Ch6 Series Solutions LE

8/9/2019 Ch6 Series Solutions LE

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CHAPTER 6

(Power Series Method)

Review of Power Series

•−−−+−+−+=−∑

∞

=

2

210

0

)()()( a xca xcca xcn

n

n

is a power series in power of (x-a),

where −−−,, 10 cc are real (complex ) constant and “a” is the centre of theseries.

•−−−+++=∑

∞

=

2

210

0

xc xcc xcn

n

n

is a power series in x with centre at 0 (Maclaurin

series).

• ∑=−=

N

n

n

n N a xc xS 0

)()(

is called nth partial sum.

• If⇒∞


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NOTE: If a series coner!es for Ra x − and for Ra x =− , it ma$ coner!e at eer$ point or on

some points or ma$ not coner!e at all.

HOW TO FIND RADIUS OF CONVERENCE R *

R!tio Test:

+ind L

c

c

a xc

a xc

n

n

nn

n

n

n

n

==−− +

∞→

++

∞→

1

1

1

limlim)(

)(

i. %he radius of coner!ence is L R

1=

ii. If Ra x =− then the test is inclusie

iii. If Ra x >− so series dier!es

i. If Ra x


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• 3 function f is anal$tic at a point a if it can e represented $ a power

series in )( a x − with a positie or infinite radius of coner!ence.

• Ter" wise !dditio# +−∑

∞

=0

)(n

n

n a xa ∑∞

=

−0

)(n

n

n a xb

'∑

∞

=

−+0

))((n

n

nn a xba

• Ter" wise "$%ti&%i'!tio#

•−∑∞

=0

)(n

n

n a xa ∑∞

=

−0

)(n

n

n a xb

'

n

nnn

n

n a xbabababa )4(5 022110

0 −+−−−+++ −−∞

=∑

• Series !dditio# in order o perform addition+−∑

∞

=l n

n

n a xc )( ∑∞

=

−k m

m

m a xb )(

!ien

series should satisf$ nmk l == and,

If the series correspond to a function sa$ 00( ) ( )nn

n f x c x x

∞

== −∑

• %hen the radius of coner!ence is the distance of the closest sin!ular point

from the centre of the series.

"xample

111

1

0

2 =⇒=−−−+++=− ∑

∞

=

R x x x x n

n

POWER SERIES METHOD

(!si' "ethod to so%e *DE)

• 6et a differential e7uation e in standard form

) 0( ( ) y y P x Q x y′′ ′+ + =.

• 3 point “ x0” is called ordinar$ point if P ( x) and Q( x) are anal$tic at “ x0”,

otherwise the point “ x0” will e called sin!ular (&e!ular, Irre!ular).

"xamples

#tandard form ) 0( ( ) y y P x Q x y′′ ′+ + =

1. 88 8 s n 0i x y y ye x+ + =

2.88 8 ln 0 x y y ye x+ + =


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• THEORM 6+,

E-iste#'e o. &ower series so%$tio#s:

If 0 x x =

is an ordinar$ point of the differential e7uation 0)()(9 =+′+ y xQ y x P y ,

e can alwa$s find two linearl$ independent solutions in the form of power

series centered at x0 that is∑

∞

=

−=0

0 )(n

n

n x xc y

.

3 series solution coner!es at least on some interal R x x


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0

/

1

0,/2

1

.


6/12

⇒=++++++ ∑ ∑ ∑∞

=

∞

=

∞

=+ 022)1)(2(1.2

1 1 1

022

k k k

k

k

k

k

k

k xcckxc xk k cc

⇒=++++++ ∑∞

=

+ 0)22)1)(2((5221

202

k

k

k k k xckck k ccc

0202 022 cccc −=⇒=+ (1)

,...,2,12

2

0)22()1)(2(

2

2

=+

−=

⇒=++++

+

+

k ck

c

ck ck k

k k

k k

(2)

&elation (2) is @nown as &ecurrence +ormula.

ii. 6et thencand c 01 10 == usin! &ecurrence formula

0A

2

>

1

>

1

0<

2

2

1

2

1

02

,1

1

2

11 >21 +−+−=⇒ x x x y

=ow, letthencand c 10 10 == usin! &ecurrence formula, we !et

2211


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[ ]

( ) [ ] ⇒=++++−++++

++++++++++

0......2

...///2

1...2012>2

2

210

2

21

2

<

2

2

xc xc xcc xc xc xcc

x x x x xc xc xcc

[ ] ⇒=++++−

++++++

+++++

t coefficiencomparin xc xc xcc

xccc xccc

xc xc xcc

0...

...)22

1()2(

...2012>2

2

210

21211

<

22

)(0122

1

)(02>

)(02

21

121

012

iiicccc

iicccc

iccc

=+++

=−++=−+

6et−−−=

−==⇒== ,0,

>

1,

2

101 210 ccccand c

−−−−+−+=⇒ 21>

1

2

11 x x y

=ow let−−−

−==−=⇒== ,

2

1,

>

1,

2

110 210 ccccand c

−−−−+−+−=⇒ 2

22

1

>

1

2

1

x x x x y

?ence, 2211 yC yC y +=


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#ection >.2

#olution aout #in!ular ;oints

Bien

) 0( ( ) y y P x Q x y′′ ′+ + =

• %he point x0 is called re!ular sin!ular point if the functions p(x) ' (x ! x0) ;(

and20( ) ( )) (" x x x Q x= − are oth anal$tic at x = x0.

