COE5310_NO_2

MICHAEL B. BAYLON MS CE

COE5310 ASSIGNMENT #2 1

COE5310: ADVANCED MATHEMATICAL METHODS

PROF. HOMER CO

ASSIGNMENT NUMBER 2:

TOPIC: POWER SERIES SOLUTION NEAR A REGULAR SINGULAR POINT

1. By power series near a regular singular point

0334 yyyx .............................................................................................................. eq. 1.0

SOLUTION:

Assume the solution of:

0n

rn

nxcy .............................................................................................. eq. 1.1

Getting the first and second derivative of the solution to substitute in the given O.D.E.:

0

1

n

rn

n xrncy ........................................................................................................ eq. 1.2a

0

21n

rn

n xrnrncy ...................................................................................... eq. 1.2b

Substitute equations 1.2a and 1.2b to the given O.D.E.:

0331400

1

0

1

n

rn

n

n

rn

n

n

rn

n xcxrncxrnrnc ............................... eq. 1.3

Shift index, i.e. from n to n-1, hence, the second term changes:

033141

1

1

0

1

n

rn

n

n

rn

n xcxrnrnrnc ........................................ eq. 1.4a

03141

1

1

0

1

n

rn

n

n

rn

n xcxrnrnc ....................................................... eq. 1.4b

Investigating eq. 1.4b for n=0: (notice that the second term starts at n=1)

arbitrary. is ;014 :0nfor 01

0 cxrrcr

The indicial equation is: 014 rr . ........................................................................................... eq. 1.4c

The indicial roots are: 0r and 4

1r .

If n1, thus the second term of eq. 1.4b will appear, thus the recurrence relation is derived:

0340314

1

2

1

1

1

nn

rn

n

rn

n

crnrnc

xcxrnrnc

114

31

nn c

rnrnc ................................................................................................. eq. 1.5



Investigating eq. 1.5, for even and odd integers of n, since there are arbitrary numbers, respectively:

For n1, r=0 For n1,

1

23

12

01

14

31

113

31

72

31

31

31

kk ckk

c

cc

cc

cc

1

23

12

01

14

31

313

31

29

31

15

31

kk ckk

c

cc

cc

cc

For the r=0 indicial root values, multiply all the terms of the LHS and equate to that of RHS:

1210

terms terms

1321141173321

33331111

k

kk

kk cccckk

ccccc

......................... eq. 1.6

Eq. 1.6 simplies to, (thus, the cn for r=0):

0

1

14!

31c

mk

ck

m

kk

k

........................................................................................................... eq. 1.7

For the r=1/4 root values, multiply all the terms of the LHS and equate to that of RHS:

1210

terms terms

1321141395321

33331111

k

kk

kk cccckk

ccccc

......................... eq. 1.8

Eq. 1.8 simplies to, (thus, the cn of the odd integers):

0

1

14!

31c

mk

ck

m

kk

k

......................................................................................................... eq. 1.9a

Changing variables from n to k, thus eq. 1.1 becomes:

0n

rn

nxcy ...................................................................................................................... eq. 1.10

Since the given O.D.E. is linear, by superposition, the general solution is: (refer to eq. 1.10)

4/10 rr yyy

0

41

0 k

k

k

k

k

k xcxcy .................................................................................................. eq. 1.11a



Substituting equations 1.7b and 1.9b to eq. 1.11a:

0

41

0

1

0

0

1

14!

31

14!

31

k

k

k

m

kk

k

k

k

m

kk

xc

mk

xc

mk

y .................................................. eq. 1.11b

Rearranging eq. 1.11b:

0

41

1

0

0

1

0

14!

31

14!

31

k

k

k

m

kk

k

k

k

m

kk

x

mk

cx

mk

cy ................................................. eq. 1.11c




052152 yyxyx ........................................................................................................ eq. 2.0

SOLUTION:


0n

rn

nxcy .............................................................................................. eq. 2.1


0

1

n

rn

n xrncy .............................................................................................................. eq. 2.2a

0

21n

rn

n xrnrncy ............................................................................................. eq. 2.2b

Substitute equations 2.2a and 2.2b to eq. 2.0:

052151200

1

0

2

n

rn

n

n

rn

n

n

rn

n xcxrncxxrnrncx .................... eq. 2.3a

Distribute x from eq. 2.3a:

0510512000

1

n

rn

n

n

rn

n

n

rn

n xcxrncxrnrnrnc ............ eq. 2.3b

Combining similar x terms in eq. 2.3b:

01253200

12

n

rn

n

n

rn

n xrncxrnrnc ......................................... eq. 2.3c


01253221

1

1

0

1

n

rn

n

n

rn

n xrncxrnrnc ..................................... eq. 2.4a

Investigating eq. 2.4a for n=0: (notice that the second term starts at n=1)


0 cxrrcr

The indicial equation is: 032 rr . .......................................................................................... eq. 2.4b

The indicial roots are: 0r and 2

3r .

