which does make = 0; and as it does not make d U dp vanish, it is a singular solution. (10) In the same way it will be seen that d Ü dy If the differential equation be of an order higher than the first, let y1 y...y, represent the successive differential coefficients of y with respect to x. U = 0 Then being the equation cleared of fractions and radicals as before, the conditions that y-X should be a singular solution of the (nm)th order are Yn-m = X, and therefore if we find a relation between x, y, and which satisfies these equations and also the given equation, it is a singular solution of the (nm)th order. (11) Let x Legendre, Mém. de l'Acad., 1790, p. 218. be the given equation. Putting write it gives y2 = x ; and, by means of this, eliminating y from the original equation we find as a singular solution of the first order. As this does not satisfy dU dy, = 0, and as there is only one factor in there is no singular solution of the final integral. d U dy2 2 = 0. y-x + 4 (1 + x2) ' from which we find the singular solution of the first order to be d U dy = xy-1=0 0 and U = 0, and as it is independent is the singular solution belonging to the (14) If the equation be of the third order Then y2+x=0 is not a singular solution, because though = 0 it also causes to vanish; and we find dy2 d U dy2 that the real value of is infinity instead of zero, as it dys would be in the case of a singular solution. Having given the general integral of a differential equation of the first order, to find the singular solutions of the equation when there are such. Let u=0 be the integral cleared of radicals. As it is supposed to be an integral of an equation of the first order, it must contain an arbitrary constant, which we shall call c. Then if the equation d U give a value of e in terms of x and y, the elimination of c = 0 will give an equation in a and y, between U = 0 and d U which is the singular solution. It is to be observed that if d U de = 0 give a constant value for e, or a value in terms of a and y, which becomes constant in consequence of the relation U= 0, the result of the elimination is not a singular solution but a particular integral. This is satisfied either by c = a or c = a ± The former gives a particular integral. The latter gives 4b2 (yax) ∞1 = 0, which is the singular solution. - (18) Let (x2 + y2 - a2) (y3 − 2cy) + (x2 − a2) c2 = 0. as the singular solution: but since this makes c = 0, it appears that it is only a particular integral found by making the arbitrary constant equal to zero. = Let U 0 be the integral of an equation of the second order, so that it contains two arbitrary constants c1, c; then, if we represent the differentiation with respect to x and y byd, and that with respect to c, and c, by d', we can obtain the dc dca singular solution by eliminating c1, C2, and between the equations U = 0, dU = 0, d'U = 0, (19) Let the given integral be so that dd'U = 0. U = { c1 x2 + C2 x + c ̧2 + c22 − y = 0. Then dU (c, x + c2) dx - dy = 0, = d'U = ( } x2 + 2c,) de1 + (x + 2c2) dc2 = 0, dd' U = (xdc, + dc) dx = 0. From the last we find dc2 = -X. de Substituting this in the preceding equation we have 4 (c1c, x) x2 = 0. Between this equation and the first two we can eliminate and we find as the singular solution of the first and C29 By a similar process to that in the last example, we find as the singular solution belonging to the first integral of the differential equation There is no singular solution belonging to the final integral, but the singular solution just found has itself a singular solution, which is x2 + 2y = 0. Singular Solutions of Partial Differential Equations. If U=0 be a partial differential equation of the first singular solution, if there be one, will be found by eliminating p and q between the three equations |