MR. J. 0. MALET ON A CLASS OF INVARIANTS. 
759 
where 
— w -^=s 
dx 
r7 2 R 
2R3 + EK -^= T 
-ph -f/rr -fiH 
R=Qi+2^7’ 
K= 
3/"3 
T=G 
y//3 
we have 
2 f'* ' ^ i y'2 2/' 
£*S 
^-T=(2S-K)R.. (12) 
Substituting this value of B in S and reducing we find for the required condition 
tfS\a 
dx) 
d? S dT 
4 8 3 +T*-(~)+2S(^-^)+SK^LS^+K(^-^)-^(T-f)=0 . (13) 
/dT 
d? S\ 
dK/ 
\dx 
dx 2 ) 
dx\ 
If K vanishes or f' 2 =Af"^, A being a constant, equation (13) is derived from (6) 
by changing H and G to S and T respectively. 
If we regard f(x) as unknown, equal u say, equation (12) is the differential equation 
of the fifth order, of which the complete solution is 
u= 
A A + %2+%3 
%i+E^ + % 
where y l9 y 2 , y% are solutions of the equation 
3+3P 1 3+8P i J+r^o 
and A, B, C, D, E, E arbitrary constants. 
To solve the cubic when condition (13) holds. From equations (9) and 12 we find 
at once Q x and Q 2 in terms of x , let their values be respectively <£(cc) and \p(x) we have 
then 
<K X ) : 
dS 
dx /" 
xfj(x) = 
2S-K a/' 
dS_ T 
n dx 1 f" f" 
Now since 
2f f * 2S-K /' 3/' 
«‘=s(S+ p ») 
5 E 
MDCCCLXXXII. 
