Differential equations derived from u = 0, it is always possible to eliminate a function of a certain function of r and Calculus. volved in u, and to obtain a relation between x, y, z, and true for every form which may be assigned A similar reasoning would prove, that, generally, m arbitrary functions may be eliminated, with the assistance d2 s de s' ds ds' and dť dt which will be found, in general, besides the two coefficients functions The same considerations will easily show what order of partial differential equations it is necessary to use, to eliminate any given number of arbitrary functions, and how many differential equations of that order may be obtained independent of those functions. In the case of m arbitrary functions to be eliminated from a given equation between three variables, it will be easy to see that the partial differential equations of the (2 m order must be used, and that m differential equations of that order may be obtained independent of those functions. 1) The order of differentiation indicated by the preceding observations, is the highest which can be required, to perform the elimination, but it may happen that such relations should exist between the terms of the proposed equations, that the arbitrary functions might be made to disappear without having recourse to it. Let us take for first example the equation z = (x + y)" $ (x2 — yo), and let us represent the differential coefficient of (x2 y) taken with respect to the function between the parenthesis, considered as a variable, be denoted by p' (x2 - y3). We shall have for the two partial differential equations of the first order, dz dx dz = n(x + y)"1 (x2 − yo) − 2 y (x + y)" p' (xa — yo). dy Eliminating '(x2 - y) between these two equations, we find y + x = n(x + y)" (xo — yo). Substituting now for (y) its value taken in the primitive equation, we shall have for the partial differential equation of the first order, independent of the function, and expressing therefore a Differential Example 3. Let the equation be Calculus. z = 4 (y + x) + x y ¥ (y − x), Between these two equations, and the primitive equation, we cannot eliminate the four quantities p, Y, P, These three new equations contain two new indeterminate functions " and y", so that we have six equations, d3 z = p"-2-(2 y − x) y" + xy Y" We have now ten equations and only eight arbitrary functions P, P', p", p'"', Y, y', y", "", therefore the eli- The other equation, independent of the arbitrary functions, would be obtained in making use of the differential coefficients of the first order, but it is much more complicated than that just obtained, and is useless to our present purpose. We shall take for the last example an equation containing two arbitrary functions, which will disappear in making use only of the partial differential coefficients of the second order Example 4. p' and ỵ may be eliminated at the same time from these two equations, by multiplying the second by 2, dz y dz = 0. dx x dy Taking the partial differential coefficients of this equation, first with respect to x, and then with respect to y, we multiplying the last equation by, and adding it with the first, we shall eliminate Ø, and find A partial differential equation of the second order, which is satisfied by the equation z = x whatever be the forms of the functions and y. Part I. Differential (54.) We have investigated the rules to determine the values of the differential coefficients of every given Part I. Calculus. explicit or implicit function of one or more variables. To complete the subject, it remains only to show how, in some cases, the value of the differential coefficients may be determined, although the relation of the function to the variables is not known either explicitly, or by unresolved equations which connect them together. This can be done in various cases by means of some circumstances which cannot be expressed analytically, and which, without any knowledge of the nature of the function, allow us to determine the limit of the ratio between its difference and the difference of the variable. We have had already an example of this method, in the manner which we have used to find the differential coefficient of sin x. Plate L Fig. 1. Fig. 2. Let us propose, for another example, to investigate the value of the differential coefficient of the area A B M P, included between the axis of the abscissæ, two ordinates A B, MP, and a curve AM. This area is clearly a function of the abscissa O P = x, since the point A being supposed a given point, the value of A B M P will be determined for every value assigned to x. Let the ordinate be called y, and let y = f (x) be the equation of the curve. If we suppose P P = h, and if we change x into x+h, the unknown function of that is represented by A B M P will become A B M'P', and the difference of the function will be M M'P P'. Let us draw the two lines M N, M' N' parallel to O P', it is obvious, that by taking h sufficiently small, the area PM PM' may always be considered as greater than the rectangular parallelogram P M N P', and less than PP' N' M'. Therefore the ratio of the difference of the unknown function to the difference of the variable, that is, PP' MM' PP'M N PP PP' is greater than and less than PP' M'N' But PP'M N =MP=y, and PP'M'N' = M'P', which is the value of the ordinate corresponding to the abscissa x+h, and consequently equal to dy dy h2 y + h + + &c., the limit of which, with respect to decreasing values of h is y. Hence the dx dr 1.2 ratio of the difference of the function A B M P, to the difference of the variable, is included between two quantities, one of which is y, and the other has for its limit y; consequently the limit of that ratio, or the differential coefficient of the area is also equal to y. We may apply the same method to a function of two variables. Let DAB, DA C, CAB be three coordinate planes, cut by a curve surface DC BHG ME F, whose equation is z = f(x, y). If by any point M of that surface, whose