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and as the difference between the circular functions contained in Uy, wy, u, respectively is a constant, it follows that either one is an integral of (174), and that therefore another particular integral of the equation is .
and therefore the general integral of the given differential equation is
--, 2 x — Y-Z _ log (yz+zx +ry)' + 37 tan 31(-25
= F(x3 + y + x3 – 3.xyz), where r is the symbol of an arbitrary function.
I may by the way observe that the solution of homogeneous equations is often facilitated by a substitution similar to that PRICE, VOL. II.
made in Art. 377. Thus we may integrate Ex. 3 in Art. 397 by assuming x = $2, y =nz; in which case the equation becomes dz (ma++1) d& + ($? +$+1) dn
= 0. Z +827 +378 + m2 +82 +n + $
399.] It is manifest from the examples of the preceding Articles, that the difficulty of determining the integrating factor is the chief obstacle, and is in most cases insurmountable: there is however another mode of solution less direct than that given above, but of which it is desirable to give a brief description, because it is the only one which has hitherto been generally applied.
Since the differential equation P dx + Q dy + R dz = 0 is a function of three variables, we may consider one of them to be dependent, and the other two to be independent; let the independent variables be x and y, so that the integral is assumed to be of the form z=f(x,y); now we may consider x and y to vary separately, and consequently z to vary owing to the variation of x or of y, when the other does not vary : suppose that in the differential equation we consider y to be constant, and the variation of z to be partial and to be due to that of x: in which case the equation becomes P dx + R dz = 0;
(175) let u be an integrating factor of this equation, when y is considered constant: and let us suppose
Tu(pdx + R dz) = F(x, y, z); then in the integration, since y has been considered constant, a function of y must be introduced ; and if y represents an arbitrary function of y, the integral is F(x, y, z) = Y:
(176) now this is manifestly such as to give the correct value of Good) in (175): y however must be so determined as to give the correct value of (): and it is also evident that if (176) satisfies these conditions it is an integral of the given differential equation. Of (176) let the total differential be taken ; then
whereby y may be determined.
But in order that y, as assumed in (176), should be a function of y only, it is necessary that - MQ should be independent of x and z: and if this is true, the x- and 2-differentials of it vanish; and therefore
de C re} = 0, de ) -we}= 0;
(177) le mie -- () - Cent) = 0; but since up = d), and we = (h), (178)
More than = n (bg) + p et = r (eng) + + Celay); substituting which in (177) and from (178) we have
which is the condition of integrability already determined; and consequently, if this condition is satisfied, this method of integration may be employed.
is u = log xyz + x +y+z+c. 400.] I will now return to the consideration of the criterion of integrability given in (166), and exhibit it from a geometrical point of view, with reference to properties of surfaces.
The differential equation Pdx + Qdy + Rdz = ( evidently expresses the condition that the line (P, Q, R) is perpendicular to a line (dx, dy, dz), joining two consecutive points ; but P, Q, R are functions of x, y, z, and vary as we pass from point to point; so that if the preceding equation expresses a property of a surface, that surface cuts orthogonally the system of straight lines whose direction-cosines are proportional to P, Q, R. It is of course conceivable that straight lines (P, Q, R) may be such that no continuous surface can cut all orthogonally, or in other words that the differential equation may not express a surface; and consequently a further condition may be required when it is capable of such an interpretation. This condition is given by the criterion of integrability, and I proceed to demonstrate that it is so.
Since the criterion involves partial differentials of P, Q, R, it evidently expresses some property of these lines in a position of displacement infinitesimally near to their former position; and as by reason of the differential equation P, Q, R are proportional to the direction cosines of the normal to the surface, the property required must depend on the position of the normals at points on the surface infinitesimally near to the point (x, y, z). Now the theorem given in Art. 340 of the present volume assigns such a relation. For since by it the radii of torsion of two geodesics on a surface intersecting at right-angles are equal at the point
of intersection, so in other words, if at a point p on a surface a normal PN is drawn, and two lines PP, and PP, of equal infinitesimal length are drawn on the surface and perpendicular to each other, the normal at P, shall make with the plane PNP, an angle equal to that which the normal at P, makes with the plane PN Pg.
I proceed to prove that this theorem interprets the criterion of integrability : let the point (x, y, z) to which reference is made in the equation Pd.x + Qdy + Rdz = 0
(179) be supposed to be on a surface, on which the line (dx, dy, dz) lies; so that the line (P, Q, R) is normal to the surface at the point (x, y, z). On the surface let two points P, and P, be taken equidistant from P, and let these points be (x +&i, y +11, 2+$i), (x + £2, Y +12, 2+$2) respectively, do being the distance of each of these from P; also let these lines make a right angle at.P. Then we have $i $2+11 12+ Š S = 0;
(180) P & +Qni+R$ = 0,
(181) P&2 + Q 2 + R2 = 0. I From the last two of which we have
(182) 7262-12 = $1$,-Ġ Sa = fi 92-97 € Also let p2 + Qo+R2 = sa.
(183) Then the direction-cosines of the normal at P1, $i, Ni, being the increments of x, y, z, are
(185) R$ drde 5 +61 de +niäz 5 +Side 5
(186) the sum of the squares of these quantities being equal to unity, when infinitesimals are neglected.
Let , be the angle which the normal at P, makes with the plane PPN. Then as $2, Naz 6 are proportional to the directioncosines of the normal of this plane, and
Ś ? +12 + = $2? + n2+ $22 = do? ; sin 8, = $, (184) + 12 (185) +62 (186),