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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
.
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⇒=++++++ ∑ ∑ ∑∞
=
∞
=
∞
=+ 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
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∑∑ ∞
=
+∞
=
==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 +=∴