If n1, thus the second term of eq. 2.4 will appear, thus the recurrence relation is derived:

01225322

0125322

1

1

1

1

rncrnrnc

xrncxrnrnc

nn

rn

n

rn

n

1322

1225

nn c

rnrn

rnc ...................................................................................................... eq. 2.5



Investigating eq. 2.5, for even and odd integers of n, since there are arbitrary numbers, respectively:

For n1, r=0 For n1,

1

23

12

01

32

125

93

55

72

35

51

15

kk ckk

kc

cc

cc

cc

033

110

021

010

11

110

23

12

01

cc

cc

cc

For the r=0 indicial root values, multiply all the terms of the LHS and equate to that of RHS:

1210

terms

1321321212753321

1253155553

k

k

kk cccckkkk

kccccc

eq. 2.6

Eq. 2.6 simplies to, (thus, the cn for r=0):

03212!53

ckkk

ck

k

.................................................................................................. eq. 2.7

For the r=-3/2 root values, cn simplifies to:

0100 10; cccc ................................................................................................................................ eq. 2.8

Changing variables from n to k, thus eq. 2.1 becomes:

0n

rn

nxcy ........................................................................................................................ eq. 2.9

Since the given O.D.E. is linear, by superposition, the general solution is: (refer to eq. 2.9)

2/30 rr yyy

0

23

0 k

k

k

k

k

k xcxcy .................................................................................................. eq. 2.10a

Substituting equations 2.7b and 2.8b to eq. 2.10a:

21

02

3

0

0

0 103212!

53xcxcxc

kkky

k

kk

.................................................. eq. 2.10b

Rearranging eq. 1.11b:

xxcx

kkkcy

k

kk

1013212!

532

3

0

0

0

......................................................... eq. 2.10c




0212 2 xyyxyx ......................................................................................................... eq. 3.0

SOLUTION:


0n

rn

nxcy .............................................................................................. eq. 3.1


0

1

n

rn

n xrncy ........................................................................................................ eq. 3.2a

0

21n

rn

n xrnrncy ...................................................................................... eq. 3.2b

Substitute equations 3.2a and 3.2b to the given O.D.E.:

0211200

12

0

2

n

rn

n

n

rn

n

n

rn

n xcxxrncxxrnrncx ................ eq. 3.3


012320

1

0

12

n

rn

n

n

rn

n xrncxrnrnc ....................................... eq. 3.4a

05223222

1

2

0

1

n

rn

n

n

rn

n xrncxrnrnc .................................... eq. 3.4b

0322 1 rnn xrnrnc Investigating eq. 3.4b for n=0 and n=1: (notice that the second term starts at n=1)


0 cxrrcr

arbitrary. is ;0121 :1nfor 11 cxrrcr

The indicial equations are:

014 rr . ....................................................................................................................... eq. 3.5

0121 rr .................................................................................................................. eq. 3.6 The indicial roots are:

For c0: 0r and 4

1r .

For c1: 1r and 2

1r .

The recurrence relation is (for n2):

2322

522

nn c

rnrn

rnc ............................................................................................... eq. 3.7



For even coefficients, n2:

For r=0 For r=1/4

222

46

24

02

342

54

96

7

54

3

12

1

kk ckk

kc

cc

cc

cc

222

46

24

02

5818

984

1925

154

1117

74

39

14

kk ckk

kc

cc

cc

cc

For r=0:

22420222642 349512642

54731

kkk cccc

kk

kccccc

02

2

2!34!2

54

ckk

m

ck

k

mk

01

234

54

!2

1c

m

m

kc

k

mkk

For r=1/4:

22420

terms

22264258191131825179

9815714444

k

k

kk cccckk

kccccc

01

25818

984 c

mm

mc

k

m

k

k

For odd coefficients, n2:

For r=-1 For r=1/2

1212

57

35

13

342

54

96

7

54

3

12

1

kk ckk

kc

cc

cc

cc

1212

57

35

13

234

122

615

52

411

32

27

12

kk ckk

kc

cc

cc

cc



For r=-1:

1275311212753 349512642

54731

kkk ccccc

kk

kccccc

12

2

12!34!2

54

ckk

m

ck

k

mk

1

1

1234

54

!2

1c

m

m

kc

k

mkk

For r=1/2:

12531

terms

121275326423415117

125312222

k

k

kk cccckk

kccccc

k

m

k cm

m

kc

1

11234

12

!

1

The solution is:

0n

rn

nxcy

0

21

0

41

0

1

0 k

k

k

k

k

k

k

k

k

k

k

k xcxcxcxcy

0

21

1

1

0

41

1

0

0

1

1

1

0 1

0

34

12

!

1

5818

984

34

54

!2

1

34

54

!2

1

k

kk

mk

kk

m

k

k

kk

mk

k

kk

mk

xm

m

kcx

mm

mc

xm

m

kcx

m

m

kcy

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COE5310_NO_2