c1 = (x − t) (t + x + y + z)3, For convenience of notation, let t + x + y + z = w3; so that by (117), the general integral is u = {(x−t) w, (y-t)w, (z-t) w} = 0. Ex. 2. Determine the form of a function of n variables which will satisfy the differential equation and if x1, x2, ... X-1 are independent variables, and x, is depen dent, it may be expressed Xn-1 Xn = Cn-1' = cni X1 Xn−1, = 0; m x1" the right-hand member of which is an homogeneous function of (n−1) variables and of in dimensions: the above is manifestly a proof of the converse of Euler's theorem. SECTION 6.-On integrating factors of differential equations of the first order and degree. 389.] We return to total differential equations, with the object of investigating the conditions, subject to which differen tial expressions of two or more variables, which are not exact differentials, and do not in themselves satisfy the criteria of integrability, may yet be made to do so by means of an integrating factor; and first we shall consider an expression of two variables of the form Pdx+Qdy = 0; (121) where P and Q are functions of x and y; and we shall shew that there is always a factor μ, which is generally a function of a and y, which will render (121) an exact differential of u, so that μ (Pdx+Qdy) = Du = 0. Suppose the general integral of (121) to be f(x, y) = c; (122) (123) where c is an arbitrary constant: then we have, by differentia respectively the x- and y- partial derived-functions of the same function f(x, y) = 0; and μ is a factor which renders (121) an exact differential. 390.] Not only is there an integrating factor of a given differential expression of the form (121), but the number of such factors is infinite. For supposing u = 0 to be such that then multiplying both members by, say, (u), an arbitrary function of u, we have μ (u) (p dx+Qdy) = 4(u) Du; (127) and as the right-hand member of this equation is an exact differential, the left-hand member also is; and as (u) is an arbitrary function of u, the number of such factors is infinite. From the form of the integrating factors it appears, that if two integrating factors of a differential expression can be found, the integral may be found without integration. For suppose μ to be one integrating factor; then (127) shews any other integrating factor to be of the form μ (u); consequently dividing the latter by the former, and equating the quotient to a constant c, we have and this is an integral of the given differential expression, inasmuch as expresses an arbitrary function. 391.] Let us however consider the question of integrating factors of a differential expression of the form (121) from a more general point of view. Let μ be an integrating factor, so that μ P dx+μQdy is an exact differential; and consequently μ P dx+μ Q dy = Du = 0; then by the condition of such exact differentials, whence (129) (130) and as this is a partial differential equation of the first order and degree in μ, it is to be integrated by the methods of the preceding Section. Let the general integral of this equation be and therefore, by virtue of the hypothesis made in (84), we have by the integration of which equations μ must be expressed as an arbitrary function of x and y. There is no general method of finding the integrals of the form (134); although, as we shall presently see, many cases admit of integration. Since the integrals of (134) involve an arbitrary function the number of factors which will render (121) an exact differential is infinite. This is the theorem which was proved in the preceding Article. 392.] The following are examples in which the integrating factors are determined by the preceding process. 2 ay dx+x (ay) dy = 0. Let be the integrating factor: then μ whence the most general value of μ may be determined; but as it takes a complicated form, let us suppose that the relation between c1 and c2 is such that c1 = c2 = 0: then ... u = 2 a log x + a log y―y = c. Ex. 2. y dx+(ax2y" —2x) dy = 0. and this is a particular value of the integrating factor; using which, the given differential equation becomes and as this integral is that of the first two members of the The integrating factor is cx-2; and the integral is 393.] Certain special cases in which the integrating factors can be determined require consideration. Let us suppose the integrating factor μ to be a function of one variable only; say, of a only; so that (du) = 0; then from (131), dy dy) - (da) dx; Q (136) If the right-hand member of this equation is an explicit function of a only, the equation is integrable; and we have Now this condition may be satisfied when Q contains a only, and P is of the form X1y+x, where x, and x, are functions of x only; in which case the equation is of the first linear form given in Art. 382; this will be fully discussed hereafter. It will also be satisfied when the functions of y which enter into (d) – (da) and into Q are the same, so that they may be divided out in the right-hand member of (136). |