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< be any
polynomial in D with constant coefficients, [psi](D)(e^ax u) = e^ax
[psi](D + a)u. Next, denoting [int] udx by D^-1 u, and any solution of
the differential equation dz/dx + az = u by z = (d + a)^-1 u, we have
D[e^ax(D + a)^-1 u] = D(e^ax z) = e^ax(D + a)z = e^ax u, so that we
may write D^-1(e^ax u) = e^ax(D+a)^-1 u, where the meaning is that one
value of the left side is equal to one value of the right side; from
this, the expression D^-2(e^axu), which means D^-1[D^-1(e^ax u)], is
equal to D^-1(e^ax z) and hence to e^ax(D + a)^-1 z, which we write
e^ax(D + a)^-2 u; proceeding thus we obtain
D^-n(e^ax u) = e^ax(D + a)^-n u,
where n is any positive integer, and the meaning, as before, is that
one value of the first expression is equal to one value of the second.
More generally, if [psi](D) be any polynomial in D with constant
coefficients, and we agree to denote by 1/[psi](D) u any solution z of
the differential equation [psi](D)z = u, we have, if v = 1/[psi](D +
a) u, the identity [psi](D)(e^ax v) = e^ax [psi](D + a)v = e^ax u,
which we write in the form
1 1
--------(e^ax u) = e^ax ------------ u.
[psi](D) [psi](D + a)
This gives us the first step in the method we are explaining, namely
that a solution of the differential equation [psi](D)y = e^ax u + e^bx
v + ... where u, v, ... are any functions of x, is any function
denoted by the expression
1 1
e^ax ------------ u + e^ax ------------ v + ....
[psi](D + a) [psi](D + b)
It is now to be shown how to obtain one value of 1/[psi](D + a) u,
when u is a polynomial in x, namely one solution of the differential
equation [psi](D + a)z = u. Let the highest power of x entering in u
be x^m; if t were a variable quantity, the rational fraction in t,
1/[psi](t + a), by first writing it as a sum of partial fractions, or
otherwise, could be identically written in the form
K_r t^-r + K_r-1 t^-r+1 + ... + K1 t^-1 + H + H1t + ... + H_m t^m + t^m+1 [p](t)/[psi](t + a),
where [p](t) is a polynomial in t; this shows tha]]>
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