DETERMINE THE FORM OF THE ROOTS OF ALGEBRAIC EQUATIONS. 737 
value. In the above example we may write 
11 = 0 $?. . +C„_ 1 3C_ 1 +C„ 
y?+y +*C ; 
m+ 1 
the last term of which becomes a vanishing fraction when m= — 1. 
form is seen to be 
The true limiting 
u = Ctfi *• • +O n - l y n i l +C n (y - 1 log#. . +y~ x log#) (8) 
This is the complete integral of (I.) when m— — 1. 
4. The theory of these failing cases may be viewed also in another aspect. When 
u=C iy ?+C 2 y?..+C n y: (9) 
is an integral, but not the general integral of the differential equation (I), it must be 
the general integral of a differential equation involved in (I), but of a lower order. We 
may in fact conclude that such reduced differential equation will be deducible from the 
higher one by a process of integration. Let us apply this consideration to the foregoing 
example. 
When m— — 1, the equation (I) becomes 
D(D-l). ,(D-»+l)»+i[2=i D-i-l]”"‘(D- M +lK«=0. 
Hence operating on both members with (D — n-\- 1) _1 , we have 
D(D-l)..(D-»+2)»+i[^=iD-i-l]"'>«=Ce<»-'>». 
It must then be possible to determine C so as to cause this differential equation to be 
satisfied by (9). First let us seek to determine C so as to cause the equation to admit 
of any of the particular integrals y ?, y ?, . . y™. Substituting for u the Lagrangian expan- 
sion reduced by making m= — 1, and giving to b any of the particular values included 
in the form 1% we shall, on equating coefficients, find 
c _ -[”- 3] n ~\ 
n 
whence it appears that if n be greater than 2, C=0. Thus the reduced differential 
equation becomes 
. .... ( 10 ) 
and this, when n is greater than 2, is satisfied by 
w=C,#- 1 +C 2 #- 1 . .+C n y-\ 
which in effect contains n— 1 arbitrary constants, and so constitutes the complete inte- 
gral of the differential equation. 
If n= 2, the differential equation becomes 
Djt+i(iD-f)e“w==V, 
( 11 ) 
