530 
ON THE THEORY OP THE SINGULAR SOLUTIONS OF 
[545 
9°. Considering (x, y) as given, the equation /=0, together with the (m — 1) fold 
relation between the constants, must constitute a m-fold relation of the order n, that is, 
must give for the constants n sets of values. We may, if we please, take f to be 
linear in regard to the constants c 1} c 2 ,..., c m , and then the condition simply is, that 
the (m— l)fold relation shall be of the order n. 
I give in regard to these definitions such explanations as seem necessary. 
2°. A one-valued function of (x, y) is either a rational function, or a function 
such as sin y, &c., which for any given values whatever, real or imaginary, of the 
variables, has only one value. A function is taken to be one-valued when, either for 
any values whatever of the variables, or for any class of values (e.g. all real values), 
we select for any given values of the variables one value, and attend exclusively to 
such one value, of the function. 
Thus, TJ a rational function of {x, y), <£ = p 2 — U — 0, <£ is one-valued in regard to 
(x, y). But if, U not being the square of a rational function, we take *J(U) to be 
a one-valued function (consider it as denoting, say for all real values of x, y for which 
U is positive, the positive square-root of U), then, <j) =p + *J(U) = 0, <f> is taken to be 
one-valued in regard to (x, y). 
3°. The meaning is, that the equation <£ = 0 is not satisfied irrespectively of the 
value of p, by any relation between the variables x, y. 
4°. The meaning is that </> is not the product of two factors, each rational and 
integral in regard to p, and being or being taken to be one-valued in regard to (x, y). 
Thus, if as before U is a rational function of (x, y) but is not the square of a 
rational function, and if we do not take any one-valued function, then the equation 
(j)z=p^—U=0 is indecomposable; but if we take \/(U) as one-valued, then we have 
(f)={p —\/(U)} {p +\/(U)} = 0, and the equation breaks up into the two equations 
p—\/(TJ) — 0 and p + \/(U) = 0. I assume as an axiom, that the curves represented by 
the indecomposable differential equation are in general indecomposable; for supposing the 
differential equation satisfied in regard to a system of curves, the general curve breaking 
into two curves, each depending on the arbitrary parameter, then we have two distinct 
systems of curves; either the differential equation is satisfied in regard to each system 
separately, and in this case they are the same system twice repeated; or the differential 
equation is satisfied in regard to one system only, and in this case the other system 
is not part of the solution, and it is to be rejected. As an instance, take the equation 
cj)=px + y = 0, (xdy + ydx = 0); if we choose to integrate this in the form x 2 y 2 —c = 0, 
this equation represents the two hyperbolas xy 4- V(c) = 0, xy — \/(c) = 0, but considering 
each of these separately, and giving to the constant theory any value whatever, we 
have simply the system of hyperbolas xy — c — 0 twice repeated. But if by any (faulty) 
process of integration the solution had been obtained in a form such as (c+x)(c—xy) = 0, 
then the differential equation is not satisfied in regard to the system c + x = 0; and 
the factor c-V x is to be rejected. Observe that it is said, that the curves are in 
general indecomposable; particular curves of the system may very well be decomposable ; 
thus in the foregoing example, where the system of curves is xy —c = 0, in the 
particular case c = 0, the hyperbola breaks up into the two lines x = 0, y = 0.