5. Professor Cayley states correctly that a particular 
solution of (14) is given by 
y = v 2 x~ l + 2vx* = v~ 2 x * + 2y~b? -i (16) 
where, v is a root of the quartic 
v 4 + 2xh 3 - 2xh - x — 0 (17) 
And, to obtain a general solution, he transforms (14) into a 
differential equation of the second order, from which a 
general solution is derived in the usual way. Such trans- 
formation is obviated by the procedure of Art. 2. Of course 
the elimination of v between (16) and (17) will give (13) 
as follows : 
y 2 = (v 2 x~% + 2vx i )(v~ 2 x i + 2v~ 1 x~ i ) 
= 2v~ 1 x% + 2vx~% + 5 
.*. y 2 -Q = 2v~ 4 x$ + 2vx~% - 1 
and, y 4 - 6y 2 = 4v~ 2 x\ + 8v~ 4 x% + iv 2 x~^ + 8vx~$ + 3 
- 4 (x + y - - 4x~ 2 xi - 8v~ 2 x% - 4v 2 x~§ - 8vx~$ 
Then, y 4 - 8y 2 - 4^x + ^y -3 = 0 
which agrees with (13) as it should do. 
6. A root of (13) may be found by changing the inde- 
pendent variable x to another independent variable 0 con- 
nected with the former by a given relation. This has been 
done by Professor Cayley in a particular case (see Messenger 
of Mathematics, page 111.) Put y = <t>(z) 
- 1 ^)W-6}- 3 
.x + 
If 0(,) 2 
m 
2 z 2 + 5z + 2 
1 z 4 + 2z 2 + 2z + 1 
m % fj z tj 2z 2 + 5 ^ + 2 
From which, x 2 = : 
3 (2 + z) 
1+22 
which agrees with Professor Cayley’s results. 
The advantage of this transformation is questionable, as 
particular values of x only can be obtained by it, except by 
the solution of the quartic 
z 4 + 2z 3 - 2 x 2 z - # 2 = 0. 
