I am trying to read through these notes on CFT, and author reaches a point in chapter $2$ saying: $$\Big(\eta_{\mu\nu}\square + (d-2)\partial_{\mu}\partial_{\nu}\Big)(\partial\cdot\epsilon) = 0\tag{2.8}$$
and upon contraction with $\eta^{\mu\nu}$ reaches $$(d-1)\square(\partial\cdot \epsilon) = 0.\tag{2.9}$$
where we also knew that $$\partial_{\mu}\epsilon_{\nu}+ \partial_{\nu}\epsilon_{\mu} = \frac{2}{d}(\partial \cdot \epsilon)\eta_{\mu\nu},\tag{2.5}$$ and we use notation $$\partial\cdot \epsilon = \partial_{\mu}\epsilon^{\mu}.$$
So far everything is clear, more or less. But then author says that second equation implies $\epsilon_{\mu} = a_{\mu}+ b_{\mu\nu}x^{\nu} + c_{\mu\nu\rho}x^{\nu}x^{\rho}$, because second equation means that epsilon is at most second order in $x^{\mu}$'s. I can't see why is that necessary, because second equation just says we have a harmonic(ish) , or wave-like , $\partial\cdot \epsilon$ and such objects are not necessarily first order in $x^{\mu}$ polynomials only. Following that, $\partial \cdot \epsilon$ if is linear , why it also implies that $\epsilon$ must be second order, since operation $\partial\cdot(\cdot)$ is not a partial derivative but a gradient(ish) object and we can add any term with zero gradient to the solution.
Also, author says second equation doesn't follow from the first one if $d=2$, but I think their results are the same, since they both say delambertian of gradient is zero. Am I missing something?
EDIT: Probably, without context this equations may not be specific enough. Here we are trying to understand if $x' = x + \epsilon$ is a conformal transformation on a flat space with metric $\eta_{\mu\nu}$, how should $\epsilon$ be, assuming $\epsilon$ is so small that $O(\epsilon^2)$ terms are negligible. Possible, this is the reason author made the second-order polynomial assumption, but not sure.
