The assumption y = (a,x") gives {n (n − 1) ... (n − r + 1) − c′′ } a„ − 0. From this it appears that a, = 0 except for those values of n which cause the other factor to vanish. These values of n are r in number; let them be n1, ng...n,, then, the corresponding values of a, being indeterminate, we have y = C1x" + C2 "9+ &c. + C, ∞".. Substituting these in the equation and equating to zero the coefficients of "-2, we have n (n - 1) a, c2an+2 = If n = 0, or n = 1 and a both vanish, and so consequently do all the superior coefficients. c2 a1 c2 a 1.2.3' + a。 (1 + + + &c.), 1.2.3 x 1.2.3.4.5 x a, and a being two arbitrary constants. Mr Leslie Ellis has given (Cambridge Mathematical Journal, Vol. 11. p. 169 and p. 193) some remarkable methods for reducing to finite functions the solutions in infinite series of certain classes of Differential Equations. Let the equation be of the form day Then on assuming y = Σ (a„x”), and substituting in the given equation we obtain as the condition for determining the coefficients = {n (n − 1) − p (p − 1)} ̧a2 + q ̊ ɑn−2 − 0................(2). Now n (n - 1) − p (p − 1) = (n − p) (n + p − 1).....(3); therefore (np) (n + p − 1) a, + q2 A2-2 = 0. Assume and (n − p + 2) (n + p − 3) b2 + q3bn-2 = 0...(4). - Again assume (n + p − 3) b„ = (n − p + 4) c2, and so on in succession. We shall thus obtain a series of equations of which the type is (n − p + μ) (n + p − μ − 1) l, + q2 ln-2 = 0................(5), μ being an even number. ...... If p be odd let pμ+ 1, then up=1. = In both cases the equation (5) becomes n (n − 1) l2 + q2ln_2 = 0. This is the relation between the coefficients which we should obtain from the equation Hence Σ (a) = C sin (qæ + a) ............... (7), that being the integral of equation (6). Now suppose (n − p + μ − 2) (n + p − μ + 1) i, + q' i„_2 = 0, (n − p + μ) (n + p − μ − 1 ) k, + q3 kn−2 = 0, to be any two consecutive equations; then but in (n + p − μ + 1) i„ = (n − p + μ) k„......................... (8), n− p + μ = n + p − μ + 1 − 2 (p − μ) −1; By the application of this formula y or (a11) may be deduced by a series of regular operations from C sin (qx+a). If be even 2 (p - μ) + 1 gives the series 1, 5, 9......... If p be odd it gives the series 3, 7, 11....................... This equation presents a peculiarity, inasmuch as if we neglect a factor, which apparently disappears, we shall have a solution which is erroneous or incomplete. then or n (n − 1) − 2} a„ + (n − 1) qan−1 = 0, (n − 2) (n + 1) a, + (n − 1) qa2-1 = 0........(1). Let (n + 1) a = (n − 1) b„ ..(2), (n − 2) (n − 1) nb, + (n − 2) (n − 1) qb,-1 = 0...(3). The factor (n-2) may be safely neglected, but (n − 1) must be retained, as it enters into the solution of the auxiliary equation d2% dz + q = 0. da and as, except when n = 1, we have nb, + qb,−1 = 0, |