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 eaxφ(x), where φ(x) is an integral polynomial in x; this includes cases in which R contains terms of the form cos bx·φ(x) or sin bx·φ(x). Denote d/dx by D; it is clear that if u be any function of x, D(eaxu) = eaxDu + aeaxu, or say, D(eaxu) = eax(D + a)u; hence D²(eaxu), i.e. d²/dx² (eaxu), being equal to D(eaxv), where v = (D + a)u, is equal to eax(D + a)v, that is to eax(D + a)²u. In this way we find Dn(eaxu) = eax(D + a)nu, where n is any positive integer. Hence if ψ(D) be any polynomial in D with constant coefficients, ψ(D) (eaxu) = eaxψ(D + a)u. Next, denoting ∫ udx by D-1u, and any solution of the differential equation dz/dx + az = u by z = (d + a)-1u, we have D[eax(D + a)-1u] = D(eaxz) = eax(D + a)z = eaxu, so that we may write D-1(eaxu) = eax(D + a)-1u, 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(eaxu), which means D-1[D-1(eaxu)], is equal to D-1(eaxz) and hence to eax(D + a)-1z, which we write eax(D + a)-2u; proceeding thus we obtain
D-n(eaxu) = eax(D + a)-nu,
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 ψ(D) be any polynomial in D with constant coefficients, and we agree to denote by [1/ψ(D)]u any solution z of the differential equation ψ(D)z = u, we have, if v = [1/ψ(D + a)]u, the identity ψ(D)(eaxv) = eaxψ(D + a)v = eaxu, which we write in the form
| 1 | (eaxu) = eax | 1 | u. |
| ψ(D) | ψ(D + a) |
This gives us the first step in the method we are explaining, namely that a solution of the differential equation ψ(D)y = eaxu + ebxv + ... where u, v, ... are any functions of x, is any function denoted by the expression
| eax | 1 | u + ebx | 1 | v + .... |
| ψ(D + a) | ψ(D + b) |
It is now to be shown how to obtain one value of [1/ψ(D + a)]u, when u is a polynomial in x, namely one solution of the differential equation ψ(D + a)z = u. Let the highest power of x entering in u be xm; if t were a variable quantity, the rational fraction in t, 1/ψ(t + a) by first writing it as a sum of partial fractions, or otherwise, could be identically written in the form
Krt-r + Kr-1t-r+1 + ... + K1t-1 + H + H1t + ... + Hmtm + tm+1φ(t)/ψ(t + a),
where φ(t) is a polynomial in t; this shows that there exists an identity of the form
1 = ψ(t + a)(Krt−r + ... + K1t−1 + H + H1t + ... + Hmtm) + φ(t)tm+1,