6 series solutions
TRANSCRIPT
Power Series Solutions of
Linear DEs
Chapter 4
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Learning Objective
At the end of the section, you should be
able to solve DE with Power Series as
solutions.
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Power Series
A power series in is an infinite
series of the form ax
...)()()( 2
210
0
axcaxccaxc n
n
n
The above power series is centered at
x = a.
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Power Series
n
n
x )1(0
center x = -1
n
n
n x
0
12 center x = 0
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Examples
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If the radius of convergence is R > 0, then
is f
• continuous
• differentiable
• integrable
over the interval (a-R, a+R).
Remark
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Example
Given
0n
n
n xcy
Find yy and
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...32 2
321
1
1
xcxccxncyn
n
n
...620 32 xccy
2
2
)1(
n
n
nxcnny
Example
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Analytic at a Point
A function is analytic at a point a
if it can be represented by a power series
in x-a:
with a positive or infinite radius
of convergence.
n
n
n axc )(0
f
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Adding two Power Series
Example
0
1
2
n
n
n xcS
Write as a single summation.
2
2
1 )1(n
n
nxcnnS
21 SS
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2 0
12
21 )1(n n
n
n
n
n xcxcnnSS
2 problems: exponents and starting indices
Example
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2 0
12)1(n n
n
n
n
n xcxcnn
Let 2 nk 1 nk
Example
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2
2
1 )1(n
n
nxcnnS
2 nk
0
21 )1)(2(S k
k
k xckk
Example
2 kn
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0
1
2
n
n
n xcS
1 nk
1
12 k
k
k xcS
Example
1 kn
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0 1
1221 )1)(2(k k
k
k
k
k xcxckkSS
Now same exponent
Yet to solve: first term!
Example
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1
122 ])1)(2[(2k
k
kk xcckkc
1 1
122 )1)(2(2k k
k
k
k
k xcxckkc
Example
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12 2
2 3)1(2)1(n
n
n
n n
n
n
n
n xncxcnnxcnn
Combine.
Exercise
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12 2
2 3)1(2)1(n
n
n
n n
n
n
n
n xncxcnnxcnn
Let nk 2 nk nk
1 2 3
Solution
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2
)1(n
n
n xcnn
nk
2
)1(k
k
k xckk
1
Solution
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2
2)1(2n
n
nxcnn
22 knnk
0
2)1)(2(2k
k
k xckk
2
Solution
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1
3n
n
n xnc
nk
1
3k
k
k xkc
3
Solution
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1
3k
k
k xkc
0
2)1)(2(2k
k
k xckk
2
)1(k
k
k xckk
Solution
Dr Faye-Jan 2014
2
1
1 3)1(3k
k
k xkcxc
2
2
1
3
0
2 )1)(2(2)6(2)2(2k
k
k xckkxcxc
2
)1(k
k
k xckk
Solution
Dr Faye-Jan 2014
22
2
2
321
3)1)(2(2
)1(1243
k
k
k
k
k
k
k
k
k
xkcxckk
xckkxccxc
Solution
Dr Faye-Jan 2014
k
k
k
kk xkcckkckk
xccc
]3)1)(2(2)1([
)123(4
2
2
312
Solution
k
k
kk xckkckk
xccc
])1)(2(2)2([
)123(4
2
2
312
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Ordinary and Singular Point
A function is analytic at if 0x
)( 0
)( xf n
f
exists for any n
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Ordinary and Singular Point
A point is said to be an ordinary
point of the DE if both and
are analytic at A point that is not an
ordinary point is said to be a singular
point of the DE.
0x
)(xP )(xQ
.0x
0)()( yxQyxPy
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Note:
If at least one of the function and
fails to be analytic at then
is a singular point. 0x
)(xP)(xQ
0x
Ordinary and Singular Point
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1) Every finite value of is an ordinary
point of the DE
x
.0)(sin)( yxyey x
Examples
0x
.0)(ln)( yxyey x
2) is a singular point of the DE
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Existence of Power Series Solutions
If is an ordinary point of the DE,
we can always find two linearly independent
solutions in the form of a power series
centered at ( ).
