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u = log (y – X) - - - + C.
Y - x
u = x sin y + y sin & + C. The conditions of integrability of a differential function of the first order between three variables such as
Pdx + Qdy + Rdz are the three following,
dP_dQdQ_dR dR dP
dy dx' dzdy' dxdx 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 w, and the integral with respect to z of the terms in R which contain z only. If we begin to integrate with respect to y or z instead of x, a corresponding change must be made in the process.
1 yda x dy xydz
- t ---
Theoretically all differential equations between two variables may be rendered differential functions by being multiplied 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 Capa dv (13) Let
If this be multiplied by x" y, it becomes
awa-1 y dx + by'-' x“ dy = cx"+" dx ; both sides of which are differential functions; and the integral is
Integrating this we find
a + x - y = C (2a - y)”.
tan-- = log (x+y)} + C.
(y’ – m?)3 b y 26 yos
6 5 23 3 2018 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 pdQ DR. dR dPdP dQ).. do-da) + Q laha - ) +RC -) = 0.
Idy dal 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 2dæ (y+x)+dy (x+3y+2x)+dz (v+y)=0.
- = 0;
X + y y + x whence 2 log (x + y) + log (y + x) = 0 (y).
2 (dx + dy) dy + dx Differentiating
a = $(y) dy,
x + y y +% or, 2dx (y + x) + dy (x + 3y + 2x) + dx (x + y) = $(y) dy;
Sect. 2. Functions of an order higher than the first.
Let v = d’u be a differential function involving ~, y, and their differentials, and let X;, &n, 13, &c. Yı, Ya, Yz, &c. be put for dx, d’x, d’x, &c. dy, d’y, d’y, &c. Then the conditions that v should be the differential of a function d"-'u, are
dv dv 19 do 23 dv.
- + &c. = 0 ... (2). dy dy
dya dyz The conditions that v should be the second differential of a function d” -? U, are
-&c. = 0 ...... dy