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
we have an equation of the first order
dx

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
dx2

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