214 Mr. Babbage on the analogy between the 
this equation can only be possible when/r .fax = 1 or when 
foe — 
and supposing foe .fax = 1, we have (pa? = consequently 
the equation becomes 
, / r , 4 -* 1 fax * 
and since the left side of this equation is symmetrical, the 
right side must also be so, consequently 
I* 
- =jf (a?, ux ) or the given equation is impossible, unless 
VJ X ft 
fxz^Vjx .f(x, ax) 
X ft 
and the solution of the equation will be found by taking the 
value of a vanishing fraction to be 
v ' fax .f(x, ax) . <px — <px + fx . (pax 
: - !_ 
T fx . <pax +fax . <px 
This method is applicable to all equations of the first degree, 
but I shall now point out a principle which extends to all 
equations of the first order, and which may probably be 
applied with some modification to those of higher orders. 
Let us consider the equation 
F jar, ax, px, pax | sap, where a* x — x 
multiply this by <p ax, px, pax^ = o 
it becomes 
F | x, ax, \J/.r, p aX J x (p | X,aX, px, \| tax | = o 
in order that this equation may be symmetrical, we must have 
F | X, aX, pax | X <p | X, ax, pX, paX j 
= F | ax, X, pax, px ] x <p | aX, X, pax, px j 
