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The assumption y = E (a,x") gives

{n (n − 1) ... (n p+1) - c"}a, = 0. From this it appears that an = 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 ni, Ny... Nin, then, the corresponding values of an being indeterminate, we have

y = C,x": + Cg x": + &c. + Cpx".

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O‘Y = (a.n+0.2M2). Substituting these in the equation and equating to zero the coefficients of w*-?, we have

n (n − 1) a, = c+20+2. If n = 0, or n =1, a, and az both vanish, and so consequently do all the superior coefficients.

c* a If n= -1, 1.20-= coa, and a -1 =

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ca,

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+ a. (1+2

1.2.3 x* * 1.2.3.4.5 a, and a, being two arbitrary constants.

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This may obviously be put under the form

y = 6 (16 + Be ;),
A= (as + ), and B = (0, -..).

Euler, Ib. p. 166.

(7) Let

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Euler, Ib. p. 167.

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will be expressed in finite terms when \ =

2r 1 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

al + q*y = p (p - 1)

(1).

Then on assuming y = (an&"), and substituting in the given equation we obtain as the condition for determining the coefficients

{n (n − 1) - p (p 1)} (x + q*an-2 = 0...... (2). Now n (n − 1) – p(p − 1) = (n P) (n + p – 1)... (3);

therefore (n p) (n + p - 1) a, + 920,-2 = 0. Assume (n + p - 1) an = (n - p + 2) bns

then n-, *?..On=2) and (n - p + 2) (1 + p 3) b, +996,-2 = 0... (4). Again assume (n + p 3) b, = (n p + 4) Cn, and so on in succession. We shall thus obtain a series of equations of which the type is

(n p+u) (n + p-M – 1) In + q*ln-2 = 0......(5), le being an even number.

If p be even let p = M, then p-M - 1 = - 1.
If p be odd let p = u + 1, then p = - 1.
In both cases the equation (5) becomes

n (n 1) In + q*ln-2 = 0. This is the relation between the coefficients which we should obtain from the equation

any + q*y = 0. .................. (6). Hence E (1.,2") = C sin (qx + a) ...... that being the integral of equation (6). Now suppose (n P + M – 2) (n + p - 1 + 1) in + q* in-2 = 0,

(n - P+u) (n + pak – 1) kn + q*kn-> = 0, to be any two consecutive equations; then

(n + p - 1 + 1) in = (n - p + ) k ,.........(8), but n- P + M = n + p - M + 1 – 2 (P M) - 1;

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By the application of this formula y or E (a,xn) may be deduced by a series of regular operations from C sin (qx + a).

If p be even 2 (p - u) + 1 gives the series 1, 5, 9.........
If p be odd it gives the series 3, 7, 11.........

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со

demy

This method may be successfully applied to reduce

1 dm-ry +q" y = p (P

me dom-2" when p or p - 1 is divisible by m.

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dy 2y (13) Let

dr . ? du 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. Assume y = (ax"), then

{n(n − 1) – 2} 2, + (n − 1) qan-1 = 0,
or (n 2)(n + 1) a, + (n 1) qan-1 = 0......... (1

Let (n + 1) a, = (n − 1) bn .........................(2), then (n 2)(n 1) nb, + (n − 2)(n − 1) qbn-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

dạx dx

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