Memoirs presented to the Cambridge philosophical society on the occasion of the jubilee of Sir George Gabriel Stokes, bart., Hon. LL. D., Hon. SC. D., Lucasian professor.

DIFFERENTIAL EQUATIONS. 45
substitute in the equations, and determine (by comparison of the coefficients) the values
of fi,. _l, g_,..., gl. Then the condition is that the coefficient of tm in
/m-1 m-1
at + 0 ( fptP, g tP, t)
-p=l p==l g
shall be zero. This statement can easily be verified.
CASE I (C): the critical quadratic has unequal roots, both of which are positive integers.
3. Let m and n be the two unequal roots, of which m is the smaller, so that
the equations may be taken in the form
dcv
td = nv + ct + 0 (u, v, t),
tdt = ~' + fit + ~b (uyt
These equations can be transformed, as in the Case I (b), by m - 1 substitutions in turn;
and ultimately they acquire the form
t d = + at + h (u,, v' t)
t -= CUv + bt + k (W, v, t)
where K, = n-m + 1, is a positive integer greater than unity, u' and v' are regular functions
of t vanishing when t = 0, and the functions h, k have the same signification as in I (b).
If the equations possess regular solutions, the latter must be of the form
-'= 2; pltl, v'= qitl;
1=1 I=1
substituting these values and equating coefficients, we have
pl=p +a, q = cq + b,
(l - 1)pi = coefficient of tl in h (u', v', t),
(l- c) ql=............. tl in k (O', v', t).
It is clear that, if a is different from zero, the first equation cannot be satisfied; and
therefore as one condition for the possession of regular integrals, a must be zero.
Assuming this satisfied, we see that p' is left undetermined: let a value a, provisionally
arbitrary, be assigned to it.
Solving now the remaining equations for the values 1= 1, 2,..., c - 1 in successive
sets, each set being associated with one value of l, we have the values of pi,..., p~-_,
q,q..., q"_c; all these in general involve a. In order that q" may have a finite value