280 Wilder’s Algebraic Solution. 
x nr ; 
here, writing ys EEE for y in (G), we have 
(ps tay'p* + 9ip ogy 
ce 
y'* — 4py'? +-Aqy' 2p? 
4 
T _ 540, 
x 
And since p, g and x are independent, we may make any 
three hypotheses we choose ; accordingly, comparing 
4 4 2. 4 
y!*—Apy'? -4qy!+2p2@—* - TP+T — 28, with 
a function y/* ++by’? +cy'’+d=0, y’' being the same in both 
functions, we have 4p=-6 : 
4q=c and 
ve} 2(d—2p?)x*+-(p4 +4q2p)x4 —q'=0, 
the same as obtained before. 
The fourth proposition gives the same result, perhaps more 
satisfactorily. Continuing to denote by « the function (bcd), 
we shall then have, = 
y= —(22°-+pt+q)s 
wom Saag 
y= —(a8x*+apt+q), 
ev en’ 
y'=—(a°x?+a*prtq), 
a*x? 
hos ate 3 
4 hace easnecrg) ; let us transpose the 
a°a 
second members, and it requires no great skill to see that 
the continued product of the four factors will be, 
y'*—Apy’? +4qy' + 2p? — (ps — Aay'p? +4q°p +24°y"") 
“7 
| 
+ - «4=0; but by the hypothesis, 
*+4q? 4 
yincapy hg ate et tl, conse- 
— oe A mp2 i) 25,/2 
quently, —Awp 4 _Y” =0; hence, the product of the 
four factors is y’* +by’*-+-cy’+d=0, j 
Let us now take y* +-by?+cy? +dy+-e=0, for the ara 
| equation. Heren=5, andif we make m=5, the function (B) 
will be proper to resolve this equation. 
