so that (241) become dz dp dz = ce dx + (cx ex + ce?") dy. :. z = cre" + ev+f(c), where f denotes an arbitrary function. Ex. 4. Find the equation of a tubular surface generated by a sphere of radius a, the centre of which moves along a directorcurve in the plane of (x,y). . The differential equation of the surface is, see (124), Art. 373, Vol. I, Ex. 5. Find the equation to the surface in which the part of the normal intercepted between the surface and the plane of (x,y) is constant and equal to unity, SECTION 10.- Various Theorems and Applications of Differ ential Equations of the First Order and of any Degree. 419.] The form of a differential equation will frequently be so much modified by a judicious substitution, or by a change of coordinates when it is presented in a geometrical form, that it is desirable to exhibit this method of integration at a greater length. No general rules however can be given, and the particular substitution must be left to the judgment of the student. The following example indicates the kind of substitution. Ex. l. Rydy +yo dx _ d.f(y). x+ y2 +a+ a? from which u is to be eliminated by means of (244), and the resulting equation will be the required solution of the given differential equation. 420.] The substitution however which is frequently very convenient is that which consists in the transformation from rectangular to polar coordinates, especially when the differential equation is a function of xdy-ydx, xdx + y dy, (x2 + y2); in this case x and y are replaced by r cos 0 and r sin 0 respectively, where r and 0 are new and independent variables. A differential expression is often much simplified hereby. Take the problem, To find the curve such that the perpendicular distance from the origin on the tangent is equal to the abscissa. In rectangular coordinates the differential equation is y dx — X dy = x ds; 421.) Sometimes the integrals of the sum of two or more expressions can be found in finite algebraical terms, although the integral of each separately would involve an elliptical or other transcendental function; the reason of course being that the transcendental parts neutralize each other : of this we have had instances in Fagnani's theorem as to elliptic arcs, and in Ex. 1, Art. 369. The following example is a remarkable illustration of this. It is required to integrate dx (245) .. (x2 - y2) dx + 2xy dy = 0; which is a homogeneous equation, and of which the integral is y2 = 2ax -22, the equation to a circle, whose radius is a. If however (245) is expressed in polar coordinates, we have x = r cos 0, dx = dr cos 0—r sin 6 do;? (246) y = r sin 0; dy = dr sin 0 + r cos 0 do; 1070 [ sin Ꮎ dᎾ . ñ = cos 0 r = 2a cos 0. which result has already been demonstrated in Ex. 12, Art. 176. It is indeed this process of transformation which has been applied in the preceding solution of homogeneous differential equations, where we have replaced y by xz. The following are other examples. m (x dy-ydx) = xdx + ydy. Substituting from (246), x dy-y dx = y2 do ; x dx + y dy = rdr; and the equation becomes my2 do = rdr; i. p = cemo. Let this process be compared with that of Ex. 1, Art. 379. Ex. 2. Find the integral of & dy-ydr = (x dr +y dy) (@? + y2 – a°)}; gu2 do = rdr (p2 – a)); 3 di, FJr Ex. 3. y dx - x dy = ds f (x2 + y2). du {2+ajx + agua + Ag.x3 +2424} {2, +any+azya +Qzy3 +Qxy*}* Let 20+ 0,2 +0.72% + Azx3 + 2424 = x, do +ay+azyê + az y3 +2444 = Y; and let each term of (247) = dt: so that dx – dy = dt; (242 :. x = aina Y = ; and consequently, if the new variable t is equicrescent, 2dx dax , 2 du dog. {1- (sin 0)2}}+{1-e? (sino)2}} = c sin (0-0), which is an integral expression satisfying (250). The preceding equation and some more general cases of a similar form are discussed by Prof. Richelot of Königsberg, in Crelle's Journal, Vol. XXIII, p. 354. The following expressions may also be integrated by a similar process : |