Post on 23-Mar-2018
© 2014, John Bird
1285
CHAPTER 83 POWER SERIES METHODS OF SOLVING
ORDINARY DIFFERENTIAL EQUATIONS
EXERCISE 312 Page 864
1. Determine the following derivatives: (a) (4)y when 2e xy = (b) (5)y when 28et
y =
(a) If eaxy = , then ( ) en n axy a= . Hence, if y = 2e x , then ( )4(4) 22 e xy = = 216e x
(b) If eaxy = , then ( ) en n axy a= . Hence, if y =128e t , then
5 1 1(5) 2 2
1 8(8) e e2 32
t ty = =
= 12
1 e4
t
2. Determine the following derivatives: (a) (4)y when y = sin 3t (b) (7)y when y = 150
sin 5θ
(a) If y = sin ax, then ( ) sin2
n n ny a ax π = +
Hence, if y = sin 3t, then ( )(4) 4 43 sin 3 81sin 3 22
y t tπ π = + = +
= 81 sin 3t
(b) If y = 150
sin 5θ, then (7) 71 7 35 sin 5 1562.5sin 550 2 2
y π πθ θ = + = +
= – 1562.5 cos 5θ
3. Determine the following derivatives: (a) (8)y when y = cos 2x (b) (9)y when y = 3 cos 23
t
(a) If y = cos ax, then ( ) cos2
n n ny a ax π = +
Hence, if y = cos 2x then (8)y = ( )8 82 cos 22
x π +
= 256 cos(2x + 4π) = 256 cos 2x
(b) If y = 23cos3
t , then 9 9
(9)8
2 2 9 2 2(3) cos cos3 3 2 3 3 2
y t tπ π = + = +
= 9
8
2 2sin3 3
t−
© 2014, John Bird
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4. Determine the following derivatives: (a) (7)y when y = 92x (b) (6)y when y = 7
8t
(a) If y = ax , then ( )
( ) !!
n a nay xa n
−=−
Hence, if y = 92x , then ( )
(7) 9 79!(2)9 7 !
y x −=−
= ( ) 29! x
(b) If y =7
8t , then
( )(6) 7 61 7!
8 7 6 !y t − = −
= 630 t
5. Determine the following derivatives: (a) (7)y when y = 1 sinh 24
x (b) (6)y when y = 2 sinh 3x
(a) If y = sinh ax, then [ ] [ ]{ }( ) 1 ( 1) sinh 1 ( 1) cosh2
nn n nay ax ax= + − + − −
Hence, if y = 1 sinh 24
x then [ ] [ ]{ }7
(7) 7 71 2 1 ( 1) sinh 2 1 ( 1) cosh 24 2
y x x = + − + − −
= 16 (2 cosh 2x) = 32 cosh 2x
(b) If y = 2 sinh 3x, then [ ] [ ]{ }6
(6) 6 63(2) 1 ( 1) sinh 3 1 ( 1) cosh 32
y x x= + − + − −
= { }63 2sinh 3 0x + = 1458 sinh 3x
6. Determine the following derivatives: (a) (7)y when y = cosh 2x (b) (8)y when y = 1 cosh 39
x
(a) If y = cosh ax, then [ ] [ ]{ }( ) 1 ( 1) sinh 1 ( 1) cosh2
nn n nay ax ax= − − + + −
Hence, if y = cosh 2x, then [ ] [ ]{ }7
(7) 7 72 1 ( 1) sinh 2 1 ( 1) cosh 22
y x x= − − + + −
= { }6 72 2sinh 2 0 2 sinh 2x x+ = = 128 sinh 2x
(b) If 1 cosh 39
x , then [ ] [ ]{ }8
(8) 8 81 3 1 ( 1) sinh 3 1 ( 1) cosh 39 2
y x x = − − + + −
© 2014, John Bird
1287
= { }364.5 0 2cosh 3x+ = 729 cosh 3x
7. Determine the following derivatives: (a) (4)y when y = 2 ln 3θ (b) (7)y when y = 1 ln 23
t
(a) If y = ln ax, then ( ) ( )1( )1 !
1 nnn
ny
x− −
= −
If y = 2 ln 3θ, then ( )( ) ( )4 1(4)4 4
4 1 ! 3!2 1 ( 2)yθ θ
− −= − = − =
4
12θ
−
(b) If y = 1 ln 23
t , then ( ) ( )7 1(7)7 7
7 1 !1 6!13 3
yt t
− − = − =
= 7
240t
© 2014, John Bird
1288
EXERCISE 313 Page 866
1. Obtain the nth derivative of: 2x y
Since y = 2x y then let v = 2x and u = y
Thus, ( ) ( ) ( 1) (1) ( 2) (2)( 1) ...2!
n n n nn ny u v nu v u v− −−
= + + +
= ( ) ( ) ( )( ) 2 ( 1) ( 2)( 1)2 22!
n n nn ny x n y x y− −−
+ +
= 2 ( ) ( 1) ( 2)2 ( 1)n n nx y nxy n n y− −+ + −
2. If y = 3 2e xx find ( )ny and hence (3)y
Since y = 3 2e xx then let v = 3x and u = 2e x and the nth derivative of 2e x is 22 en x
Thus, ( ) ( ) ( 1) (1) ( 2) (2)( 1) ...2!
n n n nn ny u v nu v u v− −−
= + + +
= ( ) ( )( ) ( )( ) ( )( )2 3 1 2 2 2 2 ( 3) 2
1 2( 1)2 e 2 e 3 2 e 6 2 e (6)2! 3!
n x n x n x n xn n nn nx n x x− − −
− −−+ + +
= ( )( )3 2 2 1 2 2 2 3 22 e 3 2 e 3 ( 1)2 e 1 2 2 en x n x n x n xx nx n n x n n n− − −+ + − + − −
or { }( ) 2 3 3 3 2 2e 2 2 3 (2) 3 ( 1) (2) ( 1)( 2)n x ny x nx n n x n n n−= + + − + − −
= { }2 3 3 2e 2 8 12 ( 1)(6 ) ( 1)( 2)x n x nx n n x n n n− + + − + − −
Hence, { }(3) 2 0 3 2e 2 8 36 3(2)6 3(2)(1)xy x x x= + + +
= { }2 3 2e 8 36 36 6x x x x+ + +
3. Determine the 4th derivative of: y = 2 3 e xx −
Since y = 32 e xx − then let v = 2 3x and u = e x− and the nth derivative of e x− is ( 1) en x−−
Thus, ( ) ( ) ( 1) (1) ( 2) (2)( 1) ...2!
