and hence an identity
u = ψ(D + a) [KrD−r + ... + K1D−1 + H + H1D + ... + HmDm] u + φ(D) Dm+1u;
in this, since u contains no power of x higher than xm, the second term on the right may be omitted. We thus reach the conclusion that a solution of the differential equation ψ(D + a)z = u is given by
z = (KrD−r + ... + K1D−1 + H + H1D + ... + HmDm)u,
of which the operator on the right is obtained simply by expanding 1/ψ(D + a) in ascending powers of D, as if D were a numerical quantity, the expansion being carried as far as the highest power of D which, operating upon u, does not give zero. In this form every term in z is capable of immediate calculation.
Example.—For the equation
| d4v | + 2 | d²y | + y = x³ cos x or (D² + 1)²y = x³ cos x, |
| dx4 | dx3 |
the roots of the associated algebraic equation (θ² + 1)² = 0 are θ = ±i, each repeated; the complementary function is thus
(A + Bx)eix + (C + Dx)e−ix,
where A, B, C, D are arbitrary constants; this is the same as