(8) This transformation fails when m + n = () while r is not equal to 0. In this case the following method may be used. The equation may evidently be put under the form d = Y = "; d then considering x as a function of and y, (2)(C) dy dia dy da dy dy dz = 1 and therefore dix By substituting these values the equation becomes (since m + n = 0) d = (-)"0"; \d Y the integral of which is X = - cyž "+ P(;'). 'd dy 2 2 VX = y The transformations in the three preceding examples are given by a writer who signs himself “G. C.” in the Cambridge Mathematical Journal, Vol. 1. p. 162. SECT. 4. Equations integrable by various methods. Lagrange’s Method. Let a partial Differential Equation between three variables be of the form where P, Q, R are functions of x, y and %; then if we can integrate two of the following equations, Pdy – Qdx = 0, Qdx – Rdy = 0, so as to obtain two integrals, ♡ (x, y, x) = B, y(x, y, z) = a, the integral of the given equation will be B = f (a). Lagrange, Mémoires de Berlin, 1774, p. 197; 1779, p. 152. For the success of this method it is necessary either that one of the three auxiliary equations should contain only the two variables the differentials of which it involves, or that by their combination such an equation should be obtained. By integrating it we obtain an equation by means of which one of the variables may be eliminated from either of the other auxiliary equations. d: dx + x = 0. dy ady - zdx = 0, xdx + ydy = 0. x + y = a'. (1) Let sin = f (y + 2), is the integral of the proposed equation. ax) dx = 0 = a. The auxiliary cquations are, 0....... (1) (x - ax) dx + (bx - ay) dy ...... (3). Multiply (2) by a, and (3) by b; subtract and divide by bx – ay: we find dx + adx + bdy = 0, or z + ax + by Again multiply (?) by X, and (3) by y; subtract and divide by bx - ay: there results xd x + ydy + zdx = 0, whence 2? + y + = ß. Therefore + y + m = f (z + ax + by) is the integral of the proposed equation. This is the general equation to surfaces of revolution. ...... The auxiliary equations are, (x + y) dy - (y – x) dx = 0 ............ (1) (x + y) dz – zdx = 0 .......... (2) (y – x) dx - xdy = 0 ....... (3). Equation (1) may be put under the form ædy - ydx + xda + ydy = 0; -1 = a. whence tan - log (x2 + y)2 y Multiplying (2) by X, (3) by y and adding, we have dx « dx + ydy w? + y (x2 + y')}} {tan log (x2 + y)}} Y is the required integral. d& dx (7) Let (x – 2y) + (2x – 3y) dv The integral is (*- y) = (x - y) This method may be extended to functions of more variables. Thus if du du du P S, dy dx and if from three equations such as +Q dx +R U = a, W = C, Pdy - Qdx = 0, Pdu - Sdx = 0, we can obtain three integrals, V = b, U = f(V, W), or p (U, V, W) = 0. du (1 + y + x) + (u + x +) + (u + x + y) = x + y +2. dx dy (1 + y + x) dy - (u + x + x) dx = 0, (u + y +) du – (x + y + x) dx = 0. Adding these three equations we have (u + y + x) (du + dx + dy + dx) – 3 (u + x + y + x) dx = 0. Putting u + x + y + z = 1', this gives dr du dz Subtracting the second equation from the first, we have dy - da or and v (y - x) = a. From the symmetry of the expressions it is obvious that we must have also v (0 - 2) = b, v (u - 2) = c. Therefore f {v(u – 2), v (- ), v (y – )}} = 0. |