n n n nn ny u v nu v u v− −−
= + + +
© 2014, John Bird
1289
= ( ) ( )( ) ( )( ) ( )( )3 1 2 2 ( 3)
1 2( 1)( 1) e 2 ( 1) e 6 ( 1) e 12 ( 1) e (12)2! 3!
n x n x n x n xn n nn nx n x x− − − − − − −
− −−− + − + − + −
Hence, ( ) ( )( ) ( )( ) ( )(4) 4 3 3 2 2 1( 1) e 2 4 ( 1) e 6 6 ( 1) e 12 4 ( 1) e (12)x x x xy x x x− − − −= − + − + − + −
= { }3 2e 2 24 72 48x x x x− − + − or { }3 22e 12 36 24x x x x− − + −
4. If y = 3 cosx x , determine the 5th derivative.
Since y = 3 cosx x then let u = cos x and v = 3x and ( ) 1 cos2
n n nu x π = +
( ) ( 1) (1) ( 2) (2)( 1) ...2!
n n n nn ny u v nu v u v− −−
= + + +
Hence, ( ) ( ) ( )( ) 3 2( 1) ( 1) ( 2)cos cos 3 cos 62 2 2! 2
n n n n n ny x x n x x x xπ π π− − − = + + + + +
( )( 1( 2) ( 3)cos 63! 2
n n n nx π− − − + +
and ( ) ( ) ( )(5) 3 25 4 5(4) 3 5(4)(3) 2cos 5 3 cos 6 cos 6 cos2 2 2! 2 3! 2
y x x x x x x xπ π π π = + + + + + + +
= 3 2sin 15 cos 60 sin 60( cos )x x x x x x x− + + + −
= ( ) ( )3 260 sin 15 60 cosx x x x x− + −
5. Find an expression for (4)y if y = e sint t− .
Since y = e sint t− then let u = sin t and v = e t− and the nth derivative of e t− is ( 1) en t−−
( ) ( 1) (1) ( 2) (2)( 1) ...2!
n n n nn ny u v nu v u v− −−
= + + +
Hence,
( ) ( )
( ) ( )
( ) ( 1)sin e sin e2 2
( 1) ( 2) ( 1)( 2) ( 3)sin e sin e2! 2 3! 2
n t t
t t
n ny t n t
n n n n n n nt t
π π
π π
− −
− −
− = + + + −
− − − − − + + + + −
( )( 1( 2)( 3) ( 4)sin e4! 2
tn n n n nt π−
− − − − + +
and ( ) ( )(4) 4 3 4(3) 2 4(3)(2)e sin 4e sin e sin e sin2 2 2! 2 3! 2
t t t ty t t t tπ π π π− − − − = + − + + + − +
© 2014, John Bird
1290
( )( )4(3)(2)(1) sin e4!
tt −+
= ( ) ( ) ( )3e sin 4e sin 6 e sin 4 e sin2 2
t t t tt t t tπ ππ− − − − − + + + − +
e sint t−+
= e sin 4e cos 6e sin 4e cost t t tt t t t− − − −+ − − e sint t−+
= 4e sint t−−
6. If y = 5 ln 2x x find (3)y
Since y = 5 ln 2x x then let u = 5x and v = ln 2x and ( )
5! 5!! (5 )!
n a n nau x xa n n
− −= =− −
( ) ( 1) (1) ( 2) (2)( 1) ...2!
n n n nn ny u v nu v u v− −−
= + + +
( ) 5 6 72
83
5! 5! 1 ( 1) 5! 1ln 2(5 )! (6 )! 2! (7 )!
( 1)( 2) 5! 23! (8 )!
n n n n
n
n ny x x n x xn n x n x
n n n xn x
− − −
−
− = + + − − − − − − + −
Hence, 3 2 3 4 52 3
5! 5! 1 3(2) 5! 1 3(2)(1) 5! 2ln 2 (3)2! 3! 2! (4)! 3! 5!
y x x x x xx x x
= + + − +
= 2 2 2 260 ln 2 60 15 2x x x x x+ − +
= 2 260 ln 2 47x x x+
i.e. ( )(3) 2 47 60ln 2y x x= +
7. Given 22 '' ' 3 0x y xy y+ + = show that ( ) ( )2 ( 2) ( 1) 2 ( )2 4 1 2 3 0n n nx y n x y n n y+ ++ + + − + =
Differentiating each term of 2 2 '' ' 3 0x y xy y+ + = n times, using Leibniz’s theorem of equation (13),
gives: ( )( 2) 2 ( 1) ( )( 1)2 2 (2) 02!
n n nn ny x n y x y+ +− + + +
+ { }( 1) ( )( ) (1) 0n ny x ny+ + + + 3{ }( )ny = 0
i.e. 2 ( 2) ( 1) ( ) ( 1) ( ) ( )2 4 2 ( 1) 3n n n n n nx y n x y n n y x y n y y+ + ++ + − + + + = 0
i.e. 2 ( 2) ( 1) 2 ( )2 (4 1) (2 2 3)n n nx y n x y n n n y+ ++ + + − + + = 0
or 2 2 ( 2) ( 1) 2 ( )(4 1) (2 3)n n nx y n x y n n y+ ++ + + − + = 0
© 2014, John Bird
1291
8. If y = ( )3 2 22 e xx x+ determine an expansion for (5)y
Since y = ( )3 2 22 e xx x+ then let u = 2e x and v = ( )3 22x x+ and 22 en n xu =
( ) ( 1) (1) ( 2) (2)( 1) ...2!
n n n nn ny u v nu v u v− −−
= + + +
Hence, ( )( ) ( ) ( )( ) 2 3 2 1 2 2 2 2( 1)2 e 2 2 e 3 4 2 e 6 42!
n n x n x n xn ny x x n x x x− −−
= + + + + +
( )3 2( 1)( 2) 2 e 63!
n xn n n−
− −+
and ( ) ( ) ( ) ( )(5) 5 2 3 2 4 2 2 3 2 2 25(4) 5(4)(3)2 e 2 (5) 2 e 3 4 2 e 6 4 2 e 62 3!
