This being a linear equation is easily integrated, and we find C'acc= dy d2y dy x + y = 0. dx dx2 dx - (C' + 1) x3 − {2y + (C2 − 1) x} 3 * There is also a singular solution y = Cx. Sometimes an equation may be considered homogeneous by reckoning a as of one dimension, y of n dimensions, and d2y da consequently of (n-1) dimensions, and of (n = 2) dy da = x-1t, dimensions. In such cases assume y=x" u, p q = "-2v; then by steps similar to those in the last case we arrive at a differential equation of the first order, between t and u, which being integrated will enable us to determine the relation between x and y. dy d2y If be reckoned of o dimensions so that y' d are of the same dimensions, a homogeneous equation may be integrated by assuming From this last if we eliminate v by means of the given equation, we have to find u in terms of a, by integrating an equation of the first order, and then by means of we can determine the relation between a and y. du v = u2 + and (a2 + a2)3 u = udx, therefore C { x + (a2 + x3)} ; whence log (C'y) = Ca3log {x + (a2 + x2)2 } + C æ { x + (a2 + x2)§}. V. Equations of the second order in which one or other of the variables is wanting. If the deficient variable be the dependent variable y, by putting dy dp between p and a, by the integration of which we obtain p in in terms of p; and then by means of the terms of x, or equation y = fpdx = xp - fxdp, we can find the relation between x and y. dx = a2 d'y 2x dx2 = a2 dp This is the equation to the elastic curve. Jac. Bernoulli, Opera, p. 576. = @ (1 + p°)3, da dx which is integrable when divided by (1 + p2). The complete integral is y = (a2 + b2 − ∞2) -- blog b + {(a2 + b2 − x2)} } where b and c are the arbitrary d'y c(x - a) constants. (28) Let ar (a2 + x2)3 + a2 The integral is dy dx dy 2 = (29) Let (x + a) += (d) - dy x dx da C' e and c' being arbitrary constants. If the independent variable (x) be wanting, we put d'y dy dp dp = = Р dx2 dx dy dy tween p and y terms of y, or y the equation , and then we have an equation be from which by integration we find p in in terms of p, and then a is known from dx (y + 1) = a linear equation in y, which being integrated gives x = dy Р y = p + C (1 + p2)3, = log p + C log {p + (1 + p2)3} + C', whence by eliminating p we obtain a relation between a |