series ordinary point

7/28/2019 Series Ordinary Point

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Series Solutions Near an Ordinary Point

We are considering methods of solving second order linear equations when thecoefficients are functions of the independent variable.

We consider the second order linear homogeneous equation

P(x)d2y

dx2+Q(x)

dy

dx+ R(x)y = 0 (1)

since the procedure for the non-homogeneous equation is similar.Many problems in mathematical physics lead to equations of this form having polynomial

coefficients; examples include the Bessel equationx

2y’’ + xy’ + (x

2– a

2) y = 0

where a is a constant, and the Legendre equation(1 – x

2) y’’ – 2xy’ + c(c + 1) y = 0

where c is a constant.

For this reason, as well as to simplify the algebraic computations, we primarily consider the case in which P, Q, and R are polynomials. However, we will see that the method can

be applied when P, Q, and R are analytic functions.For the time being, we assume P, Q, and R are polynomials and they do not have

common factors. Suppose that we wish to solve equation (1) in a neighborhood of a pointa. The solution of equation (1) in an interval containing point a is closely related with the

behavior of P(x) in the interval.

Definition: Given the equation

P(x)d2y

dx2+Q(x)

dy

dx+ R(x)y = 0

the equation

!!!!!!!!!!!!!!!!!!!!!!!!!!!! d2ydx

2+Q(x)

P(x)

dy

dx+R(x)

P(x)y = 0!or! d2y

dx2+ p(x)

dy

dx+ q(x)y = 0

where!p(x) = Q(x)

P(x)!and!q(x) = R(x)

P(x)

is called the equivalent normalized form of equation(1).

Definition: The point a is called an ordinary point of equation (1) if both of the

functions p(x) and q(x) in the equivalent normalized form, are analytic functions at the point a.

If either or both of these functions are not analytic at a, then the point a is a singularpoint of equation (1).

Examples:

1) Given y’’ + xy’ + (x2

+ 2) = 0since p(x) = x and q(x) = x2 + 2 are polynomials, then they are analytic functions

everywhere, then every real number is an ordinary point.



2) Given (x2!1)y''+ xy'+

1

xy = 0 (x2 – 1)

the equivalent normalized equation is

y ''+x

x2!1

y '+1

x x2!1( )

y = 0

where!p(x) =x

x2!1

!and!q(x) =1

x x2!1( )

then the points 1, -1, and 0 are singular points of the equation, any other real number is anordinary point of the equation.

Theorem: If a is an ordinary point of the differential equation (1),

then p(x) = Q(x)/P(x) and q(x) = R(x)/P(x) are analytic at a, and there are two nontriviallinearly independent power series solutions of equation (1) of the form

cn(x ! a)n,!

n=0

"

# dn(x ! a)n !

n=0

"

#

Furthermore, the radius of convergence for each of the series solutions is at least as largeas the minimum of the radii of convergence of the series for p(x) and q(x).

Example:

1) The point a = 0 is an ordinary point of the equationy’’ + (x

3+ 1) y’ – xy = 0

then, the differential equation has a solution in the form cnxn

n=0

!

" whose radius of

convergence is ρ = ∞.

2) The differential equation

(x2

- 4)y’’ + 3xy’ + y = 0has two singular points 2 and – 2.

Then the differential equation has a solution in the form cnxn

n=0

!

" whose radius of

convergence is at least ρ = 2.

The Method of Solution

We assume that a is an ordinary point of the equation (1), so it has a series solution near a

of the form cn (x ! a)n

n=0

"

# that converges for |x –a| < ρ .

We want to determine the coefficients c0, c1, c2, … in the series solution.Let’s differentiate the series termwise twice.



y(x) = cn (x ! a)n= c0 + c1(x ! a) + c2 (x ! a)

2+ ....

n=0

"

#

y'(x) = cnn(x ! a)n!1

= c1 + 2c2 (x ! a) + 3c3(x ! a)2+ ....

n=1

"

#

y''(x) = cnn(n !1)(x ! a)n!2

= 2c2 + 6c3(x ! a) +12c4 (x ! a)2+ ....

n=2

"

#

Substituting in the differential equation (1), we getK 0 + K 1(x - a) + K 2(x – a)

2+ ….. + K n(x –a)

n+ …. = 0 (4)

where K n is function of certain coefficients ci.

In order that series (4) be valid for all x in some interval |x – a| < ρ, we must have

K 0 = K 1 = …. = K n = …. = 0

This leads to a set of conditions that the coefficients cn must satisfy.