x x x xy x x x x x= + + + + + +
= { }2 5 3 6 2 2e 2 2 (16)15 (16)(20 ) 60 (8) (8)(40) 240x x x x x x+ + + + + +
= { }2 5 3 2e 2 304 800 560x x x x+ + +
= { }2 5 3 4 2 4 4e 2 2 (19 ) 2 (50)( ) 2 (35)x x x x+ + +
= { }2 4 3 2e 2 2 19 50 35x x x x+ + +
© 2014, John Bird
1292
EXERCISE 314 Page 869
1. Determine the power series solution of the differential equation: 2
2
d d2 0d d
y yx yx x
+ + =
using the Leibniz–Maclaurin method, given that at x = 0, y = 1 and dd
yx
= 2
2
2
d d2 0d d
y yx yx x
+ + =
(i) The differential equation is rewritten as: y′′ + 2xy′ + y = 0 and from the Leibniz theorem of
equation (13), page 865 of textbook, each term is differentiated n times, which gives:
{ }( 2) ( 1) ( ) ( )2 ( ) (1) 0 0n n n ny y x n y y+ ++ + + + =
i.e. ( 2) ( 1) ( )2 (2 1) 0n n ny x y n y+ ++ + + = (1)
(ii) At x = 0, equation (1) becomes:
( 2) ( )(2 1) 0n ny n y+ + + =
from which, ( 2) ( )(2 1)n ny n y+ = − +
This is the recurrence formula
(iii) For n = 0, ( ) ( )0 0''y y= −
n = 1, ( ) ( )0 0''' 3 'y y= −
n = 2, ( ) ( )(4) 00 05 '' 5( )y y y= − =
n = 3, ( ) ( )(5)0 0
7 '''y y= − = ( ){ } ( )0 07 3 ' 3 7 'y y− − = ×
n = 4, ( ) ( )(6) (4)0 0
9y y= − = ( ){ } ( )0 09 5 5 9y y− = − ×
n = 5, ( ) ( )(7) (5)0 0
11y y= − = ( ){ } ( )0 011 3 7 ' 3 7 11 'y y− × = − × ×
n = 6, ( ) ( )(8) (6)0 0
13y y= − = ( ){ } ( )0 013 5 9 5 9 13y y− − × = × ×
(iv) Maclaurin’s theorem is: y = ( ) ( ) ( ) ( ) ( )2 3 4
(4)0 0 0 0 0
' '' ''' ...2! 3! 4!x x xy x y y y y+ + + + +
Thus, y = ( ) ( ) ( ){ } ( ){ } ( ){ } ( ){ }2 3 4 5
0 0 0 0 0 0' 3 ' 5 3 7 '
2! 3! 4! 5!x x x xy x y y y y y+ + − + − + + ×
( ){ } ( ){ }6 7
0 05 9 3 7 11 '
6! 7!x xy y+ − × + − × ×
© 2014, John Bird
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(v) Collecting similar terms together gives:
y = ( )2 4 6 8
0
5 5 9 5 9 131 ...2! 4! 6! 8!x x x xy × × × − + − + −
( )3 5 7
0
3 3 7 3 7 11' ...3! 5! 7!x x xy x × × × + − + − +
At x = 0, y = 1 and dd
yx
= 2, hence, ( )01y = and ( )0
' 2y = .
Hence, the power series solution of the differential equation: 2
2
d d2 0d d
y yx yx x
+ + = is:
y = 2 4 6 85 5 9 5 9 131 ...
2! 4! 6! 8!x x x x× × × − + − + −
3 5 73 3 7 3 7 112 ...3! 5! 7!x x xx × × × + − + − +
2. Show that the power series solution of the differential equation: ( ) ( )2
2
d d1 1 2 0d d
y yx x yx x
+ + − − = ,
using the Leibniz–Maclaurin method, is given by: 21 e xy x −= + + given the boundary
conditions that at x = 0, y = 2 and dd
yx
= –1
( ) ( )2
2
d d1 1 2 0d d
y yx x yx x
+ + − − =
(i) The differential equation is rewritten as: (x + 1) y′′ + (x – 1)y′ – 2y = 0 and from the Leibniz
theorem of equation (13), page 865 of textbook, each term is differentiated n times, which
gives:
{ } { }( 2) ( 1) ( 1) ( ) ( )( 1) (1) 0 ( 1) (1) 0 2 0n n n n ny x ny y x n y y+ + ++ + + + − + + − =
i.e. (x + 1) ( 2) ( 1) ( )( 1) ( 2) 0n n ny n x y n y+ ++ + − + − = (1)
(ii) At x = 0, equation (1) becomes:
( 2) ( 1) ( )( 1) ( 2) 0n n ny n y n y+ ++ − + − =
from which, ( 2) ( 1) ( )(1 ) (2 )n n ny n y n y+ += − + −
This is the recurrence formula
(iii) For n = 0, ( ) ( )(2) (1) 00 02( )y y y= +
n = 1, ( ) ( )(3) (1)0 0
y y=
© 2014, John Bird
1294
n = 2, ( ) ( ) ( )(4) (3) (1)0 0 0
y y y= − = −
n = 3, ( ) ( ) ( ) ( ) ( ) ( )(5) (4) (3) (1) (1) (1)0 0 0 0 0 0
2 2y y y y y y= − − = − =
n = 4, ( ) ( ) ( ) ( ) ( ) ( )(6) (5) (4) (1) (1) (1)0 0 0 0 0 0
3 2 3 2y y y y y y= − − = − + = −
n = 5, ( ) ( ) ( ) ( ) ( ) ( )(7) (6) (5) (1) (1) (1)0 0 0 0 0 0
4 3 4 3y y y y y y= − − = − =
n = 6, ( ) ( ) ( ) ( ) ( ) ( )(8) (7) (6) (1) (1) (1)0 0 0 0 0 0
5 4 5 4y y y y y y= − − = − + = −
(iv) Maclaurin’s theorem is: y = ( ) ( ) ( ) ( ) ( )2 3 4
(1) (2) (3) (4)0 0 0 0 0
...2! 3! 4!x x xy x y y y y+ + + + +
Thus, y = ( ) ( ) ( ) ( ){ } ( ){ } ( ){ } ( ){ }2 3 4 5
(1) (1) (1) (1) (1)0 0 0 0 0 0 0
22! 3! 4! 5!x x x xy x y y y y y y+ + + + + − +
( ){ } ( ){ } ( ){ }6 7 8
(1) (1) (1)0 0 0
...6! 7! 8x x xy y y+ − + + − +
(v) Collecting similar terms together gives:
y = ( )2
01 (2)
2!xy +
( )
2 3 4 5 6 7(1)
0...
