MR. BOOLE ON A GENERAL METHOD IN ANALYSIS. 
281 
and substituting, we have 
(^fi-\-aT-\-b)u-\-{2 r 7T— a— 1)(w+A)^m+(w 2 +2wA+A 2 — q 2 )g 2 u=0. 
To take away the second term, let n= — h, then 
(V 2 + aw + b) u — q 2 g 2 u = 0 . 
Let w +a l5 w-\-a 2 be the factors of w 2 -\-aw-\-b, then 
U )(7r + c 2 )^ u b- 
To integrate this equation, assume 
V (^+a 1 )(ir+a 1 — 1)£ v b> 
nor.) 
then 
m =P 2 ^^T ?; =(^ + «i-1)(^+«i-3) . . (*+a 2 +2)v, 
which is only finite when a x —a 2 is an odd integer. 
From (107.) we have 
1 , , . 
v — 2 
S *~ Z+a^*^ 0 ’ 
from the first of which equations, 
{v-\-a^}s x —q^s x =0, 
• • \X -\-}l(p[x')Z dx J Sg<~\~Cl^Sg; — dxSx-— 0 . 
(x+a l )s x +(h—q)<p(x)s x -' i - = 0, 
Or 
q p (f? 
v 1 x-\-a l i 
=c 1 ( ? -^p 
In like manner, 
and as the constants C l5 C 2 are arbitrary, 
P(g) 
x + a/ 
t x =c 2(q+ hyp^ ; 
<p(x) 
»= {c,(? - hy+c 2 {q+hy}v ^ 
M =(^+a 1 -I)(x+a 1 -3)..( i r+« 2 +2){C 1 ( ? -A)»-+C 2 ( ? +A)*}P^g-, 
^ T" Uj 
to interpret which we have only to observe that 
<7rf(x) = ^/(x) + h(p(x]f(x — 1 ) . 
2 o 2 
