rresponding to these give the values of x +[lambda]1y and
x + [lambda]2y, from which x and y can be found as functions of t,
involving two arbitrary constants. If, however, the two roots of the
quadratic equation for [lambda] are equal, that is, if (a - b')^2 +
4a'b = 0, the method described gives only one equation, expressing x +
[lambda]y in terms of t; by means of this equation y can be eliminated
from dx/dt = ax + by + c, leading to an equation of the form dx/dt =
Px + Q + Re^(a + [lambda]a')t, where P, Q, R are constants. The
integration of this gives x, and thence y can be found.

A similar process is applicable when we have three or more dependent
variables whose differential coefficients in regard to the single
independent variables are given as linear functions of the dependent
variables with constant coefficients.

Another method of solution of the equations

dx/dt = ax + by + c, dy/dt = a'x + b'y + c',

consists in differentiating the first equation, thereby obtaining

d^2x dx dy
---- = a-- + b--;
dt^2 dt dx

from the two given equations, by elimination of y, we can express
dy/dt as a linear function of x and dx/dt; we can thus form an
equation of the shape d^2x/dt^2 = P + Qx + Rdx/dt, where P, Q, R are
constants; this can be integrated by methods previously explained, and
the integral, involving two arbitrary constants, gives, by the
equation dx/dt = ax + by + c, the corresponding value of y. Conversely
it should be noticed that any single linear differential equation

d^2x dx
---- = u + vx + w--,
dt^2 dt

where u, v, w are functions of t, by writing y for dx/dt, is
equivalent with the two equations dx/dt = y, dy/dt = u + vx + wy. In
fact a similar reduction is possible for any system of differential
equations with one independent variable.

Equations occur to be integrated of the form

Xdx + Ydy + Zdz = 0,

where X, Y, Z are functions of x, y, z. We consider only the case in
which there exists an equation [phi](x, y, z) = C whose differential

dP[phi] dP[phi] dP[phi]
-------dx + -------dy + -------dz = 0
dPx dPy dPz

is equivalent with the given differential equation; that is, [mu]
being a proper function of x, y, z, we assume that there exist
equations

dP[phi] dP[phi] v[phi]
------- = [mu]X, --