2! 3! 4! 5! 6! 7!x x x x x xy x + + + − + − + −
At x = 0, y = 2 and dd
yx
= –1, hence, ( )02y = and ( )(1)
01y = −
Hence, the power series solution of the differential equation: ( ) ( )2
2
d d1 1 2 0d d
y yx x yx x
+ + − − =
is:
y = 2{ }21 x+2 3 4 5 6 7
...2! 3! 4! 5! 6! 7!x x x x x xx − + + − + − + −
= 2 + 2x 2 – x – 2 3 4 5 6 7
...2! 3! 4! 5! 6! 7!x x x x x x
− + − + − +
= 1 + x 2 + 1 + x 2 – x – 2 3 4 5 6 7
...2! 3! 4! 5! 6! 7!x x x x x x
− + − + − +
= 1 + x 2 + 1 – x + 2 3 4 5 6 7
...2! 3! 4! 5! 6! 7!x x x x x x
− + − + − +
i.e. y = 1 + x 2 + e x− since xe− = 1 – x + 2 3 4 5 6 7
...2! 3! 4! 5! 6! 7!x x x x x x
− + − + − +
3. Find the particular solution of the differential equation: ( )2
22
d d1 4 0d d
y yx x yx x
+ + − = using the
Leibniz–Maclaurin method, given the boundary conditions that at x = 0, y = 1 and dd
yx
= 1
© 2014, John Bird
1295
( )2
22
d d1 4 0d d
y yx x yx x
+ + − =
i.e. ( )2 1x + y′′ + xy′ – 4y = 0
i.e. ( ) { }2 ( 2) ( 1) ( ) ( 1) ( )( 1)1 (2 ) (2) (1) 4 02!
n n n n n nn nx y ny x y y x ny y+ + +− + + + + + − =
i.e. ( ) ( )2 ( 2) ( 1) ( )1 2 ( ( 1) 4) 0n n nx y nx x y n n n y+ ++ + + + − + − =
At x = 0, ( )( 2) 2 ( )4 0n ny n y+ + − =
from which, ( )( 2) 2 ( )4n ny n y+ = − which is the recurrence formula
For n = 0, ( ) ( )0 0'' 4y y=
n = 1, ( ) ( )0 0''' 3 'y y=
n = 2, ( )(4)0
0y =
n = 3, ( ) ( )(5)0 0
5 '''y y= − = ( ){ } ( )( )0 05 3 ' 5 3 'y y− − = −
n = 4, ( ) ( )(6) (4)0 0
12y y= − = 12(0) 0− =
n = 5, ( ) ( )(7) (5)0 0
21y y= − = ( ){ } ( )0 021 5 3 ' 315 'y y− − × =
Maclaurin’s theorem is: y = ( ) ( ) ( ) ( ) ( )2 3 4
(4)0 0 0 0 0
' '' ''' ...2! 3! 4!x x xy x y y y y+ + + + +
Thus, y = ( ) ( ) ( ){ } ( ){ } { } ( ){ }2 3 4 5
0 0 0 0 0' 4 3 ' 0 3 5 ' 0
2! 3! 4! 5!x x x xy x y y y y+ + + + + − × + ( ){ }
7
0315 '
7!x y+
i.e. y = ( ) { }20
1 2y x+ ( )3 5 7
0' ...
2 8 16x x xy x + + − + +
At x = 0, y = 1 and dd
yx
= 1, hence, ( )01y = and ( )0
' 1y =
Hence, the power series solution of the differential equation: ( )2
22
d d1 4 0d d
y yx x yx x
+ + − = is:
y = { }21 2x+3 5 7
...2 8 16x x xx + + − + +
i.e. y = 3 5 7
21 2 ...3 8 16x x xx x+ + + − + +
4. Use the Leibniz–Maclaurin method to determine the power series solution for the differential
© 2014, John Bird
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equation: 2
2
d d 1d d
y yx xyx x
+ + = given that at x = 0, y = 1 and dd
yx
= 2
2
2
d d 1d d
y yx xyx x
+ + =
i.e. x y′′ + y′ + xy = 0
i.e. { } { } { }( 2) ( 1) ( 1) ( ) ( 1)(1) ) (1) 0n n n n nxy ny y xy ny+ + + −+ + + + + =
i.e. ( )( 2) ( 1) ( ) ( 1)1 0n n n nxy n y xy ny+ + −+ + + + =
At x = 0, ( ) ( 1) ( 1)1 0n nn y ny+ −+ + =
from which, ( 1) ( 1)
1n nny y
n+ −= −
+ which is the recurrence formula
For n = 1, ( ) ( )(2)0 0
12
y y= −
n = 2, ( ) ( )(3) (1)0 0
23
y y= −
n = 3, ( ) ( ) ( ) ( )(4) (2)0 0 0 0
3 3 1 34 4 2 8
y y y y = − = − − =
n = 4, ( ) ( ) ( ) ( )(5) (3) (1) (1)0 0 0 0
4 4 2 85 5 3 15
y y y y = − = − − =
n = 5, ( ) ( ) ( ) ( )(6) (4)0 0 0 0
5 5 3 156 6 8 48
y y y y = − = − = −
n = 6, ( ) ( ) ( ) ( )(7) (5) (1) (1)0 0 0 0
6 6 8 167 7 15 35
y y y y = − = − =
Maclaurin’s theorem is: y = ( ) ( ) ( ) ( ) ( )2 3 4
(1) (2) (3) (4)0 0 0 0 0
...2! 3! 4!x x xy x y y y y+ + + + +
Thus,
y = ( ) ( ) ( ) ( ) ( )2 3 4 5
(1) (1) (1)00 0 0 0 0
1 2 3 8( )2! 2 3! 3 4! 8 5! 15x x x xy x y y y y y + + − + − + +
( ) ( )6 7
(1)0 0
15 16 ...6! 48 7! 35x xy y − + +
i.e. y = ( ) 2 4 60
1 1 11 ...4 64 2304
y x x x − + − +
( )3 5 7
(1)0
...9 225 11025x x xy x + − + − +
At x = 0, y = 1 and dd
yx
= 2, hence, ( )01y = and ( )(1)
02y =
© 2014, John Bird
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Hence, the power series solution of the differential equation: 2
2
d d 1d d
y yx xyx x
+ + = is:
y = 2 4 61 1 11 ..4 64 2304
x x x − + − +
3 5 72 ...
