
i=0
i=0
п-2
(2p + h'
fei+8P + f6i+7h
fei+49 + fei+3k
П
fei+7P+ fei+sh) foi+39 + fei+2k,
p+h
i=0
n-2
h r
2p+h
t 11+7p+ fei+sh) (Tei+39+Ssi+2k
fBi+8p+ fei+7h
fei+49+S6i+3k
p+h
i=0
n-1
n-2
| f6i-1p+fsi-2h
I foi+6p+fei+5h
fên9+Sen-1k
Jên+19+Sönk /
i=0
n-2
h+P
i=0
n-1
| fsi+5p+ f6i+4h
foip+Sei-1h
i=0
i=0
15
n-2
fei+49 + fei+3k\
foi+39 + f6i+2k /
2р + h
(foi+sP + f6i+7h`
II
X6n-2
p+h
foi+7P + foi+sh,
i=0
n-2
A() II ( (
2p+h
P+h
foit49+S6i+3k
Sei+39+ Sei+2k
Fes+7p+fei+eh)
i=0
feng+fen-ik h
h +p(Ton+1q+fönk
n-2
foi+sP+ foi+7h
II
foi+7p+ fei+6h,
fei+49 + foi+3k\
f6i+39 + fei+2k,
2р + h
X6n-2
p+h
i=0
hr() II ( ) (f
n-2
f6i+8p+fei+7h
fei+7p+f6i+6h
i=0
f6i+49+f6i+3k
Jei+39+S6i+2k
hrpth
fönq+ fen-1k
fên +19+ fenk
h
n-2
fei+49 + fei+3k
fei+39 + f6i+2k,
2р + h
(foi+sp+ fei+7h
II
foi+7P + foi+6h,
X6n-2
p+h
i=0
п-2
2p+h
p+h
)II
f6i+sp+ fei+7h
Jei+7p+f6i+sh)
i=0
f6i+49+S6i+3k
Sei+39+f6i+2k
+-
fên+19+fénk+fenq+fôn-1k
fen+19+fenk
n-2
fei+8P + fei+7h( foi+49 + fei+3k
foi+7P + fei+6h,
2р +
П
= r
p+h
foi+39 + fei+2k,
i=0
п-2
2p+h T(fsi+8P+f6i+7h
p+h
fei+49+S6i+3k
fei+39+ fei+2k
foi+7p+fei+sh)
i=0
fön+29+fen+1k
fon+19+fenk
n-2
2р + h
II
fei+8P + fei+7h`
foi+7P + fei+6h,
fei+a9+ foi+3k
foi+39 + fei+2k
fon+19 + fönk
fen+29 + fön+1k
p+h
i=0
п-2
fei+8P+ f6i+7h
foi+7P + fei+6h,
foi+49+ fei+3k
f6i+39 + fei+2k,
fön+29 + fon+1k + fen+19 + fenk]
fon+29 + fon+1k
2р + h
= r
p+h
i=0
п-2
fei+8P + f6i+7h`
П
fei+7p + fei+6h) (foi+39 + fei+2k ) [fön+2g + fon+1k
2p +h
foi+49+ foi+3k\ [fon+39 + fon+2k]
= r
p+h
i=0
16](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2Fe94d0573-6165-4d52-b8f4-7e558dca9bca%2Fu8bz4f8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(Jên+19+S6nk )II 76ip+S6i-1h
Therefore
п-1
foi+aP + fei+3h'
I6n-3 = k11( fei+3P + fei+2h )
föiq + fei-1k
fei-19+ fei-2k
i=0
Also, from Eq.(8), we see that
X6n-4X6n-5
X6n-2 = X6n-5 +
X6n-4 + x6n-7
n-2
fei+8p+ f6i+7h'
fei+7P + fei+ch
fei+49+ fei+3k
fei+39 + fei+2k,
2р + h
p+h
i=0
n-1
n-2
( 2pth)
pth
f6i+8P+S6i+7h)((si+49+/6i+3k
i=0
i=0
+
n-2
n-
I e 6i+29+16i+1k)+pT(6it+6P+S6i+5h\(f6i+29+S6i+1*
Tôi+19+S6ik )
S6iP+S6i-1h
T6i+19+S6ik
Ssi+5P+S6i+4h ,
i=0
i=0
foi+8P + fei+7h
Jõi+7p + feineh )( 6i+49 + foi+3k
п-2
2р + h
= r
p+h
foi+39 + f6i+2k /
i=0
n-2
2p+h
S6i+8P+SBi+7h
S6i+49+S6i+3k
i=0
n-1
n-1
n-2
f6n9+Sên-1k
( S6i-1P+f6i-2hY
fên+19+f6nk)
f6ip+f6i-1h
+p
( f6i+6P+S6i+5hY
