Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

236

We will search for the second particular solution, according to Theorem 15, p.233, in the form 6 y 2 ( x )= ∞ ∑ k = 0 b k x k + J 0 log x , where b 0 = 0 . (5.56) As in the case of the first particular solution, let us first find

2 y ′′

2 y

2 + xy ′ 2 + x

L ( y 2 )= x

(5.57)

2 =

∞ ∑ k = 3

− 2

2 b

2 +

k 2 b

x k + 2 xJ ′

= b 1 x + 2

2 x

k + b k

0 + L ( J 0 ) log x .

As y 2 and J 0 are solutions of the initial differential equation ( L ( y 2 )= 0 , L ( J 0 )= 0 ) , we obtain

( s ! ) 2

x 2

∞ ∑ s = 1

∞ ∑ k = 3

( − 1 ) s 2 s

− 2

2 s

k 2 b

x k + 2 2 b

2 + b

2

k + b k

2 x

1 x = 0 .

(5.58)

+

From here coefficients b i can be determined. For odd coefficients we obtain

( 2 s + 1 ) 2 b

2 s + 1 = − b 2 s

s = 1 , 2 ,...,

− 1 ,

and as b 1 = 0, the odd coefficients are

b 2 s + 1 = 0 , s = 0 , 1 , 2 ,...

(5.59)

For even coefficients we obtain

( − 1 ) s + 1 s 2 2 s − 2 ( s ! ) 2

( 2 s ) 2 b

2 s + b 2 s

s = 2 , 3 ,...

(5.60)

− 2 =

,

It can be shown that the last relation yields b 2 s = 1 1 + 1 2 + 1 3

1 s

( − 1 ) s − 1 2 2 s ( s ! ) 2

s = 1 , 2 ,...

(5.61)

+ ··· +

,

The second particular solution of the equation (5.45) is thus also determined as

∞ ∑ s = 1

1 s

( − 1 ) s ( s ! ) 2

x 2

1 2

2 s

y 2 = J 0 log x −

1 +

(5.62)

+ ··· +

.

6 Note that for the first part of the solution we had

∞ ∑ k = 0

∞ ∑ k = 0

k

k + 1

xr 2 = x

c k x

c k x

, k + 1 = s ⇒

=

∞ ∑ s = 1

x s

xr 2 =

c s

and by substituting c s

b s we obtain

,

− 1 =

− 1

∞ ∑ s = 1

∞ ∑ s = 0

s

s , for s = 0 , b 0 = 0 .

xr 2 =

b s x

b s x

=