Example:

1) Solve (x2

– 4)y’’ + 3xy’ + y = 0 in powers of x.The only singular points of the equation are 2 and -2, so we want to find the seriessolution of the equation near zero. The series will have radius of convergence at least 2.

So y(x) = cnxn,!!y'(x) = ncnx

n!1

n=1

"

# ,!y''(x) = n(n !1)cnxn!2

n=2

"

#n=0

"

#

Substituting the series in the equation

x2

n(n !1)cnxn!2

! 4

n=2

"

# n(n !1)cnxn!2

+ 3x ncnxn!1

n=1

"

# + cnxn=

n=0

"

#n=2

"

#

n(n !1)cnxn!

n=2

"

# 4n(n !1)cnxn!2

+ 3ncnxn

n=1

"

# + cnxn=

n=0

"

#n=2

"

#

n(n !1)cnxn!

n=2

"

# 4 n + 2( )(n +1)cn+2xn+ 3ncnx

n

n=1

"

# + cnxn=

n=0

"

#n=0

"

# 0

we shift the index of summation in the second series by 2, replacing n with n + 2 and

using the initial value n = 0.Since we want to express everything in only one summation sign, we have to start the

summation at n = 2 in every series,

n(n !1)cnxn!

n=2

"

# 4 n + 2( )(n +1)cn+2xn+ 3ncnx

n

n=1

"

# + cnxn=

n=0

"

#n=0

"

#

n(n !1)cnxn !n=2

"

# 4(2)(1)c2 ! 4(3)(2)c3x ! 4(n + 2)(n +1)cnxn + 3c1x +

n=2

"

# 3ncnxn

n=2

"

# +

c0 + c1x + cnxn=!

n=2

"

# 0



(c0 ! 8c2 )+ (4c1 ! 24c 3)x ! !4(n + 2)(n +1)cn+2 + (n(n !1)+ 3n +1)cn[ ]xn=

n=2

"

#

(c0 ! 8c2 )+ (4c1 ! 24c 3)x ! (n +1)2cn ! 4(n +1)(n + 2)cn+2$

%&'x

n=

n=2

"

# !0

Nowc0 - 8c2 = 0, then c2 = c0/8

4c1 – 24c3 = 0, then c3 = 4c1/24 = c1/6

(n +1)2cn ! 4(n +1)(n + 2)cn+2 = 0,!then!cn+2 =

cn

4(n + 2),!n " 2

This last condition is called a recurrence formula, we can express each cn+2 in terms of a previous coefficient cn.

n = 2,!c4 = 3c2

4 ! 4=

3c0

42

!2 ! 4=

3c0

128

n = 3,!c5

=

4c3

4!5=

2 ! 4c1

42!3

!5=

c1

30

n = 4,!c6 = 5c4

4 !6=

3 !5c0

43

!2 ! 4 !6=

5c0

1024

n = 5,!c7 = 6c5

4 ! 7=

2 ! 4 !6c1

43

! 3 !5 ! 7=

c1

140

Notice that each even coefficient is expressed in terms of c0 and each odd coefficient isexpressed in terms of c1.

Then, the general solution is:

y(x) = c0

1+1

8x2

+3

128x4

+5

1024x6

+ ...!

" #

$

% & + c

1x +

1

6x3

+1

30x5

+1

140x7

+ ...!

" #

$

% &

2) Solve y’’ + xy’ + (x2

+ 2)y = 0 in powers of x.

There are no singular points of the equation, so we want to find the series solution of theequation near zero. The series converges for all x.


n!1

n=1

"

# ,!y''(x) = n(n !1)cnxn!2

n=2

"

#n=0

"

#


n(n !1)cnxn!2

+ x

n=2

"

# ncnxn!1

+ x2

cnxn

n=0

"

# + 2 cnxn=

n=0

"

#n=1

"

#

n(n !1)cnxn!2

+

n=2

"

# ncnxn+ cnx

n+2

n=0

"

# + 2c nxn=

n=0

"

#n=1

"

#

n + 2( )(n +1)cn+2xn+

n=0

"

# ncnxn+ cn!2x

n

n=2

"

# + 2c nxn=

n=0

"

#n=1

"

# 0



We shift the index of summation in the first series by 2, replacing n with n + 2 and usingthe initial value n = 0. We shift the index of summation in the third series by -2, replacing

n by n – 2 and using the initial value n = 2Since we want to express everything in only one summation sign, we have to start the


n + 2( )(n +1)cn+2xn+

n=0

!