coordinates M M', M'P, A P, are respectively z, y, x, two planes are drawn, F M HQ and EMG P, parallel to the coordinate planes DA B, DAC, they will form a solid DH MGA PMQ, whose volume is clearly a function of x and y. Let u be that unknown function, and let it be required to find the value dr u Let Pph and Q q = k, and by the points p and q let planes be drawn parallel to D A C and dx dy DA B, and meeting in N N'. If in the function u we change x into x+h, it becomes D Hm g! AQ m' p &c., and the partial difference of u with respect to x is M m M'm' Gg Pp of du = u + h+ dx d' u dr2 and will have for its own difference M N m n M' N'm'n', the first term of the developement of which will clearly But we may easily see that the first term of the expression of M N mn M' N'm'n' is also de u Consequently, if by the points M, N, m, n, we draw four planes parallel to the plane A B C, we shall form four rectangular parallelopipedons having the same base M' N' m' n' hk, and the first terms of the expressions of the volumes of which will all be z h k. Now, it is obvious, that the parallelopipedons are always some k of the expression d2 u dx dy greater and some less than the solid N N m n M' N' m' n', therefore the first term = z f(x, y). (55.) It does not unfrequently happen that it becomes necessary to substitute for the differential coefficients of one or several functions, with respect to one or more variables, involved in a formula, the differential coefficients of the same function, with respect to other variables connected with the first by given relations. Let us first suppose that the formula contains only the differential coefficients of y with respect to the variable ≈, and that it is required to substitute for them the differential coefficients of y with respect to another variable t, dy dy y dx' d x2 and, by taking again the differential coefficients of both sides of this equation with respect to t, we find In a similar manner, the values of the differential coefficients of higher orders may be found. Part I. With these values we shall be able to transform any analytical expression involving the differential coefficients of to y. (56.) No greater difficulty will be found to change the differential coefficients of any number of functions Y1, Y, Ya, &c., with respect to any number of variables x, x, x, &c., into the differential coefficients of the same functions with respect to an equal number of variables Z1, Z2, Z3, &c., connected with the first by as many equations as there are variables. The first partial differential coefficients will be as before, And, taking the differential coefficients of each of those with respect to each of the new variables, the values of We have now explained all the general rules of the Differential Calculus, and sufficiently illustrated the meaning PART II INTEGRAL CALCULUS. Integral (57.) WE have before stated, that the Integral Calculus was the inverse of the Differential Calculus, and had for Part II. Calculus. its object to determine the value of a function, the differential coefficient of which is known, or, more generally, to discover the relations which exist between the variables and the functions, from given equations between them and their differential coefficients. (58.) We shall first consider the simplest case; which is, to find the value of a function of one variable, when the first differential coefficient is given explicitly in terms of that variable. dy dx Let X be the given differential coefficient, and let y designate the unknown function, then X, or dy = Xdx. The required function is generally represented by X dr, the characteristic denoting an operation precisely the inverse of that indicated by d in the differential calculus. Hence, if the two characteristics and d were prefixed to the same function u, they would neutralize each other, and we would have ƒ du = u. It follows also from (30) that d1 X d x would signify the same thing as Xda; and consequently that we might dispense with the use of a new sign. But as it is universally employed, we shall retain it here. In the sequel we shall have occasion to mention the origin of this notation, and also of the name integral, applied to ƒ X d x, or to the function whose differential coefficient is X. The operation, by means of which the integral of a given differential is determined, is called integration. To integrate a differential, is to find the value of its integral. These definitions and notations understood, we shall deduce without any difficulty from the observations and rules stated in the differential calculus, the following results. (59.) If y represent a function of x, and if d y = X dx, then, from (18), Xdxy + A, where A is an arbitrary constant. Hence we may always add to the integral of a given differential a constant quantity, whose value remains in general indeterminate. If however the value of the integral corresponding to a particular value of, happen to be known, then the constant may be determined. Let us suppose, for instance, that we know that the integral becomes equal to B, when x is assumed equal to b. Then, if we designate by C the value of y corresponding to the same supposition, we must have C + A = B, and consequently A= B ̊ – C. (60.) We shall also have, by (18), M being a constant, fMXdx = MSXdx = My + A. Hence, when a constant factor multiplies a given differential function, it may be written out of the sign of = = J (X, d x + X, d x − X, d x) = f X, d x + √ X, d x − f X, dx = y1 + Y2 - Y2 + A. Hence the integral of the sum, or difference, of the several differential functions of the same variable is equal to the sum or difference of the integrals of these differentials. (62.) The rule given (22) to find the differential coefficient of the product of two functions of the same variable, will give f y1 X, d x = y1 Y» — f y2 X, d x, or ƒ y1 d y2 = Y1 Y2 - S y, d y1. This result shows, that when the differential function may be decomposed into two factors y, and X, dø, and |