46 
MR. HARGREAVE ON THE SOLUTION OF 
To verify this, we have 
and the equation is verified, if vanishes between the limits. Whether the limits 
can be so taken depends upon the form of 
Apply this to xTj''u-\-ii=0. 
Here <pz=z‘^ -4/2=1, 
and 
and 
Z' 
n -\- 1 
2-n+\ 
u = hfz~”z -n+\z^^dz . 
The limits 0 and — oc cause g"^^2-n+i to vanish and satisfy the equation. So that 
/^O w+1 
U — k -»+> 
is a partial solution ; which may be completed by writing for 2, a2, (iz, &c. where 
1, a, j3, &c. are the roots 
Again, if we apply the above solution to 
D“w— <rM=0, 
1 
we have <p2 = — 1, -4/2=2”, x^ — ' 
_zn+l 
n-\-\ 
and 
u 
=£/’ ^ 
£ n+l 
's^^'dz. 
zn-V\ 
Here, there do not appear to be any limits which will make £“ n+i 2®* vanish ; but if we 
take for the limits 0 and oc we have JTu—xu—k. This last equation is also solved 
^«+i 
by u—ka. J ^ dz, a being a root of ^”+^=1. 
Therefore the original equation is solved by 
u = kj ^ £“^+r(£*‘* — a£““)6/2, 
which may be completed by the use of the other roots*. 
Let us apply this method to the solution of 
and we have at once 
bn 
where 
+ . . . bnZ^ + hn-\Z^~'^ + . . 
B 
ttnZ"^ Un-xZ"'-^ an{z — a){z — 
z — a.' z — fS 
* See Moigno’s Calcul. vol. ii. p. 644. 