" ncnxn+ cn#2x

n

n=2

!

" + 2cnxn=

n=0

!

"n=1

!

"

2c2 + 6c3 + n + 2( )(n +1)cn+2xn+!

n=2

!

" c1x + ncnxn+ 3c1x +

n=2

!

" cn#2xn

n=2

!

" +

2c0 + 2c1x + cnxn=!

n=2

!

" 0

(2c0 + 2c2 )+ (3c1 + 6c3)x ! (n + 2)(n +1)cn+2 + (n + 2)cn + cn!1[ ]xn=

n=2

"

# !0

Now2c0 + 2c2 = 0, then c2 = -c0

3c1 + 6 c3 = 0, then c3 = -1/2 c1

(n + 2)(n +1)cn+2 + (n + 2)cn + cn!2 = 0,!then!cn+2 = !(n + 2)cn + cn!2

n +1( )(n + 2),!n " 2

This last condition is called a recurrence formula, we can express each cn+2 in terms of a previous coefficient cn-2 and cn.

n = 2,!c4 = !

4c2 + c0

12= !

!4c0 + c0

12=3c0

12=1

4c0

n = 3,!c5 = !

5c3 + c1

20

= !

5 !1

2( )c1 + c1

20

=3c1

40

n = 4,!c6 = !

6c4 + c2

5 "6= !

6 1

4( )c0 ! c0

30= !

c0

60

n = 5,c7 = !

7c5 + c3

6 " 7= !

7 3

40( )c1 !1

2c1

42= !

c1

1680



y(x) = c0 1! x2

+

1

4 x4

!1

60 x6

+ ...

" # $

% & ' + c1 x !

1

2 x3

+

3

4 x5

!1

1680 x7

+ ...

" # $

% & '

3) Solve y’’ - xy’ - x2y = 0 in powers of x.There are no singular points of the equation, so we want to find the series solution of the

equation near zero. The series converges for all x.


n!1

n=1

"

# ,!y''(x) = n(n !1)cnxn!2

n=2

"

#n=0

"

#




n(n !1)cnxn!2

! x

n=2

"

# ncnxn!1

! x2

cnxn

n=0

"

#n=1

"

# =

n(n !1)cnxn!2

!n=2

"

# ncnxn! cnx

n+2

n=0

"

#n=1

"

# =

n + 2( )(n +1)cn+2xn!

n=0

"

# ncnxn! cn!2x

n

n=2

"

# =!n=1

"

# 0

We shift the index of summation in the first series by 2, replacing n with n + 2 and using

the initial value n = 0. We shift the index of summation in the third series by -2, replacingn by n – 2 and using the initial value n = 2

Since we want to express everything in only one summation, we have to start thesummation at n = 2 in every series,

n + 2( )(n +1)cn+2xn!

n=0

"

# ncnxn! cn!2x

n

n=2

"

# =

n=1

"

#

2c2 + 6c3x + n + 2( )(n +1)cn+2xn!!

n=2

"

# c1x ! ncnxn+

n=2

"

# cn!2xn

n=2

"

# =

2c2 + (6c3 ! c1)x + (n + 2)(n +1)cn+2 ! ncn ! cn!2[ ]n=2

"

# xn= 0

Now

2c2 = 0, then c2 = 06c3 – c1 = 0, then c3 = -1/6 c1

cn+2=

ncn + cn!2

(n + 2)(n +1),!n"

2

This last condition is called a recurrence formula, we can express each cn+2 in terms of a

previous coefficient cn-2 and cn.

n = 2,!c4 = 2c2 + c0

12=c0

12

n = 3,!c5 = 3c3 + c1

20=

1

2c1 + c1

20=3c1

40

n = 4,!c6 = 4c4 + c2

30=

4 1

12( )c030

=c0

90

n = 5,!c7 = 5c5 + c3

42=

340 c1 + 16 c1

42=13c1

1080





y(x) = c0 1+1

12x4

+1

90x6

+ ...! " #

$ % & + c1 x +

1

6x3

+3

40x5

+13

1080x7

+ ....! " #

$ % &

4) Solve (x2

+ 1)y’’ - y’ + y = 0 in powers of x.