Ctherwise irre!ular sin!ular point

De.i#itio#

3 function, f#x$, is called !#!%/ti' at x= x0 if the %a$lor series for f#x$ aout x= x0has a positie radius of coner!ence (series exist), if x0 ' 0 we hae Maclaurin

series.

• %he point x = x0 is a re!ular sin!ular point if (x ! x0) has at most power 1 in

the denominator of P ( x) and at most power 2 in the denominator of Q ( x).

) 0( ( ) y y P x Q x y′′ ′+ + ="xample


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2

2

0

first put the"7.in stander form

0

0

0 poin

.

t

y y y

y y y

y y y

x x

x

x

x

x x x

x x

irreular

′′ ′+ + =

′′ ′+ + =

′′ ′+ + =

⇒ =

DDDDDDDDDDDDD..

.int,0

0)(

poreular x

y y x x

−=⇒=−′′+

DDDDDDDDDDDDD.

iii. irreular xreular x

y x y x y x x


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NOTE If r * and r 2 are the roots of indicial e7uation then

C!se , 2121 r r and r r −≠ is ne!atie inte!er then solution will e

+

= ∑∑

∞

=

∞

= 02

0

121

n

n

n

r

n

n

n

r xb xc xa xc y

C!se 2 2121 r r and r r −≠

is a positie inte!er then solution will e

+= = ∑∑

∞

=

∞

= 012

0

1121 ln

n

nn

r

n

nn

r xb x xCy yand xa xc y

y=c1 y

1+c

2 y

2

C!se 3 r r r == 21 then

+=

≠

=

∑

∑∞

=

∞

=

1

12

0

0

1

ln

0,

n

n

n

r

n

n

n

r

xb x x y y

a xa x y

y=c1 y

1+c

2 y

2

=C%" If 1 y

is @nown, "ethod o. red$'tio# o. order ma$ e used to !et

∫

∫

=

−

21

12 y

e

y y

pdx

"xample

0)1()1( =+′−+′′− y y x y x xSol+e

#ol 3ssume


11/12

∑∑ ∞

=

+∞

=

==00 n

r n

n

n

n

n

r xa xa x y

is a solution.n r

n

n r

n

$ a (x r)x E$ a (n r)(n r )x

+ −

+ −′ = +

′′ = + + −∑∑

*

2*

x(x )$ ( x )$ $′′ ′− + − + =* ' * 0

∑ ∑∑∑∑

⇒=++−++

−++−−+++−++

−++

0)()(

)1)(()1)((

1

1

r n

n

r n

n

r n

n

r n

n

r n

n

xa xar n xar n

xar nr n xar nr n

{ }{ } )(0)()1)((

1)()1)((1 , xar nr nr n

xar nr nr nr n

n

r n

n

−−−−−−=++−++−+++−++

∑∑ −++

{ }

{ } )(0)()1)((

1)()1)((

1 , xar nr nr n

xar nr nr n

r n

n

r n

n

−−−−−−=++−++−

+++−++

∑∑

−+

+

Indicial "7uation %a@e n ' 0 and select coefficient (except na ) of the smallest

power of x e et 0)1( =+− r r r the indicial e7uation. &oots of indiciale7uation are r ' 0 , 0 (doule root)

=ow we need &ecurrence +ormula

In (3), ta@e r ' 0 ⇒

{ }

{ } )(0)()1)((

1)()1)((

1 , xar nr nr n

xar nr nr n

r n

n

r n

n

−−−−−−=++−++−

+++−++

∑

∑−+

+

{ } { } 0)()1)((1)()1)(( 1 =+−−++− ∑∑ −nnnn xannn xannn

%a@in! n ' @ in 1st term and n -1 ' @ in 2nd term, we !et


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{ } { }@ @ @ @ (@)(@ ) (@) a x (@ )(@) (@ ) a x∑ ∑ +− + + − + + + =** ' * * * 0@ @

a a a

a a+= = =−−−−−−= ⇒

=*

*0 * 2

r n

n$ x a x x ( x x x ) if x

x∑= = + + + + − − − = <

−0 2 '

*

** *

*

=ow*2 = y

:se the reduction formula

∫ ∫ =−

2

1

12 y

e y y

pdx

x(x )$ ( x )$ $′′ ′− + − + =* ' * 0

∫ −=⇒−+=−−

=

2

)1)(ln(1

21

)1(

1

)( x x pdx x x x x

x

x P

x

xdx

x x

dx x x

x

xdx

x

e

x y

e y y

x x pdx

−

=

−

=−

−−

=

−−

=∫

=

∫

∫ ∫ ∫ −−−

1

ln1

1

1

)1(

)1(

1

1

)1(

11

12

2

2

)1)(ln(

2

1

12

2

2211 yc yc y +=∴

Ch6 Series Solutions LE

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