0xx
0xx
0
0 )(n
n
n xxcy
Theorem
,0 Rxx
Each series solution converges at least on
some interval defined by where R is
the distance from to the closest singular point 0x
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Example
Find a power series solution centered at 0 for the following DE
.0 xyy
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.0 xyy
Ordinary points: All real numbers x.
Example
Since there are no finite singular points,
The previous Theorem guarantees two power
series solutions centered at 0, and
convergent for .x
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.0 xyy
Let the solution be n
n
n
n
n
n xcxcy
00
)0(
1
1
n
n
nnxcy
2
2
)1(
n
n
n xnncy
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0)1(
0
2 0
2
n n
n
n
n
n xcxxnnc
xyy
2 0
12 0)1(n n
n
n
n
n xcxnnc
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0)1(2 0
12
n n
n
n
n
n xcxnnc
2 nk1 nk
0)1)(2(0 1
12
k k
k
k
k
k xcxkkc
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0)1)(2(0 1
12
k k
k
k
k
k xcxkkc
0)1)(2(21 1
12
0
2
k k
k
k
k
k xcxkkcxc
0])1)(2([21
122
k
k
kk xckkcc
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0)2)(1(.2
02.1
12
2
kk cckk
c
for ,...3,2,1k
Using the Identity Property:
The (recursive) relation generate consecutive
coefficients of the solution.
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0)2)(1( 12 kk cckk
60)3)(2( ,1 0
303
cccck
120)4)(3( ,2 1
414
cccck
020
0)5)(4( ,3 2525
cccck
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0)2)(1( 12 kk cckk
504420)7)(6( ,5 14
747
ccccck
180300)6)(5( ,4 03
636
ccccck
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...5
5
4
4
3
3
2
210 xcxcxcxcxccy
...180
0126
0 60413010 x
cx
cx
cxccy
n
n
n xcy
0
..
504
1
12
1..
180
1
6
11 74
1
63
0 xxxcxxcy
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...180
1
6
11)( 63
1 xxxy
...504
1
12
1)( 74
2 xxxxy
)()()( 2110 xycxycxy
where
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Example
.0)1( 2 yyxyx
Find a power series solution centered at 0 for the following DE
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.0)1( 2 yyxyx
.0)1(
1
)1( 22
y
xy
x
xy
The standard form:
Ordinary Points: All real numbers x.
Singular point: None.
Example
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Let the solution be n
n
nxcy
0
1
1
n
n
nnxcy
2
2
)1(
n
n
n xnncy
.0)1( 2 yyxyx
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.0
)1()1(
0
1
1
2
22
n
n
n
n
n
n
n
n
n
xc
xncxxcnnx
.0)1( 2 yyxyx
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0
)1()1(
01
2
2
2
n
n
n
n
n
n
n
n
n
n
n
n
xcxnc
xcnnxcnn
nk 2 nk
nk nk
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0
)1)(2()1(
01
0
2
2
k
k
k
k
k
k
k
k
k
k
k
k
xcxkc
xckkxckk
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0][1
)1)(2(62)1(
2
1
0
0
2
1
2
2
2
3
0
2
k
k
k
k
k
k
k
k
k
k
k
k
xcxcxcxkcxc
xckkxcxcxckk
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0
)1)(2()1(
62
22
2
2
2
32110
k
k
k
k
k
k
k
k
k
k
k
k
xcxkc
xckkxckk
xccxcxcc
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0
)1)(2()1(
62
22
