=-3m3k [sin (go-y) - sin {(2-2m-g) 0-2B+y}]. In all these expressions we have rejected those terms of the third and higher orders which, according to Art. (29), would not influence the second order.* 46. We must now substitute these values in the differential equations for u and s, and then integrate, omitting the complementary term A cos(0–B); for though, by the theory of differential equations, this would form a necessary part of the solution, we have seen Art. (35), that it cannot in this shape form a part of the correct value of u or s, but will be comprised in the terms whose arguments are cơ – a and go-y. 47. Since the form of the solution is known, the actual expressions for u and s will be obtained with more facility P h2u3' * Instead of the forces which really act on the moon, we originally substituted three equivalent ones, P, T, S; these again are, by the preceding expressions, replaced by a set of others. For, we may conceive each of the terms in &c. to correspond to a force, -a component of P, T, or S; each force having the same argument as the term to which it corresponds, and therefore going through its cycle of values in the same time. Now, by Art. (29), when the coefficient of 0 in the argument is near unity, the term becomes important in the radius vector, and when near zero, in the longitude: hence, a force whose period is nearly the same as that of the moon, produces important effects in the radius vector; and a force whose period is very long will be important in its effect on the longitude. See Airy's Tracts, Planetary Theory, p. 78. by assuming them with arbitrary coefficients, the values of which are afterwards determined by substitution. We must remember, however, that the coefficients of cos (co — a) in u and of sin (g0 — y) in s must be assumed the same as in the first approximate solutions; and that these assumptions will enable us to obtain the values of c and g to the same order of approximation as that to which we are working, Art. (38). 48. Considering, firstly, the equation in u, we have The last two terms would not be retained if we wished to find the value of u only, but, in finding t afterwards, they will become of the second order. Assume ua+B cos {(2 − 2m) 0 – 2ẞ} +C cos {(2 - 2m − c) 0 − 2B+ a} +D cos(m✪ + B−5) Assume S= - §m3k sin (g0 — y) + m23k sin ((2-2m-g) 0-2B+y}. (k sin(g0 − y) (+ A sin {(2 — 2m −g) 0 −2B+y}. Then, by substitution, we get 50. We can now find the connexion between the longitude and the time to the second order, = = (2, being the sum of all the terms of the second, and 2, those of the third order in u,) 1 ha2 [1 − 2e cos (c0 − a) − 2Σ„ − 2Σ ̧ + že2 + že2 cos2 (c0—a) + 45me" cos {(2 − 2m — 2c) 0 − 2ß +2a}] ́1+ že2 + §k2 + m2 - 2e cos (ca) +ge cos2 (c0-α) +1k2 cos 2 (g0- y) — 2m2 cos {(2 – 2m) 0 — 2ß} — 15me cos {(2—2m−c)0−2B+a} +3m2e' cos (mo+B-5) · [m2 d0 = 1 − 3m2 cos { (2 — 2m) 0 — 2ß} и 3 +15me2 cos((2- 2m - 2c) 0 - 2ẞ+2a}, neglecting the other term of the third order, the coefficient - of the argument not being small. We have now to multiply these results together, and we see that the term having for argument (2—2m-2c)0−2B+2a will disappear in the product. If we trace this term, we shall de, from retaining originally terms of the fourth order, but in it arises from combining terms hu2 not exceeding the third order. If, therefore, we had rejected terms beyond the third order indiscriminately, the expression dt 1 would have contained this term, introduced by and 21 hu2 do in t it would have been raised to the second order, and therefore formed an important part of its value instead of disappearing altogether from the expression. Hence the necessity for retaining such terms of the fourth order in |