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, WHICH OCCUR IN THE LUNAR THEORY. 101
We saw earlier that the equations (2) admit of a first integral
5= C',
and that this should be derivable from the integral
F=C,
of the non-linear equations when the former are considered as derived from the latter.
The constant C' should therefore in this case be zero. It is easy to see that the constant
is zero when we substitute in b the solutions us, s. or u2, s2 or u3, s. For the solution
U4, S4, the constant takes the value C12 which is not zero. Hence though (u4, s4) belongs
to the linear equations (2) it plays no part in the non-linear equations from which these
were derived.
The solutions u,, s, and u2, s2 are those used in developing the Lunar Theory; they
contain the terms dependent on the first power of the lunar eccentricity. It is necessary
to see why the solutions u3, s3 and U4, S4 are not used in the development.
The particular solution of the original equations of which use was made was,=Uo), S= So
where no = Eiai 2i+1 = iai exp. (2i + 1) (n - n') (t - t).
If we add a small quantity Sto to to (which is an arbitrary constant of this solution)
the resulting expression will still be a solution. Expand in powers of St0 neglecting squares
and higher powers. The additions to u0, s, will be
u = Sto = -Duo. Sto, Ss = o to = - DSo. Sto.
ato ato
These values when substituted for u, s in (2) must satisfy them independently of the
value of Sto. Hence t =kDu, s= kDs is a solution obtained merely by altering the arbitrary
to and is therefore unnecessary for the development of the Lunar Theory.
The other arbitrary constant in Uq is n, and the coefficients ai are functions of n. If
we make a small addition Sn to n and proceed as before we see that
au, 0 as
u=k, s=n o
an an
is a solution of the linear equations (2). It is only necessary to identify this with U4, s4.
The forms for both are evidently the same. For we have
a — i {m a + (2i + 1) (t-to) ai exp. (2i + 1) (n - ) (t - to)
aai
= Si a exp. (2i + 1)(n ')( - t) + (t — to ) Duo.
The terms with t as factor agree (to was put zero in the expression for u4) when
the proper constant factor is introduced, and the remaining parts are of the same form.
As no linear relation can exist between the first three solutions and either of the forms