9 225 1025x x xx + − + − +
and it may be shown that this is equivalent to:
y = 2 4 62 2 2 2 2 2
1 1 11 ...2 2 4 2 4 6
x x x − + − + × × ×
3 5 7
2 2 2 2 2 22 ...
3 3 5 3 5 7x x xx + − + − × × ×
© 2014, John Bird
1298
EXERCISE 315 Page 875
1. Produce, using Frobenius’ method, a power series solution for the differential equation:
2
2
d d2 0d d
y yx yx x
+ − =
2
2
d d2 0d d
y yx yx x
+ − = may be rewritten as: 2xy′′ + y′ – y = 0
(i) Let a trial solution be of the form y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr+…}
where a0 ≠ 0,
i.e. y = a0 xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +…
(ii) Differentiating gives:
y′ = a0cxc–1 + a1(c + 1)xc + a2(c + 2)xc+1 + … + ar(c + r)xc+r–1 + …
and y′′ = a0c(c – 1)xc–2 + a1c(c + 1)xc–1 + a2(c + 1)(c + 2)xc + …. + ar(c + r – 1)(c + r)xc+r–2 +
…
(iii) Substituting y, y′ and y′′ into each term of the given equation 2xy′′ + y′ – y = 0 gives:
2xy′′ = 2a0c(c – 1)xc–1 + 2a1c(c + 1)xc + 2a2(c + 1)(c + 2)xc+1 + …
+ 2ar(c + r –1)(c + r)xc+r–1 + … (a)
y′ = a0cxc–1 + a1(c + 1)xc + a2(c + 2)xc+1 + …. + ar(c + r)xc+r–1 + …
(b)
–y = –a0xc – a1xc+1 – a2xc+2 – a3xc+3 – … – arxc+r –… (c)
(iv) The sum of these three terms forms the left-hand side of the equation. Since the right-hand side
is zero, the coefficients of each power of x can be equated to zero.
For example, the coefficient of xc–1 is equated to zero, giving: 2a0c(c – 1) + a0c = 0
or a0c [2c – 2 + 1] = a0c(2c – 1) = 0 (1)
Equation (1) is the indicial equation, from which, c = 0 or c = 12
The coefficient of xc is equated to zero, giving: 2a1c(c + 1) + a1(c + 1) – a0 = 0
i.e. a1 (2c2 + 2c + c + 1) – a0 = a1(2c2 + 3c + 1) – a0 = 0
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or a1(2c + 1)(c + 1) – a0 = 0 (2)
Replacing r by (r + 1) will give:
in series (a), 2ar+1(c + r + 1)(c + r)xc+r
in series (b), ar+1(c + r + 1)xc+r
in series (c), –arxc+r
Equating the total coefficients of xc+r to zero gives:
2ar+1(c + r + 1)(c + r) + ar+1(c + r + 1) – ar = 0
which simplifies to: ar+1{(c + r + 1)(2c + 2r + 1)} – ar = 0 (3)
(a) When c = 0:
From equation (2), if c = 0, a1(1 × 1) – a0 = 0, i.e. a1 = 0a
From equation (3), if c = 0, ar+1(r + 1)(2r + 1) – ar = 0, i.e. ar+1 = ( 1)(2 1)
rar r+ +
r ≥ 0
Thus, when r = 1, 1 02
(2 3) (2 3)a aa = =× ×
since 1 0a a=
when r = 2, 2 03
(3 5) (2 3)(3 5)a aa = =× × ×
when r = 3, 3 0 04
(4 7) (2 3)(3 5)(4 7) (2 3 4)(3 5 7)a a aa = = =× × × × × × × ×
and so on
The trial solution is: y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}
Substituting c = 0 and the above values of a1, a2, a3, … into the trial solution gives:
y = 0 0 00 2 3 40 0 ...(2 3) (2 3)(3 5) (2 3 4)(3 5 7)
a a ax a a x x x x
+ + + + + × × × × × × ×
i.e. y = ( ) ( )( ) ( )( )
2 3 4
0 1 ...2 3 2 3 3 5 2 3 4 3 5 7x x xa x
+ + + + + × × × × × × × (4)
(b) When c = 12
:
From equation (2), if c = 12
, a1 ( ) 322
– a0 = 0, i.e. a1 = 0
3a
From equation (3), if c = 12
, ar+1 ( )1 1 1 2 12
r r + + + +
– ar = 0,
i.e. ar+1 ( )3 2 22
r r + +
– ar = ar+1(2 2r + 5r +3) – ar = 0,
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i.e. ar+1 = (2 3)( 1)
rar r+ +
r ≥ 0
Thus, when r = 1, 1 02
(2 5) (2 3 5)a aa = =× × ×
since a1 = 0
3a
when r = 2, 2 03
(3 7) (2 3 5)(3 7)a aa = =× × × ×
when r = 3, 3 04
(4 9) (2 3 4)(3 5 7 9)a aa = =× × × × × ×
and so on
The trial solution is: y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}
Substituting c = 12
and the above values of a1, a2, a3, … into the trial solution gives:
y = 1
0 0 0 02 3 42 0 ...3 2 3 5 (2 3 5)(3 7) (2 3 4)(3 5 7 9)a a a ax a x x x x
+ + + + + × × × × × × × × × ×
i.e. y = 1 2 3 420 1 ...
(1 3) (1 2)(3 5) (1 2 3)(3 5 7) (1 2 3 4)(3 5 7 9)x x x xa x
+ + + + + × × × × × × × × × × × × × (5)
Let 0a = A in equation (4), and 0a = B in equation (5)
Hence, y = ( ) ( )( ) ( )( )
2 3 41 ...
2 3 2 3 3 5 2 3 4 3 5 7x x xA x
+ + + + + × × × × × × ×
+ 1 2 3 42 1 ...
