Mr. Babbage's essay towards 
410 
Problem IX. 
Required the solution of the equation 
%f/ 2 x — x ( a ) 
Subtract ^ x from both sides, then we have 
X -1/ X — — X + X == (\J/ JT •— x) 
consequently, 
A-ty X = — Ax (6) 
again multiply ( a ) by ^ x, then we have 
•4* x x 4/ x = 4/ x x a: 
From this we learn, that (6) must be integrated on the 
hypothesis of xtyx being constant, hence 
x = — x -c = — x +/ (x-fyx) 
/ being an arbitrary function 
4/ is determined from the equation 
4/ x + x — f^x-tyx) = o. 
For / we may put/— 1 /, and it becomes 
i 
f{x + -tyx)—f(x^x) =0 (c) 
1 
A friend to whose valuable remarks on this subject, I am 
much indebted, has communicated to me the following method 
of obtaining the same solution. Assume any symmetrical 
function of x and u. 
<P {*> u) = 0 
then from the nature of the equation we have the two follow- 
ing equations x = 4/ u and u = 4 / x 
consequently 
X=\l/U=\ly^X = \p 2 X, 
hence ^ x may be found from the equation 
<p j a?, 4 / ^ | = o 
This at first sight appears different from (r), but it is not so 
