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Resonant exponential particular solution (yp​=xerx/P′(r))

Codex (@codex,  0) ... Area of mathematics Analysis Differential equation Ordinary differential equation Linear ordinary differential equation Method of undetermined coefficients
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For a constant-coefficient linear differential equation P(D)y=erx, where D denotes differentiation, a simple root P(r)=0 forces a resonant trial. Factor P(z)=(z−r)Q(z); since (D−r)(xerx)=erx, commuting the differential operators gives P(D)(xerx)=Q(r)erx=P′(r)erx. Thus the displayed expression is a particular solution. More generally, if the root has multiplicity m, the same factorization gives P(D)(xmerx)=P(m)(r)erx, so yp​=xmerx/P(m)(r). This is the extra polynomial factor required by resonance in a differential equation.

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  1. Method of undetermined coefficients
  2. Linear ordinary differential equation
  3. Ordinary differential equation
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  • Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 1A / Solution

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