(1 3) (1 2)(3 5) (1 2 3)(3 5 7) (1 2 3 4)(3 5 7 9)x x x xB x
+ + + + + × × × × × × × × × × × × ×
2. Use the Frobenius method to determine the general power series solution of the differential
equation: 2
2
d 0d
y yx
+ =
The differential equation may be rewritten as: y′′ + y = 0
(i) Let a trial solution be of the form y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…} (1)
where a0 ≠ 0,
i.e. y = a0xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +… (2)
(ii) Differentiating equation (2) gives:
y′ = a0cxc–1 + a1(c + 1)xc + a2(c + 2)xc+1 + …. + ar(c + r)xc+r–1 + …
© 2014, John Bird
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and y′′ = a0c(c – 1)xc–2 + a1c(c + 1)xc–1 + a2(c + 1)(c + 2)xc + … + ar(c + r – 1)(c + r)xc+r–2 +
…
(iii) Replacing r by (r + 2) in ar(c + r – 1)(c + r)xc+r–2 gives: ar+2(c + r + 1)(c + r + 2)xc+r
Substituting y and y′′ into each term of the given equation y′′ + y = 0 gives:
y′′ + y = a0c(c – 1)xc–2 + a1c(c + 1)xc–1 + [a2(c + 1)(c + 2) + a0]xc + …
+ [ar+2(c + r + 1)(c + r + 2) + ar] xc+r + … = 0 (3)
(iv) The indicial equation is obtained by equating the coefficient of the lowest power of x to zero.
Hence, a0c(c – 1) = 0 from which, c = 0 or c = 1 since a0 ≠ 0
For the term in xc–1, i.e. a1c(c + 1) = 0
With c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1 combined
with the zero value of c would make the product zero
For the term in xc, a2(c + 1)(c + 2) + a0 = 0 from which, 02
( 1)( 2)aa
c c−
=+ +
(4)
For the term in xc+r, ar+2(c + r + 1)(c + r + 2) + ar = 0
from which, 2( 1)( 2)
rr
aac r c r
+−
=+ + + +
(5)
(a) When c = 0: a1 is indeterminate, and from equation (4)
0 02
(1 2) 2!a aa − −
= =×
In general, 2( 1)( 2)
rr
aar r
+−
=+ +
and when r = 1, 1 1 13
(2 3) (1 2 3) 3!a a aa − − −
= = =× × ×
when r = 2, 2 04
3 4 4!a aa −
= =×
when r = 3,
1
3 15
3!4 5 4 5 5!
aa aa
−−−
= = =× ×
Hence, y = 0 1 0 10 2 3 4 50 1 ...2! 3! 4! 5!a a a ax a a x x x x x + − − + +
from equation (1)
= 2 4 3 5
0 11 ... ...2! 4! 3! 5!x x x xa a x − + − + − + −
Since 0a and 1a are arbitrary constants depending on boundary conditions, let 0a = A and
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1a = B, then: y = 2 4 3 5
1 ... ...2! 4! 3! 5!x x x xA B x − + − + − + −
(6)
(b) When c = 1: a1 = 0, and from equation (4), 0 02
(2 3) 3!a aa − −
= =×
Since c = 1, 2( 1)( 2) ( 2)( 3)
r rr
a aac r c r r r
+− −
= =+ + + + + +
from equation (5)
and when r = 1, 13
(3 4)aa −
=×
= 0 since a1 = 0
when r = 2,
0
2 04
3!(4 5) 4 5 5!
aa aa
− − − = = =× ×
when r = 3, 35 0
(5 6)aa −
= =×
Hence, when c = 1, y = 0 01 2 40 ...3! 5!a ax a x x − + +
from equation (1)
i.e. y = 3 5
0 ...3! 5!x xa x − + +
Again, 0a is an arbitrary constant; let 0a = K,
then y = 3 5
...3! 5!x xK x − + −
However, this latter solution is not a separate solution, for it is the same form as the second series in
equation (6) above. Hence, equation (6) with its two arbitrary constants A and B gives the general
solution.
Hence the general power series solution of the differential equation: 2
2
d 0d
y yx
+ = is given by:
y = 2 4 3 5
1 ... ...2! 4! 3! 5!x x x xA B x − + − + − + −
or y = P cos x + Q sin x from the series expansions of cos x and sin x
3. Determine the power series solution of the differential equation: 2
2
d d3 4 0d d
y yx yx x
+ − =
using the Frobenius method.
2
2
d d3 4 0d d
y yx yx x
+ − = may be rewritten as: 3xy′′ + 4y′ – y = 0
© 2014, John Bird
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(i) Let a trial solution be of the form y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}
i.e. y = a0xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +…
(ii) Differentiating gives:
y′ = a0cxc–1 + a1(c + 1)xc + a2(c + 2)xc+1 + … + ar(c + r)xc+r–1 + …
and y′′ = a0c(c – 1)xc–2 + a1c(c + 1)xc–1 + a2(c + 1)(c + 2)xc + … + ar(c + r – 1)(c + r)xc+r–2 +
…
(iii) Substituting y, y′ and y′′ into each term of the given equation 3xy′′ + 4y′ – y = 0 gives:
3xy′′ = 3a0c(c – 1)xc–1 + 3a1c(c + 1)xc + 3a2(c + 1)(c + 2)xc+1 + …
+ 3ar(c + r –1)(c + r)xc+r–1 + … (a)
4y′ = 4a0cxc–1 + 4a1(c + 1)xc + 4a2(c + 2)xc+1 + … + 4ar(c + r)xc+r–1 + … (b)
–y = –a0xc – a1xc+1 – a2xc+2 – a3xc+3 – … – arxc+r – …
(c)
(iv) The coefficient of xc–1 is equated to zero giving: 3a0c(c – 1) + 4a0c = 0
or a0c [3c – 3 + 4] = a0c(3c + 1) = 0
This is the indicial equation, from which, c = 0 or c = 13
−
The coefficient of xc is equated to zero giving: 3a1c(c + 1) + 4a1(c + 1) – a0 = 0
i.e. a1 (3c(c + 1) +4(c+1)) – a0 = a1(c + 1)(3c + 4) – a0 = 0
or a1(c + 1)(3c + 4) – a0 = 0 (1)
Equating the total coefficients of xc+r to zero gives:
3ar+1(c + r)(c + r + 1) + 4ar+1(c + r + 1) – ar = 0
i.e. ar+1(c + r + 1)(3c + 3r + 4) – ar = 0
which simplifies to: 1( 1)(3 3 4)
rr
aac r c r
+ =+ + + +
(2)
(a) When c = 0:
From equation (1), if c = 0, a1(4) – a0 = 0, i.e. a1 = 0
4a
© 2014, John Bird
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From equation (2), if c = 0, 1( 1)(3 4)
rr
aar r
+ =+ +
r ≥ 0
Thus, when r = 1, 1 02
(2 7) (2 4 7)a aa = =× × ×
since 01
4aa =
when r = 2, 2 0 03
(3 10) (3 10)(2 4 7) (1 2 3)(4 7 10)a a aa = = =× × × × × × × ×
when r = 3, 3 0 04
