2 21 
the calculus of Junctions . 
and putting <p ' x for x 9 and (p*y for y , it becomes 
F {$ x,$ f'’*{x,y)}=¥ {<$' 
this is identical if we assume / so that the two conditions 
fiy a [ x> y) = y and f 2 > 1 (x 9 y) = x may be fulfilled. 
The same method is applicable to the equation 
{^•'[x,y) — x\F{x,y,i/(x,y) & c.) = f 4' 1 ’ * (x, y ) ( 
b (^x 9 y 9 if ( y ) > &C. ) 
i' 
for the two factors which multiply F and F vanish on account 
of the value of f. 
Problem XXIV. 
Given the equation 
x^ z (x 9 y) = a ^ 2 > 1 (x, y) 
this equation, by means of the substitution already so frequently 
employed, becomes 
x§ f~ z (<p*z, <py) = a<p / 2 > 1 ((px, <py) 
and putting (p'x for x t and y for y we have 
(p 1 x.(p l (x,y) == a f 2,1 (x,y) 
An artifice somewhat similar to the one already employed, 
will afford the solution of this equation : if we can find such 
a value of/(jr, y,) that /^ ( x , y) = c and also / 2 * 1 (aqy) = * 
the equation will become identical by making c = (pm Such 
a value of/ is/ (a?, y) = £ ~ for 
c “ 
f 2 > 2 (x,y) = c—j — candf 2 >* (/,/) = c i = # 
X X 
hence the general solution of the given equation is 