i=0
i=0
i=0
п-2
(2p + h'
fei+8P + f6i+7h
fei+49 + fei+3k
П
fei+7P+ fei+sh) foi+39 + fei+2k,
p+h
i=0
n-2
h r
2p+h
t 11+7p+ fei+sh) (Tei+39+Ssi+2k
fBi+8p+ fei+7h
fei+49+S6i+3k
p+h
i=0
n-1
n-2
| f6i-1p+fsi-2h
I foi+6p+fei+5h
fên9+Sen-1k
Jên+19+Sönk /
i=0
n-2
h+P
i=0
n-1
| fsi+5p+ f6i+4h
foip+Sei-1h
i=0
i=0
15
n-2
fei+49 + fei+3k\
foi+39 + f6i+2k /
2р + h
(foi+sP + f6i+7h`
II
X6n-2
p+h
foi+7P + foi+sh,
i=0
n-2
A() II ( (
2p+h
P+h
foit49+S6i+3k
Sei+39+ Sei+2k
Fes+7p+fei+eh)
i=0
feng+fen-ik h
h +p(Ton+1q+fönk
n-2
foi+sP+ foi+7h
II
foi+7p+ fei+6h,
fei+49 + foi+3k\
f6i+39 + fei+2k,
2р + h
X6n-2
p+h
i=0
hr() II ( ) (f
n-2
f6i+8p+fei+7h
fei+7p+f6i+6h
i=0
f6i+49+f6i+3k
Jei+39+S6i+2k
hrpth
fönq+ fen-1k
fên +19+ fenk
h
n-2
fei+49 + fei+3k
fei+39 + f6i+2k,
2р + h
(foi+sp+ fei+7h
II
foi+7P + foi+6h,
X6n-2
p+h
i=0
п-2
2p+h
p+h
)II
f6i+sp+ fei+7h
Jei+7p+f6i+sh)
i=0
f6i+49+S6i+3k
Sei+39+f6i+2k
+-
fên+19+fénk+fenq+fôn-1k
fen+19+fenk
n-2
fei+8P + fei+7h( foi+49 + fei+3k
foi+7P + fei+6h,
2р +
П
= r
p+h
foi+39 + fei+2k,
i=0
п-2
2p+h T(fsi+8P+f6i+7h
p+h
fei+49+S6i+3k
fei+39+ fei+2k
foi+7p+fei+sh)
i=0
fön+29+fen+1k
fon+19+fenk
n-2
2р + h
II
fei+8P + fei+7h`
foi+7P + fei+6h,
fei+a9+ foi+3k
foi+39 + fei+2k
fon+19 + fönk
fen+29 + fön+1k
p+h
i=0
п-2
fei+8P+ f6i+7h
foi+7P + fei+6h,
foi+49+ fei+3k
f6i+39 + fei+2k,
fön+29 + fon+1k + fen+19 + fenk]
fon+29 + fon+1k
2р + h
= r
p+h
i=0
п-2
fei+8P + f6i+7h`
П
fei+7p + fei+6h) (foi+39 + fei+2k ) [fön+2g + fon+1k
2p +h
foi+49+ foi+3k\ [fon+39 + fon+2k]
= r
p+h
i=0
16
![Brn-1In-2
YIn-1 + &xn-4
In+1 = a1n-2 +
n = 0, 1,...,
(1)
1
The following special case of Eq.(1) has been studied
In-1In-2
Int1 = In-2+
(8)
Tn-1 + In-4
where the initial conditions r-4, I-3, T-2, T-1,and xo are arbitrary non zero real
numbers.
Theorem 4. Let {zn}-4 be a solution of Eq.-(8). Then for n = 0, 1,2, ...
fep + fes-ih ( foi+29 + foiik
Jei-1p+ fei-zh)
fo419 + Sauk)
farq + Sai-1k
feirap + forah ) To-19 + foi-ak ,
(Sei+aP + fot+3h)
fei-2P + fei+1h (foi+19 + Sei+ak
= rT
fei+1P+ fesh ) Joi+39 + fei+2k )
(fei+ep + fes+sh\ (foi+29 + fei+ik
Soi-sp + feisah)
fei+ap + fei+ah\ ( feiseq + fei-sk
Sai+ap + fei+2h ) Jai459 + Seisak )
fei+19 + fask
2p +h
II ap+ fesch) (Tars9 + Suvzk,
foi+sP + fouth ( Seisa9 + Soirak
In+1 =
p+h
where r-4 = h, r-3 = k, z-2 = r, r-1 = P, 2o = q, {fm}-1 = {1,0,1,1,2, 3, 5, 8, .}.
