(8) Let (sin y + y cos x) dæ + (sin x + x cos y) dy : = du; The integral is u = v sin y + y sin a + C. The conditions of integrability of a differential function of the first order between three variables such as The integral will then be found by adding together the integral of P with respect to a, the integral with respect to y of the terms in Q which do not contain a, and the integral with respect to ≈ of the terms in R which contain only. If we begin to integrate with respect to y or ≈ instead of æ, a corresponding change must be made in the process. (x2 + y2 + x2) § This satisfies the criteria of integrability, and [Pdx = (x2 + y2 + ≈3)§ + tan 20 <-1 taking the terms involving ≈ only. Hence u = (x2 + y2 + x2) + tan-1 20 + + C. (11) Let du = (y + ≈) d x + (≈ + x) dy + (x + y) dz. Then u = xy + yz + zx = C. adx-bdy Theoretically all differential equations between two variables may be rendered differential functions by being multi plied by an integrating factor, but the investigation of the proper factor is a problem of as high an order of difficulty as the solution of the equations, and no general method can be given for finding it. In some cases, however, the factor is seen without difficulty on a consideration of the form of the equation, and in these cases the method may be used with advantage. A few examples of such equations are subjoined. a dx bdy cx dx (13) Let + If this be multiplied by ay, it becomes axa-1y' dx + by-1 xa dy = cx2+” dx; both sides of which are differential functions; and the integral is (16) Let dæ + (a dx + 2by dy) (1 + x2)3. The integrating factor is (1 + a2), and the integral is (17) Let (2x − y) dy + (2a − y) dx = 0. The factor is (2a - y)-3; multiplying by it we have (2x−y)dy + (2 a−y) dx (2 a−y) (dx-dy)+2(a+x−y) dy = 0. Integrating this we find a+x-y=C(2a - y)3. (18) Let yda – xảy = d +ydy. The factor is (x2+ y2)-1, and the integral is The integrating factor is (y2 2), and the integral is 5 x5 3 x3 When the equation is between three variables and of the form Pdx+Qdy + Rdx = 0, the condition that it should be made a differential function. by means of a multiplier, is The method of integration is to assume one of the variables as constant, and then to integrate the remaining terms as an equation between two variables, adding, instead of a constant, a function of the third variable, which is determined by comparing the differential of the integral with the given equation. (20) Let 2dx (y+z)+dy(x+3y+2x)+dz(x+y)= 0. whence 2 log (x + y) + log (y + x) = $ (y). or, 2dx (y + x) + dy (x + 3y + 2 ≈) + dx (x + y) = $'(y) dy ; comparing this with the given equation we find '(y) = 0, and therefore (y) = C. Hence the integral is (x + y)2 (y + x) = C. (21) Let dx (ay−bz)+dy (cz− a x) + d z (b x − c y) = 0. (y2 + y z + x2) dx + (x2 + x z + ≈2) dy + (x2 + xy + y2) d≈ = 0. SECT. 2. = Functions of an order higher than the first. Letv du be a differential function involving x, y, and their differentials, and let x1, x2, x3, &C. Y1, Y2, Yз, &c. be put for da, d2x, d3x, &c. dy, d'y, d3y, &c. Then the conditions that v should be the differential of a function d"-lu, are dv dx dv |