du 413.] The differential equation which is integrated by the above method admits easily of a geometrical interpretation. Since tan-1 ay is the angle between ay is the angle between the axis of x and the tangent to a plane curve at the point (a, y), equation (220) is that to the tangent of a curve, in which the intercept by the tangent of the axis of y is expressed as a function of an, or, in other words, no = f (6%); thus the general integral is the equation of any tangent line, the arbitrary constant contained in which is the tangent of the angle between it and the axis of x; and the singular solution is the curve-envelope of all such tangents, and is found in fact by a method which is identical with those of the last section: viz. by eliminating either c between the general integral and its c-differential, or the between the differential equation and its ay - differential. The most general geometrical problem which involves a differential equation of Clairaut's form is that wherein the length of the perpendicular from the origin on the tangent to a curve is a function of the angle which the perpendicular from the origin makes with the axis of x : and as problems of this kind are numerous, and often elegant, two or three are subjoined. Ex. 1. To find the equation to the curve, the perpendicular from the origin on the tangent of which is of constant length a. The differential equation of the curve is plainly y dx - x dy = a ds; (224) which is the general integral; and is the equation to a straight line inclined at tan-lc to the axis of x, and the perpendicular distance from the origin on which is equal to a. This line evidently satisfies the conditions of the problem. Also from the second factor of (225), we have (a’ — 22) substituting which in the given differential equation, 22 + y2 = a’; the equation of a circle whose radius is a, and which is the singular solution, being the envelope of all the lines whose equations are given by the general integral. The following process is worth noticing: let us differentiate (224) taking neither x nor y to be equicrescent; y do q– đoy = a dos :(y-a te) da x–(2+ady) do y = 0; and since dax and dụy are arbitrary, dix .. x2 + y2 = a. Ex. 2. The product of two ordinates of the tangent of a curve drawn at two given points on the axis of x is constant; it is required to find the equation of the curve. Let the origin, see fig. 51, be taken at the point of bisection of the line AB which joins the two points A and B at which the ordinates AQ, BR are drawn : and let x and y be the current coordinates to the tangent line: let on = 0B = a; and let AQ XBR = ké: then the equation to RQ is which is the general integral : also from the second factor of (228) y = --- kæ __; a{a— x2}}' and therefore from (226), which is the singular solution; and is an ellipse of which AB is the major axis. Or thus; differentiating (227), and equating to zero the coefficients of dưx and d’y, kdx - ady k kdx2 + a2 dy}}' a {k2dx? + aż dy2}}' then squaring and adding, az + = 1. 2012 Ex. 3. The triangle contained between the rectangular axes and the tangent of a plane curve is of constant area and = 5 shew that the equation of the tangent of the curve is ydx— «dy = k(dy dx)}; and that the singular solution is xy = a. Ex. 4. TRQ being a tangent to a curve in fig. 51, perpendiculars AY, BZ are drawn to it from the points A and B, and the included area ABZY is constant : find the equation to the line TRQ; and shew that the singular solution of it is the equation to a parabola. Ex. 5. Find the equation to the curve the portion of whose tangent intercepted between the coordinate axes is of constant length. Ex. 6. Determine the curve whose tangent cuts off from the coordinate axes parts the sum of which is constant. 414.] Let us now return to the more general form given in (219); and differentiate it; then we have y = + (%)+{x®"(y) +f"(y)}.; (229) {$(3) —;} + x(4)+f(y) = 0; (230) which is a linear differential equation of the first order and degree, and can be integrated by the method of Art. 382 ; whereby we shall obtain y' in terms of x; and thus y may be eliminated from the given differential equation, and the resulting equation in terms of x and y will be the required solution. Ex. 1. y + xy'= ay2; .:. 24+ (r—2 ay) dy = 0; 2y'dx + x dy = 2 ay' dy' ; • xy = e; by means of which and the original equation, ý' may be eliminated, and the resulting equation will be the required solution. Ex. 2. y = xy^2 + 2y'; : y = x^2+2(a +1) 4 , 2 which is a linear equation, and of which the integrating factor is (y' — 1)2; :: (* — 1)dx + 2 x (1 – 1)dy'= -247 dy', x(y' - 1)2 = log y2– 2y +c, between which and the given equation we may eliminate y', and so obtain the required result. Ex. 3. Prove that the curve in which the length of the perpendicular from the origin on the tangent is equal to the abscissa of the point of contact is the circle, the origin being on the circumference. Ex. 4. Determine the plane curve such that the normal makes equal angles with the radius vector and the x-axis. 415.] Also if the differential equation can be put into the form y = f (2, fi) = f(x, y), (231) then differentiating we have ý'= f'(x, y)+ 4.f(x,y) dy'. dy'. dx which is a differential equation involving two variables x and y', the integral of which will be of the form F(x, y,c) = 0, where c is an arbitrary constant: and if ý can be eliminated by means of this equation and (231), the resulting equation will contain x, y, and c, and will be the integral of (231). Similarly, if an equation is capable of being put into the form w=f(y,y'); (232) then by differentiation (233) and since (233) becomes which involves only y and y'; whence the integral of it is of the form (y, y', c) = 0, and thus if y' is eliminated by means of this and of (232), the result will be the integral of (232). The following examples are illustrative of these processes. |