Ex. 6. Determine the curve of which the radius of curvature

(1) Let k = 1; that is, the radius of curvature is equal to

y2+(x− a)2 = c2;

the equation of a circle, whose centre is on the axis of x.

(3) Let k = 2; that is, the radius of curvature is equal to twice the normal.

(x-a)2 = 4c (y-c);

the equation of a parabola whose axis is the y-axis.

the equation of a cycloid, whose starting point is the origin, and whose base is the axis of x.

Ex. 7. Determine the curve whose radius of curvature varies inversely as the abscissa.

dy

d.

k

k

dx2

an equation which does not admit of further integration, but which represents the elastic curve. Also see Art. 329.

Ex. 8. Determine the equation of the curve of which the radius of curvature varies as the cube of the normal.

the equation of an ellipse. The curve is a hyperbola if k2 is replaced by 2; and is a parabola if no constant is introduced at the first integration.

Ex. 9. Find the equation to the curve in which

where ds is the length-element of the curve.

Ex. 10. A form of differential equation which frequently occurs in subsequent investigations is

d2y
dx2± k2y = 0.

Let both terms be multiplied by dy; then integrating,

and according as the upper or the lower sign is taken, we have

464.] Next, let us consider homogeneous equations of the second order: the principle of homogeneity being estimated in the following manner; the variables x, y, and their differentials dx, dy, d2y are considered to be factors of the first degree; and each term of the equation is of the same degree in respect of them; thus the equation, 3dy-(y dx-xdy)2 = 0, is homogeneous and of the fourth degree. Now in such an equation let the following substitutions be made; viz.

and it is manifest that x will enter in the same power into all the terms, and therefore may be divided out; this property in fact is the characteristic of the equation; and thus the resulting

dy

equation will contain z, v and ; for convenience of notation,

dx

v

=

dz

Р 2

v

dx

:

and v may be expressed in terms of z and p by means of the given equation, and therefore by the last two members of the equality we shall have a differential equation of the first order in terms of p and z, whereby p may be expressed in terms of and therefore from the first two members of (153) we shall obtain a differential equation of the first order in terms of x and z; and this after resubstitution will give the required integral. = (y dx-xdy)2.

Ex. 1. x3 d2y

=

log

=

=-log (c+e");

.*. v = (z−p)2 ;

dp
(p—~) 2 ;

and this is the required integral.

Also differential equations which become homogeneous, if we

d2y consider to be of one dimension, y of n, of n-1, and of

dy da

dx2 n-2 dimensions respectively, may be integrated by a similar process by assuming

It is to Euler that we are indebted for these processes; other examples will be found in his works, and in the ordinary collections of such problems; and particularly in the Integral Calculus of M. Moigno.

465.] The following differential equation is also capable of integration by either of the two following processes;

where x and y are functions respectively of x and y only divide

Also the integration may be effected by the variation of parameters. Omitting the last term, the equation is

where a is an arbitrary constant; so that from (156),

dy

the form of which is the same as that of (155).

466.] One other property of differential equations of a higher order, to which allusion has already been made, deserves explanation in this place.

Let there be a differential equation of the form

ƒ (x, y, y', y′′, ... y(")) = 0;

and suppose its integral to be

(158)

(159)

y = F(x, C1, C2, ... Cn);

then y and its derived-functions depend not only on &, but also on the values of the n undetermined constants; but a may be considered independent of them. Let us suppose any one, say C, of these constants to vary; then the variation of (158) is

df

(af) dy

dy

(1/4) d/% + (df) dx +
dy do dy' dc

dy dc

dy' dz dy" d2z

Let = 2; then =

(160) becomes

dc dx' dc

dz df d2z

=

,

d"z df

= (1/() + 1 = ( d ) + 1 = ( 1 ) + + d() = 0; (161) (df)

dy dx dy dx dy

now by reason of (159) y, y',... y(") are functions of a and of

C1, C, C: if then we substitute these in (158), (d),

df df
dy

become functions of x, C1, C2, ... C; and therefore the coefficients of z and of its derived-functions in (161) are variable, and the equation is linear; and we know that a particular integral of it

because the equation was found by making = 2:

dy dc

and as for the general value c we may substitute each of the c's, so the general integral of (161) is