Proof: For n -0 the result holds. Now suppose that n>0 and that our assumption
holds for n - 2. That is;
feirap + foi+ah
Jei+ap + fei42h) J6i-19 + fei-2k
fauq + foi-1 k
i-0
fai+1p + fesh ) fet439 + feirak)
Joi+ep + fei+sh (fei+29 + foi+zk
foi+sp + fes+ah)
Iom-6 = 9TT(fei-4p + foi+3h (fei+09 + fei-sk
+ fei+zh) \Torr39 + fei+ak ) *
(2p + h ( foi-sp+ fesuzh (foir49 + foi+ak
Ji+7p + Soisgh) Tausg + Soi+zk
p+h
Now, it follows from Eq.(8) that
IGn-4 = Iộn-7 +
11
feisep + fo+sh ( foi+29 + fe+ik
Sunsp + Seissh ) (Tus1g+ fesk)
= p||
•II )( )II( ) )
(foi+op + fou-sh ( fei-29 + fei+ikY
1foi+sp + foi+sh)
foi+19 + fosk
+ A
PIT fasop + foish) ( for-29 + Sousik
Joi4sp+fostah)
n-2
= p
Fei+sp + feirah)
feir19 + faik
pg
( foi+6P + fei+sh) ( fei+29 + fei-ik
=PlI(asp+ fash)( Far1q + Sak )
12
PII
´ fouvap + faunh\ ( fa-29 + feieik)
Sos+19 + Souk
PIIusp + Jourah)
1+ (
PII( ) )
-PT (furep + fouosh (fau29 + fuosk
Sei-sp + foi-gh )
fois19 + fuk)
n-2
PII( (unt
Jeusp + Sanah)
Sesig+ fauk
(An
( fesep + fursh ( fev29 + Jairak1+a+ fon-ek
- PII
Sen-o9 + fon-zk]
n-2
Sen-s9 + Son -ak-"
Sei+ep + Sei+sh) ( Su-29 + Sai+ik [ Sen-sq + Sen-sk
= PII(sp+ fursh)Tuer9+ feck ) [Ton-39 + Sen-sk]
Therefore
R-1
Also, from Eq.(8), we see that
Tusap + Seurzli ) ( Jonnog + Senek )
( )--II( )( )
13](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2Fe94d0573-6165-4d52-b8f4-7e558dca9bca%2Fb44uf1f_processed.png&w=3840&q=75)
Transcribed Image Text:Brn-1In-2
YIn-1 + &xn-4
In+1 = a1n-2 +
n = 0, 1,...,
(1)
1
The following special case of Eq.(1) has been studied
In-1In-2
Int1 = In-2+
(8)
Tn-1 + In-4
where the initial conditions r-4, I-3, T-2, T-1,and xo are arbitrary non zero real
numbers.
Theorem 4. Let {zn}-4 be a solution of Eq.-(8). Then for n = 0, 1,2, ...
fep + fes-ih ( foi+29 + foiik
Jei-1p+ fei-zh)
fo419 + Sauk)
farq + Sai-1k
feirap + forah ) To-19 + foi-ak ,
(Sei+aP + fot+3h)
fei-2P + fei+1h (foi+19 + Sei+ak
= rT
fei+1P+ fesh ) Joi+39 + fei+2k )
(fei+ep + fes+sh\ (foi+29 + fei+ik
Soi-sp + feisah)
fei+ap + fei+ah\ ( feiseq + fei-sk
Sai+ap + fei+2h ) Jai459 + Seisak )
fei+19 + fask
2p +h
II ap+ fesch) (Tars9 + Suvzk,
foi+sP + fouth ( Seisa9 + Soirak
In+1 =
p+h
where r-4 = h, r-3 = k, z-2 = r, r-1 = P, 2o = q, {fm}-1 = {1,0,1,1,2, 3, 5, 8, .}.
Proof: For n -0 the result holds. Now suppose that n>0 and that our assumption
holds for n - 2. That is;
feirap + foi+ah
Jei+ap + fei42h) J6i-19 + fei-2k
fauq + foi-1 k
i-0
fai+1p + fesh ) fet439 + feirak)
Joi+ep + fei+sh (fei+29 + foi+zk
foi+sp + fes+ah)
Iom-6 = 9TT(fei-4p + foi+3h (fei+09 + fei-sk
+ fei+zh) \Torr39 + fei+ak ) *
(2p + h ( foi-sp+ fesuzh (foir49 + foi+ak
Ji+7p + Soisgh) Tausg + Soi+zk
p+h
Now, it follows from Eq.(8) that
IGn-4 = Iộn-7 +
11
feisep + fo+sh ( foi+29 + fe+ik
Sunsp + Seissh ) (Tus1g+ fesk)
= p||
•II )( )II( ) )
(foi+op + fou-sh ( fei-29 + fei+ikY
1foi+sp + foi+sh)
foi+19 + fosk
+ A
PIT fasop + foish) ( for-29 + Sousik
Joi4sp+fostah)
n-2
= p
Fei+sp + feirah)
feir19 + faik
pg
( foi+6P + fei+sh) ( fei+29 + fei-ik
=PlI(asp+ fash)( Far1q + Sak )
12
PII
´ fouvap + faunh\ ( fa-29 + feieik)
Sos+19 + Souk
PIIusp + Jourah)
1+ (
PII( ) )
-PT (furep + fouosh (fau29 + fuosk
Sei-sp + foi-gh )
fois19 + fuk)
n-2
PII( (unt
Jeusp + Sanah)
Sesig+ fauk
(An
( fesep + fursh ( fev29 + Jairak1+a+ fon-ek
- PII
Sen-o9 + fon-zk]
n-2
Sen-s9 + Son -ak-"
Sei+ep + Sei+sh) ( Su-29 + Sai+ik [ Sen-sq + Sen-sk
= PII(sp+ fursh)Tuer9+ feck ) [Ton-39 + Sen-sk]
Therefore
R-1
Also, from Eq.(8), we see that
Tusap + Seurzli ) ( Jonnog + Senek )
( )--II( )( )
13
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