20 
ON THE THEORY OF THE SINGULAR SOLUTIONS OF 
20 ON THE THEORY OF THE SINGULAR SOLUTIONS OF [636 
636] 
The theorem just referred to, that the system of algebraic curves U=0 has 
always an envelope, is an interesting theorem, which I proceed to prove. Assume 
that in general, that is, for an arbitrary value of the parameter, the equation U = 0 
represents a curve of the order m, with 8 nodes and k cusps (and therefore of the 
class n, with i inflexions and t double tangents, the numbers m, 8, k, n, t, i being 
connected by Plticker’s equations); for particular values of the parameter, the values 
of 8 and k may be increased, or the curve may break up, but this is immaterial. 
The consecutive curve U+8cd c U=0 is a curve of the same order m, with 8 nodes 
and k cusps, consecutive to the nodes and cusps of the original curve U, and the two 
curves intersect in m 2 points; but of these, there are 2 coinciding with each node, 
and 3 coinciding with each cusp of the curve U = 0, as at once appears by drawing 
a curve with a node or a cusp, and the consecutive curve with a consecutive node 
or cusp; the number of the remaining intersections is = m 2 — 28 — 3k, and the envelope 
is the locus of these m 2 — 28 — 3k points. Observe that the two curves have in common 
n 2 tangents; but of these, 2 coincide with each double tangent and 3 coincide with 
each stationary tangent of the curve U = 0, viz. the number of the remaining common 
tangents is = n 2 —2r—Si (which is =m 2 — 2S — 3/c): and that these n 2 — 2r — Si common 
tangents are indefinitely near to the m 2 — 28 — 3k common points respectively, and are 
in fact the tangents of the envelope at the m 2 — 28 — 3k points respectively. Now in 
an algebraic curve we have m+ n = m 2 — 28 — Sk, viz. the number m 2 — 28 — 3/c cannot 
be =0, and we have therefore always an envelope the locus of the system of the 
m 2 — 28 — 3k points. It might be thought that the conclusion extends to transcendental 
curves; if this were so, the result would prove too much, viz. it would follow that a 
differential equation (L, M, N\p, 1) 2 = 0 without a singular solution had no general 
integral; but it will appear by an example that the theorem as to the envelope does 
not extend to transcendental curves. 
by chan; 
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Ex. 
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Ex. 1. 
p 2 — (1 — y 2 ) = 0, that is, dy 2 — (1 — y 2 ) dx 2 = 0. 
The 
function 
Here there is no algebraical integral, but there is a quasi-algebraical integral of 
the form (P, Q, R\c, 1) 2 = 0; viz. starting with the form y — sin (x -f C) and expressing 
sin G and cos C rationally in terms of a new parameter, this is 
In fact, 
c 2 (y + cos x) — 2c sin x + (y — cos x) = 0, 
where the coefficients are one-valued functions of (x, y). The discriminant of the 
differential equation in regard to p and that of the integral equation in regard to c 
are each =y 2 — 1, and we have a true singular solution y 2 — 1 = 0. 
and thei 
Ex. 2. 
(1 — x 2 )p 2 — (1 — y 2 ) = 0, 
that is, 
(1 — x 2 ) dy 2 — (1 — y 2 ) dx 2 = 0. 
that is, l 
It i 
We have here an algebraic integral of the proper form, which is at once derived 
from the circular form 
C = cos -1 x + cos -1 y 
point of 
curve co