2
2
2
320
k
k
k
k
k
k
k
k
k
k
k
k
xcxkc
xckkxckk
xccc
combine
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0]
)1)(2()1([
62
2
2
320
k
kk
k
kk
xckc
ckkckk
xccc
0])1)(2()1)(1[(
62
2
2
320
k
k
kk xckkckk
xccc
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0])1)(2()1)(1[(
62
2
2
320
k
k
kk xckkckk
xccc
0)1)(2()1)(1(
006
2
102
2
33
0220
kk ckkckk
cc
cccc
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kk
kk
ck
kc
ckk
kkc
c
cc
)2(
)1(
)1)(2(
)1)(1(
0
2
1
2
2
3
02
,...5,4,3,2k
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kk ck
kc
ccc
)2(
)1(
0;2
1
2
302
,...5,4,3,2k
0244.2
1
4
1,2 ccck
05
2,3 35
cck
0466.4.2
3
6
3,4 ccck
07
4,5 57
cck
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...5
5
4
4
3
3
2
210 xcxcxcxcxccy
...6.4.2
3.1
4.2
1
2
1 6
0
4
0
2
010 xcxcxcxccy
)()(
][...]6.4.2
3.1
4.2
1
2
11[
2110
1
642
0
xycxycy
xcxxxcy
n
n
n xcy
0
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xxy
xxxxy
)(
...6.4.2
3.1
4.2
1
2
11)(
2
642
1
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Example
.0)1( yxy
Find a power series solution centered at 0 for the following DE
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Example
0)1( yxy
0)1()1(02
2
n
n
n
n
n
n xcxxcnn
0)1(0
1
02
2
n
n
n
n
n
n
n
n
n xcxcxcnn
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Example
0)1)(2(1
1
00
2
k
k
k
k
k
k
k
k
k xcxcxckk
0)1(0
1
02
2
n
n
n
n
n
n
n
n
n xcxcxcnn
2 nknk
1 nk
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Example
0)1)(2(1
1
00
2
k
k
k
k
k
k
k
k
k xcxcxckk
0
)1)(2(2
1
1
1
0
0
1
2
0
2
k
k
k
k
k
k
k
k
k
xc
xcxcxckkxc
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Example
0
)1)(2(2
1
1
11
202
k
k
k
k
k
k
k
k
k
xc
xcxckkcc
0])1)(2[(2 1
1
202
k
kk
k
k xccckkcc
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Example
,...3,2,1,)2)(1(
0)2)(1(
2
102
12
12
0202
kkk
ccc
ccckk
cccc
kkk
kkk
Using Identity Property:
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Example ,...3,2,1,)2)(1(
2
1
12
02
k
kk
ccc
cc
kkk
3.2,1 01
3
ccck
4.3,2 12
4
ccck
5.4,3 23
5
ccck
6.5,4 34
6
ccck
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Example 02
2
1cc
3.23.2,1 001
3
cccck
4.3.24.34.3,2 0212
4
ccccck
5.3.25.4.25.4.3.25.4,3 00023
5
cccccck
6.4.3.26.5.3.26.5.4.3.26.5,4 00034
6
cccccck
Case 1: 0,0 10 cc
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Example 0
2
102 cc
3.23.2,1 101
3
cccck
4.34.3,2 112
4
cccck
5.4.3.25.4,3 123
5
cccck
6.5.46.5.3.26.5.4.36.5,4 01134
6
cccccck
Case 2: 0,0 10 cc
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Example
...5
5
4
4
3
3
2
210 xcxcxcxcxccy
From case 1:
...5.3.24.3.23.22
504030200 x
cx
cx
cx
ccy
...]5.3.2
1
4.3.2
1
3.2
1
2
11[ 5432
0 xxxxcy
...5.3.2
1
4.3.2
1
3.2
1
2
11)( 5432
1 xxxxxy
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Example
...5
5
4
4
3
3
2
210 xcxcxcxcxccy
From case 2:
...6.5.44.33.2
5141311 x
cx
cx
cxcy
...]6.5.4
1
4.3
1
3.2
1[ 543
1 xxxxcy
...6.5.4
1
4.3
1
3.2
1)( 543
2 xxxxxy
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End
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