There are no singular points of the equation, so we want to find the series solution of the

equation near zero. The series converges for all x.


n!1

n=1

"

# ,!y''(x) = n(n !1)cnxn!2

n=2

"

#n=0

"

#


x2

n(n !1)cnxn!2

+

n=2

"

# n(n !1)cnxn!2

! ncnxn!1

n=1

"

# + cnxn=

n=0

"

#n=2

"

#

n(n !1)cnxn+

n=2

"

# n(n !1)cnxn!2

! ncnxn!1

n=1

"

# + cnxn=

n=0

"

#n=2

"

#

n(n !1)cnxn +

n=2

"

# n + 2( )(n +1)cn+2xn ! n +1( )cnxn

n=0

"

# + cnxn =

n=0

"

#n=0

"

# 0

we shift the index of summation in the second series by 2, replacing n with n + 2 and

using the initial value n = 0, and we shift the index of summation in the third series by 1,replacing n with n + 1 and using the initial value n = 0.

Since we want to express everything in only one summation sign, we have to start thesummation at n = 2 in every series,

n(n !1)cnxn+

n=2

"

# n + 2( )(n +1)cn+2xn! n +1( )cn+1x

n

n=0

"

# + cnxn=

n=0

"

#n=0

"

#

2c2+ (3)(2)c

3

x + (n + 2)(n +1)cn+

2

xn+ n(n !1)c

n

xn

n=2

"

#! c

1

! 2c2

x +

n=2

"

#n +1( )c

n+

1

xn

n=2

"

#+

c0 + c1x + cnxn=!

n=2

"

# 0

(2c2 ! c1 + c0 ) + (6c3 ! 2c2 + c1)x + (n + 2)(n +1)cn+2 + n(n !1)cn ! (n +1)cn+1 + cn[ ]xn=

n=2

"

#

(2c2 ! c1 + c0 ) + (6c3 ! 2c2 + c1)x + (n + 2)(n +1)cn+2 + n2 ! n +1( )cn ! (n +1)cn+1

$%

&'x

n=

n=2

"

# !0 Now

2c2 –c1 = c0 = 0, then c2 = (c1 - c0)/26c3 -2c2 + c1= 0, then c3 = (2c2 - c1)/6 = -c0/6

!cn+2 =

(n +1)cn+1 ! n2! n +1( )cn

(n + 2)(n +1),!n " 2



This last condition is called a recurrence formula, we can express each cn+2 in terms of a previous coefficient cn+1 and cn.

n = 2,!c4 = 3c3 ! 3c2

4 " 3=c3 ! c2

4=

! 1

6c0 ! 1

2c1 ! c0( )

128=2c0 ! 3c1

24

n = 3,!c5 = 4c4 ! 7c3

5 " 4=

4

2c0 ! 3c1

24

# $ %

& ' ( + 7

c0

6

5 " 4=3c0 ! c1

40

n = 4,!c6 == 36c1 !17c0720


y(x) = c0 + c1x +c1 ! c0

2x2 !

c0

6x3

+2c0 ! 3c1

24x4

+3c0 ! c1

40x5

+36c1 !17c0

720+ ....=

c0 1!1

2

x2 !1

6

x3+

1

12

x4+

3

40

x5+ ...

" # $

% & ' + c1 x +

1

2

x2 !1

8

x4 !1

40

x5+ ...

" # $

% & '

5) Solve the non-homogeneous equation

y’’ – 2x2y’ + 4xy = x2 + 2x + 2in powers of x.



n!1

n=1

"

# ,!y''(x) = n(n !1)cnxn!2

n=2

"

#n=0

"

#


n(n !1)cnxn!2

! 2x2

n=2

"

# ncnxn!1

+ 4x cnxn! x

2! 2x ! 2 = 0

n=0

"

#n=1

"

#

n(n !1)cnxn!2

!n=2

"

# 2nc nxn+1

+ 4cnxn+1

! x2! 2x ! 2 = 0

n=0

"

#n=1

"

#

n + 2( )(n +1)cn+2xn!

n=0

"

# 2(n !1)cn!1xn+ 4c n!1x

n! x

2! 2x ! 2 =

n=1

"

#n=2

"

# 0

We shift the index of summation in the first series by 2, replacing n with n + 2 and usingthe initial value n = 0. We shift the index of summation in the second series by -1,

replacing n by n – 1 and using the initial value n = 2. We shift the index of summation inthe third series by -1, replacing n by n – 1 and using the initial value n = 1.