(4 13) (4 13)(3 10)(2 4 7) (2 3 4)(4 7 10 13)a a aa = = =× × × × × × × × × ×
and so on
The trial solution is: y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}
Substituting c = 0 and the above values of a1, a2, a3, … into the trial solution gives:
y = 0 0 0 00 2 3 40 ...4 (1 2)(4 7) (1 2 3)(4 7 10) (2 3 4)(4 7 10 13)a a a ax a x x x x
+ + + + + × × × × × × × × × × ×
i.e. y = ( ) ( )( )
2 3 4
0 1 ...(1 4) 1 2)(4 7 1 2 3 4 7 10 (2 3 4)(4 7 10 13)
x x x xa + + + + + × × × × × × × × × × × ×
(3)
(b) When c = 13
− :
From equation (1), if c = 13
− , a1 ( )2 33
– a0 = 0, i.e. a1 = 0
2a
From equation (2), if c = 13
− , ( )
1 12 (3 2)( 1)(3 2)3( 1)3 333
r r rr
a a aar rr rr r
+ = = =+ + + ++ +
r ≥ 0
Thus, when r = 1, 1 0 02
2(5 2) (2 5) (1 2)(2 5)a a aa = = =× × × ×
since a1 = 0
2a
when r = 2, 2 03
(8 3) (1 2 3)(2 5 8)a aa = =× × × × ×
when r = 3, 3 04
(11 4) (1 2 3 4)(2 5 8 11)a aa = =× × × × × × ×
and so on
The trial solution is: y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…}
Substituting c = 13
− and the above values of a1, a2, a3, … into the trial solution gives:
y = 1
0 0 0 02 33 0 ...2 (1 2)(2 5) (1 2 3)(2 5 8) (1 2 3 4)(2 5 8 11)a a a ax a x x x−
+ + + + + × × × × × × × × × × × ×
i.e. y = 1 2 3 430 1 ...
(1 2) (1 2)(2 5) (1 2 3)(2 5 8) (1 2 3 4)(2 5 8 11)x x x xa x−
+ + + + + × × × × × × × × × × × × × (4)
Let 0a = A in equation (3), and 0a = B in equation (4)
© 2014, John Bird
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Hence, y = ( ) ( )( ) ( )( )
2 31 ...
1 4 1 2 4 7 1 2 3 4 7 10x x xA
+ + + + × × × × × × ×
+ 1 2 33 1 ...
(1 2) (1 2)(2 5) (1 2 3)(2 5 8)x x xB x−
+ + + + × × × × × × ×
4. Show, using the Frobenius method, that the power series solution of the differential equation:
2
2
d 0d
y yx
− = may be expressed as y = P cosh x + Q sinh x, where P and Q are constants. [Hint:
check the series expansions for cosh x and sinh x on page 221.]
The differential equation may be rewritten as: y′′ – y = 0
(i) Let a trial solution be of the form y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr +…} (1)
where a0 ≠ 0,
i.e. y = a0xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +… (2)
(ii) Differentiating equation (2) gives:
y′ = a0cxc–1 + a1(c + 1)xc + a2(c + 2)xc+1 + … + ar(c + r)xc+r–1 + …
and y′′ = a0c(c – 1)xc–2 + a1c(c + 1)xc–1 + a2(c + 1)(c + 2)xc + … + ar(c + r – 1)(c + r)xc+r–2 +
…
(iii) Replacing r by (r + 2) in ar(c + r – 1)(c + r)xc+r–2 gives: ar+2(c + r + 1)(c + r + 2)xc+r
Substituting y and y′′ into each term of the given equation y′′ – y = 0 gives:
y′′ – y = a0c(c – 1)xc–2 + a1c(c + 1)xc–1 + [a2(c + 1)(c + 2) – a0]xc + …
+ [ar+2(c + r + 1)(c + r + 2) – ar] xc+r + … = 0 (3)
(iv) The indicial equation is obtained by equating the coefficient of the lowest power of x to zero
Hence, a0c(c – 1) = 0 from which, c = 0 or c = 1 since a0 ≠ 0
For the term in xc–1, i.e. a1c(c + 1) = 0
With c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1 combined
with the zero value of c would make the product zero
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For the term in xc, a2(c + 1)(c + 2) – a0 = 0 from which, 02
( 1)( 2)aa
c c=
+ + (4)
For the term in xc+r, ar+2(c + r + 1)(c + r + 2) – ar = 0
from which, 2( 1)( 2)
rr
aac r c r
+ =+ + + +
(5)
(a) When c = 0: a1 is indeterminate, and from equation (4)
0 02
(1 2) 2!a aa = =×
In general, 2( 1)( 2)
rr
aar r
+ =+ +
and when r = 1, 1 1 13
(2 3) (1 2 3) 3!a a aa = = =× × ×
when r = 2, 2 04
3 4 4!a aa = =×
when r = 3,
1
3 15
3!4 5 4 5 5!
aa aa = = =× ×
Hence, y = 0 1 0 10 2 3 4 50 1 ...2! 3! 4! 5!a a a ax a a x x x x x + + + + +
from equation (1)
= 2 4 3 5
0 11 ... ...2! 4! 3! 5!x x x xa a x + + + + + + +
Since 0a and 1a are arbitrary constants depending on boundary conditions, let 0a = A and
1a = B, then: y = 2 4 3 5
1 ... ...2! 4! 3! 5!x x x xA B x + + + + + + +
(6)
(b) When c = 1: a1 = 0, and from equation (4), 0 02
(2 3) 3!a aa −
= =×
Since c = 1, 2( 1)( 2) ( 2)( 3)
r rr
a aac r c r r r
+ = =+ + + + + +
from equation (5)
and when r = 1, 13
(3 4)aa =×
= 0 since a1 = 0
when r = 2,
0
2 04
3!(4 5) 4 5 5!
aa aa
= = =
× ×
when r = 3, 35 0
(5 6)aa = =×
Hence, when c = 1, y = 0 01 2 40 ...3! 5!a ax a x x + + +
from equation (1)
i.e. y = 3 5
0 ...3! 5!x xa x + + +
© 2014, John Bird
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Again, 0a is an arbitrary constant; let 0a = K,
then y = 3 5
...3! 5!x xK x + + −
However, this latter solution is not a separate solution, for it is the same form as the second series in
equation (6) above. Hence, equation (6) with its two arbitrary constants A and B gives the general
solution
Hence the general power series solution of the differential equation: 2
2
d 0d
y yx
+ = is given by:
y =2 4 3 5
1 ... ...2! 4! 3! 5!x x x xA B x + + + + + + +
or y = P cosh x + Q sinh x from the series expansions of cosh x and sinh x
© 2014, John Bird
1308
EXERCISE 316 Page 879
1. Determine the power series solution of Bessel’s equation: ( )2
2 2 22
d d 0d d
y yx x x v yx x
+ + − =
when v = 2, up to and including the term in 4x
The complete solution of Bessel’s equation: ( )2
2 2 22
d d 0d d
y yx x x v yx x
+ + − = is:
y = 2 4 6
2 4 61 ...
