1903 - 4 .] Solution of the Differential Equation of J. 
275 
[»] 
From 
we obtain 
( W l)a=-W = 0 
(o) 2 )a=-n = [2] W I^ 2 ([ w + 1]) E J[ n ]( x ) • • • • 
n- 1 
r=l 
[2 m ] ■ x [2r-n] 
[2] [i] • • [2r] • [2 - 2 n] • • • [2r - 2 k] 
(S, 
- v log p log X ( - IV y" +2ra[W+2,,] 
_ V~ 1 8 Zj ' [r]l[» + r]l(2M2KH- 
r=l 
E r-iZj ( “ } im + W [2 
P- 
r=l 
,2 £,4 
[2] + M 
pZn+Z 
[2 r] [2n + 2] 
rp%Yl-{-%T ^ ^[7l-f*2f] 
[2re + 2r] J [r]![w + r]!( 2 ) r ( 2 )»+^ 
If we give C the value - so that E = 75V pb,, , i T \ 
[»] ( 2 )n-iI J) (L ?i + 'll 
or what is equivalent 
E= 
[2]“" 1 IJ([k+1]) 
we obtain an integral from ( 3 ) and ( 4 ) which may be termed 
Wi + W 2 . If p = 1 , this integral reduces to that given on p. 102 , 
vol. iii., Forsyth’s Theory of Differential Equations. 
In the case when n = 0 , the integral is 
cJ C 0] {p,x)\o%x-e ^ j r|n + rrn + - • 
;=1 
M 
• • + [2s] I [ 2 ] 2 [ 4 ] 2 • • • [2s] 2 
2^2s 
X™ 
• • ( 5 ) 
These functions satisfy the same recurrence equations as the 
