The Chemistry Maths Book, Second Edition

13.3 The Frobenius method 373

EXAMPLE 13.3Indicial equations

(i)

We haveb

0

1 = 1 − 122 ,c

0

1 = 1122 , and the indicial equation isr

2

1 − 13 r 221 + 11221 = 10 with

rootsr

1

1 = 11 andr

2

1 = 1122.

(ii)x

2

y′′ 1 + 1 xy′ 1 + 1 (x

2

1 − 112 4)y 1 = 10

This is the Bessel equation (see Table 13.1) forn 1 = 1122. We haveb(x) 1 = 111 = 1 b

0

and

c(x) 1 = 1 x

2

1 − 1124 so thatc

0

1 = 1 − 124. Thenr

2

1 − 11241 = 1 (r 1 − 112 2)(r 1 + 112 2) 1 = 10 and the

indicial parameters arer 1 = 1 ± 12 2.

(iii)xy′′ 1 + 12 y′ 1 + 14 xy 1 = 10

We havex

2

y′′ 1 + 12 xy′ 1 + 14 x

2

y 1 = 10. Therefore,b(x) 1 = 121 = 1 b

0

and c(x) 1 = 1 x

2

so that

c

0

1 = 10. The indicial equation isr

2

1 + 1 (b

0

1 − 1 1)r 1 + 1 c

0

1 = 1 r

2

1 + 1 r 1 = 10 , with rootsr

1

1 = 10

andr

2

1 = 1 −1.

0 Exercises 6–9

Once the values of the indicial parameter have been determined, the solution of

the differential equation can proceed as in the power-series method. The general

solution is

y(x) 1 = 1 c

1

y

1

(x) 1 + 1 c

2

y

2

(x) (13.9)

in which c

1

and c

2

are arbitrary constants, andy

1

(x)andy

2

(x)are independent

particular solutions. At least one of y

1

and y

2

has the form (13.4), and the other

depends on the solutions,r

1

andr

2

, of the indicial equation. We have the following

three cases.

1 Distinct roots not differing by an integer

Bothy

1

andy

2

have the form (13.4),

y

1

(x) 1 = 1 x

r

1

(a

0

1 + 1 a

1

x 1 + 1 a

2

x

2

1 +1-) (13.10a)

y

2

(x) 1 = 1 x

r

2

(A

0

1 + 1 A

1

x 1 + 1 A

2

x

2

1 +1-) (13.10b)

2 Double root

Ifr

1

1 = 1 r

2

1 = 1 r, one solution has the form (13.4),

y

1

(x) 1 = 1 x

r

(a

0

1 + 1 a

1

x 1 + 1 a

2

x

2

1 +1-) (13.11a)

and the second solution is

y

2

(x) 1 = 1 y

1

(x) 1 ln 1 x 1 + 1 x

r+ 1

(A

0

1 + 1 A

1

x 1 + 1 A

2

x

2

1 +1-) (13.11b)

xy xy y

2

1

2

1

2

′′− ′+= 0