/ /
y = (A + Bx)e^[t]^x + e^[t]^x | | Re^-[t]^x dxdx,
_/ _/

or, say, y = (A + Bx)e^[t]^x + U, where A, B are arbitrary constants,
and U is a function of x not present at all when R = 0. The portion
Ae^[t]^x + Be^[p]^x or (A + Bx)e^[t]^x of the solution, which is known
as the _complementary function_, can clearly be written down at once
by inspection of the given differential equation. The remaining
portion U may, by taking the constants in the complementary function
properly, be replaced by any particular solution whatever of the
differential equation

d^2v dy
---- + P -- + Qy = R;
dx^2 dx

for if u be any particular solution, this has a form

u = A0 e^[t]^x + B0 e^[p]^x + U,

or a form

u = (A0 + B0x)e^[t]^x + U;

thus the general solution can be written

(A - A0)e^[t]^x + (B - B0)e^[p]^x + u,

or

{A - A0 + (B - B0)x}e^[t]^x + u,

where A - A0, B - B0, like A, B, are arbitrary constants.

A similar result holds for a linear differential equation of any
order, say

d^n y d^n-1 y
----- + P1 ------- + ... + P_n y = R,
dx_n dx^n-1

where P1, P2, ... Pn are constants, and R is a function of x. If we
form the algebraic equation [t]^n + P1[t]^n-1 + ... + P_n = 0, and all
the roots of this equation be different, say they are [t]1, [t]2, ...
[t]n, the general solution of the differential equation is

y = A1 e^[t]1^x + A2 e^[t]2^x + ... + A_n e^[t]_n^x + u,

where A1, A2, ... An are arbitrary constants, and u is any particular
solution whatever; but if there be one root [t]1 repeated r times, the
terms A1 e^[t]1^x + ... + A_r e^[t]_r^x must be replaced by (A1 + A2x
+ ... + A_r x^r-1)e^[t]1x where A1, ... An are arbitrary constants;
the remaining terms in the complementary function will similarly need
alteration of form if there be other repeated roots.

To complete the solution of the differential equation we need some
method of determining a particular integral u; we explain a procedure
which is effective for this purpose in the cases in which R is a sum
of terms of the form e^ax[p](x), where [p](x) is an integral
polynomial in x; this includes cases in which R contains terms of the
form cos bx.[p](x) or sin bx.[p](x). Denote d/dx by D; it is clear
that if u be any function of x, D(e^ax u) = e^ax Du + ae^ax u, or