chapter 83 power series methods of solving ordinary...

© 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

1286

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


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!


= + + +

= ( ) ( )( ) ( )( ) ( )( )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!


= + + +

© 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!


= + + +

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!


= + + +

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!


= + + +

( ) 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!


= + + +

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 69

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

1293

(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 + + + − + − + −


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 + + − + +


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

1296

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 + − + − +


yx

= 2, hence, ( )01y = and ( )(1)

02y =

© 2014, John Bird

1297

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


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

© 2014, John Bird

1299

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


, 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,

© 2014, John Bird

1300

i.e. ar+1 = (2 3)( 1)

rar r+ +

r ≥ 0


(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


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

1301

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

© 2014, John Bird

1302

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

1303

(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:



…

(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

1304

From equation (2), if c = 0, 1( 1)(3 4)

rr

aar r

+ =+ +

r ≥ 0


(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


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

− :


− , a1 ( )2 33

– a0 = 0, i.e. a1 = 0

2a


− , ( )

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


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

1305

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:



…

(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

© 2014, John Bird

1306

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

1307

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


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

1309

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


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 − +


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 − + − = − +


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− +

chapter 83 power series methods of solving ordinary...

Documents