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arbitrary constants being thus 2n + 4. But there are relations subsisting between the constants which reduce the number of independent constants to 2n. In the first place, the constant ß will only alter gi, hi, &z , &c., and it may therefore be neglected, so that the number of arbitrary constants is reduced to 2n + 3. Again, since

go? = U2 + 02 + x2 + &c., we have by squaring and adding equations (13)

1 = cos' PE(go) + sino p?(ho) + 2 sin cos 08 (gh).

In order that this equation may subsist for all values of we must have the conditions

(g) = 1, (h) = 1, S(gh) = 0.

These three conditions reduce the number of arbitrary constants to 2n.

It is to be observed that the integrals for determining t and are not independent: for if we assume a function

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By a singular solution of a differential equation, is meant a certain relation between the variables which satisfies the differential equation, but does not satisfy the general integral. Solutions of this kind have long attracted the attention of mathematicians, and the memoirs in which they are discussed are very numerous. Their existence was first pointed out by Taylor, in his Methodus Incrementorum, p. 27, and afterwards they were noticed by Clairaut, in the Mémoires de l'Académie des Sciences for 1734. But Euler, in the Mémoires de l'Académie de Berlin for 1756, was the first who considered the subject in its bearing on the general Theory of Integration ; and in his Integral Calculus, Vol. 1. Sect. 2, Chap. IV., he gave a test for discovering whether a given solution be or be not included in the general integral. Lagrange, in the Mémoires de l'Académie de Berlin, and afterwards in his Théorie des Fonctions, and his Calcul des Fonctions, discussed the theory of these solutions, and shewed the connection between them and the general integral, and their relative geometric interpretations. Other points of the theory have been elucidated by Laplace (Mémoires de l'Académie des Sciences, 1772), Legendre (Ib. 1790), and Poisson, Journal de l'Ecole Polytechnique, Cahier xili.

Having given a differential equation, to find its singular solutions if it have any. Let

U = 0 be a differential equation of the first order between x and y

I dy , cleared of radicals and fractions, then if we represent by p, the relations between w and y found by eliminating p between

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U = 0, and a = 0, are singular -solutions of U = 0, provided they satisfy that equation, and do not at the same time make * = 0. We

dy might also deduce the singular solutions from eliminating


. U = 0, and - = 0,

dpi da

. where p =

., provided that they do not at the same time

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and eliminating p between this and the preceding equation, we find

ye – 4m x = 0, as the singular solution.

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be the given equation : the singular solution is

y = 1 + x?. • Laplace, Mémoires de l'Académie, 1772.

This is the equation with respect to which Taylor first made the remark that it admitted of a solution not involved in the general integral—“singularis quædam solutio," as he terms it. See his Methodus Incrementorum, p. 27.

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y– 4x = 0); but as this does not satisfy the given equation it is not a singular solution.

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(1 + p) Eliminating p by means of this equation we find as the singular solution

21+ y} = a).

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Here = = 0 gives us p =-; and the result of the

dp elimination of p is

ax + by - x2 = 0;

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but as this does not satisfy the given equation it is no solution at all. (8) Let the equation be

(v – 9) (* - 29) + x2 = 0. The singular solution is

y – 4203 = 0. When a solution of an equation is given, and it is required to find whether it be a singular solution or a particular integral, we must deduce from it the value of p= , and

dU see whether when substituted in it make it vanish, and do not at the same time make a vanish. If this be the case it is a singular solution, otherwise it is a particular integral

(9) Are y' = 2 x + 1, and y? + 19 = 0, singular solutions or particular integrals of the equation



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Now from the first of the given solutions we find


which does not make vanish: it is therefore a particular

ake dp van integral.

From the second of the given solutions we find

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