Since we want to express everything in only one summation sign, we have to start thesummation at n = 3 in every series (notice the term x

2by itself),



n + 2( )(n +1)cn+2xn!

n=0

"

# 2(n !1)cn!1xn+ 4cn!1x

n! x

2! 2x ! 2 =

n=1

"

#n=2

"

#

2c2 + 6c3x + n + 2( )(n +1)cn+2xn!

n=2

"

# 2(n !1)cn!1xn

n=2

"

# + 4c0x + 4cn!1xn! x

2! 2x ! 2 =

n=2

"

#

2c2 ! 2( )+ 6c3 + 4c0 ! 2( )x + !1+12c4 + 2c1( )x2+ (n + 2)(n +1)cn+2 ! 2(n !1)cn!1 + 4cn!1[ ]xn

n=3

"

#

Now2c2 – 2 = 0, then c2 = 1

6c3 + 4 c0 – 2 = 0, then c3 = 1/3 – 2/3 c0 -1 + 12 c4 + 2c1 = 0, then c4 = 1/12 – 1/6 c1

(n + 2)(n + 1) cn+2 – 2(n -3) cn-1 = 0, n ≥ 3

cn+2 =2(n ! 3)

(n +1)(n + 2)cn!1,!n " 3

n = 3, c5 = 0

n = 4, c6 = 2/30 c4 = 1/15(1/3 – 2/3c0) = 1/45 – 2/45c0 n = 5, c7 = 4/42 c4 = 1/126 – 1/63 c1

y(x) = c0 + c1x + x2+ (1

3!

23c0 )x

3+

112!

16c1( )x4

+145!

245c0( )x6

+ . 1126

!

163( )x7

.....=

c0 1! 23x3!

245x6+ .....( )+ c1 x ! 1

6x4!

1126

x7.....( )+ (x

2+

13x3+

112x4+

145x6+

1126

x7......)

6) Solve the non-homogeneous equation

y’’ – xy’ + 2y = e-x

in powers of x.



n!1

n=1

"

# ,!y''(x) = n(n !1)cnxn!2

n=2

"

#n=0

"

# ,!e!x = (!1)n xn

n!n=0

"

#


n(n !1)cnxn!2

! x

n=2

"

# ncnxn!1

+ 2 cnxn! (!1)

n xn

n!=

n=0

"

#n=0

"

#n=1

"

#

n(n !1)cnxn!2

!n=2

"

# ncnxn+ 2c nx

n!

n=0

"

#n=1

"

# (!1)n x

n

n!=

n=0

"

#

n + 2( )(n +1)cn+2xn!

n=0

"

# ncnxn+ 2c nx

n! (!1)

n xn

n!=

n=0

"

#n=0

"

#n=1

"

# !0

We shift the index of summation in the first series by 2, replacing n with n + 2 and using

the initial value n = 0. Since we want to express everything in only one summation sign,we have to start the summation at n = 1 in every series,



n + 2( )(n +1)cn+2xn !

n=0

"

# ncnxn+ 2cnx

n ! (!1)nxn

n!=

n=0

"

#n=0

"

#n=1

"

# !2c2 + 2c0 !1( )+ n + 2( ) n +1( )cn+2 ! (n ! 2)cn !

(!1)n

n!

$

%&

'

()x

n= 0#

Now,2c2 + 2c0 – 1 = 0, then c2 = ½ - ½ c0

(n + 2)(n +1)cn+2 ! (n ! 2)cn !(!1)n

n!= 0,!n " 1

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!cn+2 =(n ! 2)cn +

(!1)n

n!

(n + 2)(n +1),!n " 1

n = 1,!c3 = !c1 !1

3 #2= !

c1

6!1

6

n = 2,!c4 =12

4 # 3=

124

n = 3,!c5 = c3 ! 16

5 # 4=

c3

20!

1

120=

1

20!c1

6!1

6

$ % &

' ( ) !

1

120= !

c1

120!

1

60

n = 4,!c6 = 2c4 +124

6 #5=c4

15+

1

720=

1

15 #24+

1

720=

3

720=

1

240

n = 5,!c7 = 3c5 ! 1120

7 #6=c5

14!

1

5040=

1

14!

c1

120!

1

60

$ % &

' ( ) !

1

5040= !

c1

1680!

7

5040= !

c1

1680!

1

720

y(x) = c0 + c1x + (1

2!

1

2c0 )x

2+ (!

c1

6!

1

6)x

3+

1

24x4+ (!

c1

120!

1

60)x

5+

1

240x6+

(!c1

1680!

1

720)x7

+ ........ =

c0 (1!1

2x2 )+ c1(x !

