762 
MR. J. 0. MALET ON A CLASS OF INVARIANTS. 
If V\—y$c—y?x 2 —then we have 
H=0, G=0, 1=0 
and the complete solution is 
y = e~\ Vidx {K.x z + ~Bx 2 + Gx-\- D) 
It is obvious that when H, G, and I instead of being each equal to 0 are each equal 
to a given constant the equation can be at once solved. 
If we seek the conditions for 
y*=yM x )> ys=y±x( x ) 
where \Js(x), ^(x) are given functions of x; then changing y to yy^ if the resulting 
equation be written 
<&!+ 4 Qi + 6 Qs vl+° 
dx 2 
dx 
we have 
</> (lv) +IQ i^ w + 6Q2^+ 4Q 3 <^'= 0 
^ (iv) + iQir+ 6 Q#'+ 4Qsf = 0 
X (i,) +4QV'+6Q 2X "+4Q 3X '=0 
from which equations we find Q 1( Q 2 , Q s are terms of x. Let these values be F^ic), 
F 2 (a:), and F a (*), and the required conditions are 
H=F 2 (z) —(F^))’—F/Or) 
G= 2(F 1 (*))3-3F 1 ( a: )F 2 (*)+2F 3 (ir) - F/'(x) 
I=-6(F>))Hl2(F 1 (aO)^*)-4F 1 («)F,(*)-8CE , i (»))*-Fr(«) 
To find the solution in this case, we have 
Qi*=Sjt+ p i y* 
Hence 
y A =el (qi ~ l>l)dx 
Now if we let 
f f x' 
r r x" 
r r x"' 
= A 
we find from the previous equations 
dA 
