 Resonant exponential particular solution

For a constant-coefficient linear differential equation $P(D)y=e^{rx}$, where $D$ denotes differentiation, a simple root $P(r)=0$ forces a resonant trial. Factor $P(z)=(z-r)Q(z)$; since $(D-r)(x{e}^{rx})=e^{rx}$, commuting the differential operators gives $P(D)(x{e}^{rx})=Q(r)e^{rx}=P^{\prime }(r)e^{rx}$. Thus the displayed expression is a particular solution. More generally, if the root has multiplicity $m$, the same factorization gives $P(D)(x^m{e}^{rx})=P^{(m)}(r)e^{rx}$, so $y_p=x^m{e}^{rx}/P^{(m)}(r)$. This is the extra polynomial factor required by resonance in a differential equation.