2 ( 1) 2 2!( 1)( 2) 2 3!( 1)( 2)( 3)v x x xA x
v v v v v v − + − + + × + + × + + +
+ 2 4 6
2 4 61 ...
2 ( 1) 2 2!( 1)( 2) 2 3!( 1)( 2)( 3)v x x xB x
v v v v v v− + + + + − × − − × − − −
and y = 2 4 6
2 4 61 ...
2 ( 1) 2 2!( 1)( 2) 2 3!( 1)( 2)( 3)v x x xA x
v v v v v v − + − + + × + + × + + +
when v is a
positive integer
Hence, when v = 2, y = 2 4
22 4
1 ...2 (2 1) 2 2!(2 1)(2 2)
x xA x − + + + × + +
i.e. y = 2 4 4 6
2 21 ... or ...12 384 12 384x x x xA x A x − + − − + −
2. Find the power series solution of the Bessel function: ( )2 2 2'' ' 0x y xy x v y+ + − = in terms of the
Bessel function 3( )J x when v = 3. Give the answer up to and including the term in 4x
( )vJ x = 2 4
2 4
1 ...2 ( 1) 2 (1!) ( 2) 2 (2!) ( 3)
vx x xv v v
− + − Γ + Γ + Γ + provided v is not a negative integer
Hence, when v = 3, 3 ( )J x = 3 2 4
2 4
1 ...2 (3 1) 2 (1!) (3 2) 2 (2!) (3 3)x x x − + − Γ + Γ + Γ +
i.e. 3 ( )J x = 3 2 4 3 5 7
2 5 5 8
1 ... or ...2 4 2 5 2 6 8 4 2 5 2 6x x x x x x − + − − + − Γ Γ Γ Γ Γ Γ
3. Evaluate the Bessel functions 0 ( )J x and 1( )J x when x = 1, correct to 3 decimal places.
© 2014, John Bird
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0 ( )J x = ( )
2 4 6
22 2 6 241 ...
2 (1!) 2 (3!)2 2!x x x
− + − +
and when x = 1, 0 ( )J x = ( )
2 4 6
22 2 6 24
1 1 11 ...2 (1!) 2 (3!)2 2!
− + − +
= 1 – 0.25 + 0.015625 – 0.000434 + …
= 0.765 correct to 3 decimal places
1( )J x = 3 5 7
3 5 7...
2 2 (1!)(2!) 2 (2!)(3!) 2 (3!)(4!)x x x x− + − +
and when x = 1, 1( )J x = 3 5 7
3 5 7
1 1 1 1 ...2 2 (1!)(2!) 2 (2!)(3!) 2 (3!)(4!)− + − +
= 0.5 – 0.0625 + 0.002604 – 0.000054
= 0.440 correct to 3 decimal places
© 2014, John Bird
1310
EXERCISE 317 Page 883
1. Determine the power series solution of the Legendre equation: ( )21 '' 2 ' ( 1) 0x y xy k k y− − + + =
when (a) k = 0 (b) k = 2, up to and including the term in 5x The power series solution of the Legendre equation is:
y = 2 40( 1) ( 1)( 2)( 3)1 ...
2! 4!k k k k k ka x x+ + − + − + −
+ 3 51( 1)( 2) ( 1)( 3)( 2)( 4) ...
3! 5!k k k k k ka x x x− + − − + + − + −
(a) When k = 0, y = { }0 1 0 0 ...a − + − + 3 51( 1)( 2) ( 1)( 3)( 2)( 4) ...
3! 5!a x x x− + − − + + − + −
i.e. y = 0a + 3 5
1 ...3 5x xa x + + +
(b) When k = 2, y = 2 402(3) 2(3)(0)(5)1 ...2! 4!
a x x − + −
+ 3 51(1)(4) (1)( 1)(4)(6) ...
3! 5!a x x x− − + −
i.e. y = ( )20 1 3a x− + 3 512 1 ...3 5
a x x x − − −
2. Find the following Legendre polynomials: (a) 1( )P x (b) 4 ( )P x (c) 5 ( )P x
(a) Since in 1( )P x , n = k = 1, then from the second part of equation (47), page 881 of textbook, i.e.
the odd powers of x:
y = { }1 0a x − = 1a x
1a is chosen to make y = 1 when x = 1
i.e. 1 = 1a
Hence, 1( )P x x=
(b) Since in 4 ( )P x , n = k = 4, then from the first part of equation (47), page 881 of textbook, i.e. the
even powers of x:
y = 2 404(5) 4(5)(2)(7)1 02! 4!
a x x − + +
= 2 40351 103
a x x − +
0a is chosen to make y = 1 when x = 1
i.e. 1 = 0 0 035 2 81 10 1 10 113 3 3
a a a − + = − + =
, from which, 0a = 38
© 2014, John Bird
1311
Hence, 4 ( )P x = 2 43 351 108 3
x x − +
or 4 ( )P x = ( )4 21 35 30 38
x x− +
(c) Since in 5 ( )P x , n = k = 5, then from the second part of equation (47), i.e. the odd powers of x:
y = 3 51( 1)( 2) ( 1)( 3)( 2)( 4) ...
3! 5!k k k k k ka x x x− + − − + + − + −
i.e. y = 3 5 3 51 1(4)(7) (4)(2)(7)(9) 14 210
3! 5! 3 5a x x x a x x x − + − = − +
1a is chosen to make y = 1 when x = 1
i.e. 1 = 1 1 114 21 15 70 63 813 5 15 15
a a a− + − + = =
from which, 1158
a =
Hence, 5 ( )P x = 3 515 14 218 3 5
x x x − +
or 5 ( )P x = ( )5 31 63 70 158
x x x− +