LINEAR DIFFERENTIAL EQUATIONS. 
ol 
To verify this, we have 
= = suppose, 
(p(D)u=2{.,..(p{z — 2).(p{z— l).<p2.£^'^) = P^ ; 
consequently 
the last term being taken between proper limits. 
Let a, /3, &c. be the roots of (pz=0 ; then if at one limit ^ have the values l+os, 
1+/3, &c. successively, Pj,_j will vanish; and z— — (x will give a vanishing value at 
the other limit. The form of the function %a7 must be ascertained by verification. 
Let ^ <pz=z^+azP-^-\-.. = (z-a){z-(i)(z-y).... 
<p(n)....(p{z-2).<p{z-l) = {(^n-ci)....{z-l-cc)}{(7i-(i)...(z-l-l5)}.... 
_ r(z-cc) T{z-^) 
r{n—ct) r(n— /3) 
1 
r(M— «)r(ra— /3) 
z-|3- 
^ 12 £^VjV^.. — l’ 
which taken between the limits — oc and l+a is 
e(“+i)%“-0y3“-y... 
and, if this be integrated successively with respect to Vj ^ 2 ... and taken between the 
proper limits, we get 
- Ln-oc- 1] [n-(3-~ l]...M=£^“+i)-{ [a_/3] [^-7]..+ [«-i3+ 1] [a-y+ !]..£' 
+ [2] [a— /3+2] [a — y-|-2]...£^'‘'-f-...}. 
If w=a4-l, we have 
“=-2'“'"^"{l + («-/3-hl)(a-yH-l)..2"+1.2(a-/3+l)(a-^-f-2)(a-y+l) 
(a-y+2)...£^"+...}. 
To verify this, we have 
s""w=-£“"{l + (a-)3+l)(a-y-i- !)..£"+..} 
?)(D)m=-£“" {(a_/3+l)(a—y+l). 
£“%—9(D)m = — £*•!■. 
If all the roots of <pz are zero, we have for a partial solution of 
£-^M— D”m=c, 
u=ce^rdv, rdv,...rdv I 
= C£^(1+£^+(L2)V"+(1.2.3)V"+....). 
