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Detuning measures the small mismatch from a resonance condition. When the mismatch and nonlinear forcing are both $O(ϵ)$, they enter the same slow time equation. A small shift of the wavenumber can therefore balance a quadratic interaction and permit a locked periodic travelling wave.

With $z=(D-\beta )y$ and $\gamma =\alpha$, the first equation is $z^{\prime }-\alpha z=k{e}^{\alpha x}$. Multiplication by its integrating factor gives $(e^{-\alpha x}z{)}^{\prime }=k$, henceThis holds whether or not $\beta =\alpha$. If $\beta \ne \alpha$, solving $y^{\prime }-\beta y=z$ by a second integrating factor givesThe extra factor $x$ is the response to the resonance condition at the forcing exponent; the coincident-root case requires the next part.

On the forcing $e^{-i\omega t}$,The other three factors have roots with real decay rate $-1/\tau$, so a real-frequency gas wave can resonate only with the undamped factor $\mathcal{D}$. The resonance condition is thereforeAt resonance, $\mathcal{D}$ annihilates the forcing, so a particular solution acquires one power of time:This is a streaming instability resonance: the gas wave's phase velocity along $\mathbf{k}$ matches the dust drift projected along the same wavevector.

Use ${\cos }^{2}x=(1+\cos 2x)/2$. The constant forcing has particular solution $1/8$, while the $\cos 2x$ term satisfies the resonance condition and needs the trial $Cx\sin 2x$. Since $(d^2/d{x}^2+4)(x\sin 2x)=4\cos 2x$, the forced term is $x\sin 2x/8$. Adding the homogeneous solution and imposing the initial conditions gives

A resonance condition specifies a match between forcing parameters and a homogeneous mode. For the constant-coefficient system $x^{\prime }-Ax=e^{\lambda t}v$, an exponential trial $x=e^{\lambda t}u$ requiresIf $\lambda$ is not an eigenvalue, this is invertible. If it is an eigenvalue, solvability depends on the direction $v$, not merely on the parameter match. For a diagonalizable matrix, every left eigenvector $\ell$ with eigenvalue $\lambda$ requires $\ell v=0$ for a same-shape trial; a nonzero projection drives a polynomial-exponential response.

For $A=\mathrm{diag}(1,-1)$ and $\lambda =1$, forcing $e^t(1,0)$ gives a $t{e}^t$ term, whereas forcing $e^t(0,1)$ has the ordinary particular solution $e^t(0,1/2)$. Thus a spectral match is a potential resonance, and coupling to that mode determines whether the resonant response actually appears.