Proof. Write \(p(\lambda ) = (\lambda - \mu )^mp_1(\lambda )\) with \(p_1(\mu ) \neq 0\). By Lemma 6.1.1, \(p(D)(e^{\mu t}w) = e^{\mu t}p_1(D + \mu )D^mw\), so it suffices to solve \(p_1(D + \mu )D^mw = q\) with \(w = t^mr\). The operator \(p_1(D + \mu )\) maps polynomials of degree at most \(d\) into themselves and has constant term \(p_1(\mu ) \neq 0\), so on that finite-dimensional space it is triangular with non-zero diagonal, hence invertible: there is a polynomial \(v\) of degree at most \(d\) with \(p_1(D + \mu )v = q\). Then \(w\), an \(m\)-fold antiderivative of \(v\), can be chosen as \(t^mr\) with \(\deg r \leq d\), dropping the lower powers, which are killed by \(D^m\). For the trigonometric case take real or imaginary parts. □