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. 61
of t, OZ, z2, which will be proved to be such that 01=0, 02=B when t=0, B being an
arbitrary constant.
The first stage of the proof will establish the existence of the parts zp, z-r: the
second stage will establish the existence of the parts z^1, z2^2. It may be added that,
had it been deemed desirable, a selection from t, and t2 of terms that do not involve z,
might first have been made: the forms of ti and t2 would then have been
tl =- 2pi + Zii, t2 = 2Ti7 + Z,19,
where pl-=0, -T1 = B when t = O, and pl, ri are regular functions of t and z,2: also '1 = A,
*.2 = 0 when t=0, and Ti, I2 are regular functions of t, zl, z. Further, it will be seen
from the forms of the functions that p, T, i, T3 all vanish when A =0: and that 01, 03,
pl, rT all vanish when B=O.
10. It is clear that if the equations under consideration possess integrals of the
form
tl = pzX, t2 = TZ1,
where p and T are to be regular functions of z and zl, then, taking account of the forms
of 1i and 02, the quantities p and T must satisfy the equations
ta ap = t d pt)
aT ar dr
t + el - =t (2- )T + (p 1 t)
The functions k1i '2 are regular in their arguments: both of them vanish when p =0,
=0: in each of them, every term, which is of dimensions X in p and T combined,
possesses a factor z,-1: and no term is of dimensions less than 2 in p, T, t combined.
Because p and r are to be regular functions of t and zl, they will be expressible in
the forms
p = ZEkmnzitn, T = lmnnzf, ntîl;
substituting these values and equating coefficients on the two sides of both equations, we
find
(n + mlf2) cn,, = Fmn )
{n + (m + 1) ei - e2} mln = l mn
where k'n, and l'mn are linear in the coefficients of frî and 42 respectively, and are
rational integral functions of the coefficients km',', I,' in p and T such that m'ém, n'én,
m' + n'< m + n.
From the forms of the functions ~1 and f2, we have k'oo=, 1'oo=0. Hence when
m = O0, n = 0, the first of the coefficient-equations leaves k00 undetermined: we therefore
make it an arbitrary (finite) quantity A: the second of the coefficient-equations gives
100 = 0, for 1 and:2 are unequal.