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Particular Solution of an Inhomogeneous Linear ODE


Definition

Write a linear ordinary differential equation (ODE) of order nn as L[x]=gL[x]=g, where

L[x]=x(n)+an−1x(n−1)+⋯+a1x′+a0x,L[x]=x^{(n)}+a_{n-1}x^{(n-1)}+\dots +a_1x'+a_0x,

x′(t)=dxdt(t)x'(t)=\dfrac{\mathrm{d}x}{\mathrm{d}t}(t), and gg is a given forcing term with g≢0g\not\equiv0 (the inhomogeneous case). A particular solution is any one function xpx_p that satisfies L[xp]=gL[x_p]=g on the interval considered. The coefficients may be constants or continuous functions of tt.


Interpretation

The particular solution is the response to the forcing gg; the solutions of the homogeneous equation L[x]=0L[x]=0 describe the free motion of the system. The word "particular" means "one specific", not "special": xpx_p is not unique, since adding any homogeneous solution gives another particular solution.

Two particular solutions of x'' + 3x' + 2x = 4t: the line x_p = 2t - 3 and the curve 2t - 3 + 2e^(-t), whose vertical gap 2e^(-t) shrinks as t grows.

The figure uses the polynomial-forcing example below. Besides xp=2t−3x_p=2t-3, the function 2t−3+2e−t2t-3+2\mathrm{e}^{-t} also satisfies x′′+3x′+2x=4tx''+3x'+2x=4t, since 2e−t2\mathrm{e}^{-t} solves the homogeneous equation (r=−1r=-1 is a root of r2+3r+2r^2+3r+2). The gap between the two curves is exactly this homogeneous solution; here it dies out, so both curves follow the same forced response for large tt.

This mirrors linear algebra: for a homogeneous and inhomogeneous system Ax=bA\boldsymbol{x}=\boldsymbol{b}, one solution of the inhomogeneous system plus the null space gives all solutions. Here LL is linear, so the same structure holds; see general solution of a linear ODE.

For first-order equations a particular solution can always be computed with an integrating factor. For constant coefficients and forcing of polynomial, exponential, sine or cosine type, the method of undetermined coefficients is quicker: guess a function of the same type as gg with unknown coefficients and match terms. If the guess already solves the homogeneous equation (resonance), it must be multiplied by tt; forgetting this leads to the impossible equation 0=g0=g.


Key Results

  • If xp,1x_{p,1} and xp,2x_{p,2} are particular solutions, then xp,1−xp,2x_{p,1}-x_{p,2} solves L[x]=0L[x]=0.
  • Superposition: if L[x1]=g1L[x_1]=g_1 and L[x2]=g2L[x_2]=g_2, then L[x1+x2]=g1+g2L[x_1+x_2]=g_1+g_2.
  • Constant coefficients and constant forcing g≡cg\equiv c with a0≠0a_0\neq0: xp=c/a0x_p=c/a_0, the equilibrium of the equation.
  • Exponential forcing g=kestg=k\mathrm{e}^{st} with constant coefficients: if p(s)≠0p(s)\neq0 for the characteristic polynomial pp, then xp=kp(s)estx_p=\frac{k}{p(s)}\mathrm{e}^{st}; if ss is a simple root of pp, then xp=kp′(s)testx_p=\frac{k}{p'(s)}t\mathrm{e}^{st}.
  • In general, if the exponent ss (or α±iβ\alpha\pm\mathrm{i}\beta for eαtcos⁡βt\mathrm{e}^{\alpha t}\cos\beta t, eαtsin⁡βt\mathrm{e}^{\alpha t}\sin\beta t) is a root of multiplicity mm, the trial function is multiplied by tmt^m.
  • Vector case: for x′=Ax+b\boldsymbol{x}'=A\boldsymbol{x}+\boldsymbol{b} with constant b\boldsymbol{b} and invertible AA, the constant xp=−A−1b\boldsymbol{x}_p=-A^{-1}\boldsymbol{b} is a particular solution.

Examples

Polynomial forcing. For the second-order differential equation x′′+3x′+2x=4tx''+3x'+2x=4t, try xp=At+Bx_p=At+B: then 3A+2At+2B=4t3A+2At+2B=4t, so A=2A=2 and B=−3B=-3. Check: 0+3⋅2+2(2t−3)=4t0+3\cdot2+2(2t-3)=4t.

Exponential forcing. x′′−x′−6x=etx''-x'-6x=\mathrm{e}^{t} with p(r)=r2−r−6p(r)=r^2-r-6: p(1)=−6≠0p(1)=-6\neq0, so xp=−16etx_p=-\tfrac{1}{6}\mathrm{e}^{t}.

Resonance. x′′−x′−6x=e3tx''-x'-6x=\mathrm{e}^{3t}: 33 is a simple root of pp, so try xp=Ate3tx_p=At\mathrm{e}^{3t}. Then xp′=A(1+3t)e3tx_p'=A(1+3t)\mathrm{e}^{3t}, xp′′=A(6+9t)e3tx_p''=A(6+9t)\mathrm{e}^{3t} and L[xp]=A[(6+9t)−(1+3t)−6t]e3t=5Ae3tL[x_p]=A\big[(6+9t)-(1+3t)-6t\big]\mathrm{e}^{3t}=5A\mathrm{e}^{3t}, so A=15A=\tfrac{1}{5}, in line with p′(3)=5p'(3)=5.

Sinusoidal forcing. x′′+x=cos⁡2tx''+x=\cos2t: try Acos⁡2tA\cos2t; −4A+A=1-4A+A=1 gives xp=−13cos⁡2tx_p=-\tfrac{1}{3}\cos2t. For x′′+x=cos⁡tx''+x=\cos t the frequency matches the roots ±i\pm\mathrm{i}, and xp=t2sin⁡tx_p=\tfrac{t}{2}\sin t (check: xp′′=cos⁡t−t2sin⁡tx_p''=\cos t-\tfrac{t}{2}\sin t).


Prerequisites



Sources

Aliases: particular solution, particular integral Keywords: x_p, method of undetermined coefficients, trial solution, resonance, forcing term Symbols: alpha, beta

Last updated: 22 October 2026 19:04