u + x + y + z = y2ƒ {x (u − y), x (y − x)}. where P, Q, R are functions of x, y, z, and if from these two systems we can find two integrals Ua, V=b, then V = f (U) is the first integral of the proposed equation; and the integral of this is the complete integral of the proposed equation. It is generally more convenient (when possible) to find another first integral, of the from V = f(U), and between these to eliminate por q so as to obtain an equation involving only one differential coefficient, and which is therefore easily integrable. Monge, Mémoires de l'Académie des Sciences, 1784, p. 118. From the second, since de pdx+qdy, we have dz = = 0, or = a. since constant, and therefore (~) the equation pcq = 0, we easily obtain ≈ = f(x + cy) = ƒ {x + y $(≈)}, which is the required integral. (2ya) (dp-dq) + 2 (p − q) dy + 2dx = 0; From and substituting this in the equation just found, it becomes This is a linear equation, and is therefore easily integrated. The result is where x + y is to be substituted for a, after integration. If we put p+q=a, this takes the form da2 + (q − p) a dx dy (1 + pa) = 0. dy is (1 + pa) = 0; dp (1+qa) - dq (1 + pa) = 0 (1), ... dy dx = 0; The second equation of (2) gives p + q = b or a = b. gives therefore x + y + (p + q) ≈ = a; x + y + (p + q) ≈ = ¢ (p + q). The first equation of (1) gives y p-q=B, we have x = a, and putting dp = } (da + dB), dq = } (da – dB), and therefore the second equation of (1) may be put under the form αβ ada whence ẞb, (2 + a2)3, В and therefore == 2+ a 2 This first integral will enable us to determine the second integral. Putting p + q = a, pq = ẞ, we have 2= ß, : = 1 (a + ẞ) d x + 1 (a− ẞ) dy = ↓ a (dx+dy) + 1⁄2 ß (dx−dy); or, putting for ẞ its value (x − y) (2 + a2)3, dz = 1⁄2 a (dx + dy) + 1 (dx − dy) f (x − y) (2 + a2)1. This is integrable if we suppose a to be constant, and gives ≈ + 4 (a) = ¦ a (x + y) + √, (x − y) (2 + a2)1; which, combined with represents the integral of the proposed equation. Poisson has shewn how to obtain a particular integral of equations of the form. where P is a function of p, q, r, s, t, homogeneous with respect to the last three quantities, and Q is a function of x, y, ≈, and the differentials of x, which does not become infinite when rt - s2 = 0. * Correspondance sur l'Ecole Polytechnique, Vol. 11. p. 410. If we assume q=f(p), we have s = rf' (p), t = sf'(p) = r {ƒ′(p)}2 ; and on substituting in it the values of q, 8, and t, the quantity r will divide out, as P is homogeneous in r, s, and t, and the equation is reduced to the form F{p, f(p)ƒ' (p)} = 0, which is an ordinary differential equation, and being integrated determines the form of ƒ (p) involving an arbitrary constant. The partial differential equation q = ƒ (p) can always be integrated, and furnishes a value of ≈ involving an arbitrary function and an arbitrary constant. This process comes to the same as finding what developable surfaces satisfy the equation (1). C being an arbitrary constant. On integrating this we find ≈ = Cx ± 4 (y ± x) as a particular integral of the given equation. (14) Lett+2ps + (p2 − a3) r = 0. ...............(1) In this case Q = 0, and on putting q = f (p) we have, after dividing by r, |