1

6x3

!

1

120x5

!

1

1680x7

+ .....) + (1

2x2

!

1

6x3

+1

24x4

!

1

60x5

!

1

720x7

+ ...)

Translated Series Solutions

If we look for a solution of the equation P(x)

d2y

dx2 +Q(x)

dy

dx + R(x)y = 0 of the form

y = cn (x ! a)n

n=0

"

# , we can make a change of variable x – a = t, obtaining a new

differential equation for y as a function of t, and then look for solutions of this new



equation of the form y(t) = cntn

n=0

!

" . When we have finished the calculations, we replace

t by x –a.

Example:

Solve 2y’’ + xy’ + y = 0 in powers of x - 1. Notice that x = 1 is an ordinary point of the equation. We look for a solution in the form

y = cn(x !1)n

n=0

"

# .

We can simplify the computation of the coefficients cn by translating the center of theexpansion from x = 1 to t = 0.

This is accomplished by the substitution x – 1 = t or t = x + 1, then dt/dx = 1 or dt = dx.Since y(x) = y(t + 1), using chain rule:

dy

dx=

dy

dt

dt

dx=

dy

dt!1 =

dy

dt

d2y

dx2=

d

dx

dy

dx

"

#$%

&'=

d

dx

dy

dt

"

#$%

&'=

d2y

dt2

dt

dx=

d2y

dt2

So the equation transforms: 2d2y

dt2+ (t !1)

dy

dt+ y = 0 .

We seek a solution in the form y(t) = cntn

n=0

!

" where the coefficients cn are the same that

appear in solution (2).

So y(x)=

cntn

,!!y'(x) = ncntn!1

n=1

"

# ,!y''(x) = n(n !1)cntn!2

n=2

"

#n=0

"

#


2 n(n !1)cntn!2

+

n=2

"

# t ncntn!1

! ncntn!1

n=1

"

# + cntn=

n=0

"

#n=1

"

#

2n(n !1)cntn!2

+

n=2

"

# ncntn! ncnt

n!1

n=1

"

# + cntn=

n=0

"

#n=1

"

#

2(n + 2)(n +1)cn+2tn+

n=0

"

# ncntn! (n +1)cn+1t

n

n=0

"

# + cntn=

n=0

"

#n=1

"

# 0

We shift the index of summation in the first series by 2, replacing n with n + 2 and usingthe initial value n = 0. We shift the index of summation in the third series by 1, replacing

n by n + 1 and using the initial value n = 0Since we want to express everything in only one summation sign, we have to start the




2(n + 2)(n +1)cn+2tn+

n=0

!

" ncntn# (n +1)cn+1t

n

n=0

!

" + cntn=

n=0

!

"n=1

!

"

4c2 + 2(n + 2)(n +1)cn+2tn+

n=1

!

" ncntn

n=1

!

" # c1 # (n +1)cn+1tn

n=1

!

" + c0 + cntn=

n=0

!

"

4c2 # c1 + c0( )+ ! 2(n + 2)(n +1)cn+2 + (n +1)cn # (n +1)cn+1[ ]n=1

!

" tn= 0

Now,4c2 – c1 + c0 = 0, then c2 = ¼(c1 – c0)

2(n+2)(n+1)cn+2 + (n+1)(cn –cn+1) = 0, n ≥ 1

cn+2 = !(n +1)(cn ! cn+1)

2(n + 2)(n +1)= !

cn ! cn+1

2(n + 2),!n " 1

n = 1,!c3 = !c1 ! c2

2#

3

= !

c1 !c1 ! c0

4

6

= !3c1 + c0

24

n = 2,!c4 = !c2 ! c3

2 # 4= !

c1 ! c0

4+3c1 + c0

24

8= !

9c1 ! 5c0

192

y(t) = c0 + c1t !c1 ! c0

4t2 !

3c1 + c0

24t3 !

9c1 ! 5c0

192t4 ! ........ =

c0 1+1

4t2 !

1

24t3 +

5

192t4 + ....

" # $

% & ' + c1 t !

1

4t2 !

1

8t3 !

9

192t4 ! .....

" # $

% & '

y(x) = c0 1+1

4(x !1)2 !

1

24(x !1)3 +

5

192(x !1)4 + ....

"

# $

%

& ' +

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!c1 (x !1)!1

4(x !1)2 !

1

8(x !1)3 !

9

192(x !1)4 ! .....

" # $

% & '

series ordinary point

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