and from (4) it appears that y 1+ 9X This appears to be the solution of the equation, but it does not satisfy it unless C, = 0, when it becomes 2 y = C2 (1+ e-9, 90 which is only a particular integral, and therefore incomplete. This arises from our implying in the use of equation (4) that nb, + qon-1 = 0 is generally true, whereas the equation (n − 1) (nbn + qon-1) = 0, derived from the auxiliary equation shews that b, is not necessarily connected with bu, since it may be satisfied by n = = 1. To complete the solution, we have from (2) which is always true ao bos and from (4) which is true for n = 1, we have These quantities are independent of an, Q2, &c., therefore writing C, for a, as it is an arbitrary constant, 2 y = C, (1 qxd is a particular integral of the proposed equation, and the complete solution is P 1-1 d x - 1 (14) To integrate pm day dom- 1 d 1 (a,x") = 2*(p+1)-1 (6,2"), "a, = (n - pm + 1)... (n - m + 1) br But from the given equation n (n − 1) ... (n − m + 2) {n- m (P + 1) + 1} an + k" an-- = 0, from which m (m - 1)... ( − m + 2) ... (7 – m + 1)(x + ko b, m = 0. But this is the equation which would result from substituting (6") in day + ky = 0; d x therefore E (1,2") is the solution of this last equation, and is therefore known. Calling it X, we have d X y= (@,2") = dx Let m = 2, p = 2, then the integral of d'y 4 dy + kʼy = 0 dr 1 = x^(p+1)-1 P x dx or y=C{(3 – k* m?) cos (k «c + a) + 3kx sin (k & + a)}. Ellis, Cam. Math. Jour. Vol. 11. p. 202. CHAPTER VI. PARTIAL DIFFERENTIAL EQUATIONS. Secr. 1. Linear Equations with Constant Coefficients. By the method of the separation of symbols the integration of Linear Partial Differential Equations is reduced to the same processes as those for the integration of ordinary differential equations of the same class. Hence the theory which is given in the beginning of Chap. iv. is equally applicable to the present subject, and it is unnecessary to repeat it here; I shall therefore content myself with referring to what has been previously said in the Chapter alluded to, adding that every differential equation of this class between two variables has an exact analogue among partial differential equations of the same class, and that the form of the solution of the latter is the same as that of the former. On this point one remark may be made which is of considerable importance in the interpretation of our results. As in the solution of ordinary differential equations we continually meet with expressions of the form Ce", so in partial differential equations we shall find expressions of the form d a (y), in which the arbitrary function takes the place of the arbitrary constant. Now as the preceding formula is the symbolical expression for Taylor's Theorem, we know that d a dy E (y) = P(y + aw). Hence, in the solution of partial differential equations, arbitrary functions of binomials play the same parts as arbitrary constants multiplied by exponentials do in equations between two variables. dy d Now supposing « to be the independent variable, and a constant, with respect to it, by the Theorem given in Ex. (11), Chap. xv. of the Diff. Calc. this is equivalent to a or, effecting the integration, and adding an arbitrary function of Y, instead of an arbitrary constant, or, as the form of ø is arbitrary, we may for b It is obvious that if we had taken y for our independent d variable, and considered as a constant with respect to it, we should have had + Ø (bx – ay). b d d But by Taylor's Theorem y cos ry = cos r(y – a x); therefore * = f*** Sdxem* cos r(y – a x) ; and, integrating with respect to X, ,{mcosr(y - ax) – arsinr(y – ax)} €"* ten kdy p (y); mo + ao p? (m cos ry ar sin ry) or, + P(y + ax). mo + a po? The same method is applicable to any number of independent variables. ai dy = If we expand the operating factor in ascending powers of d d we shall have d d d d d XYZ; \dx) dy dx the other terms being neglected, because when the operations |