How to find the particular solution for Augmented Thin Plate Splines in the context of the Dual Reciprocity Boundary Element Method In the dual reciprocity boundary element method (DRBEM) the non-homogeneous terms are expanded in terms of radial basis functions. This expansion involves approximating the solution to the linear operator in terms of the radial basis function being used.
For example with a Laplacian/Poisson operator and augmented thin plate splines (ATPS):
The radial basis function approximation of function $g(\mathbf{x})$ looks like $$ g(\mathbf{x_i})=\Sigma_{j=1}^{N}\alpha_j]\phi(||\mathbf{x_i}-\mathbf{x_j}||)_2+\Sigma_{j=1}^{M}\beta_jp_j(\mathbf{x_j})]$$
$$ \mathbf{F}=\left[
    \begin{matrix}
    \phi(||\mathbf{x_1}-\mathbf{x_1}||)_2 & \cdots & \phi(||\mathbf{x_1}-\mathbf{x_N}||)_2 \\
    \vdots & \ddots & \vdots \\
    \phi(||\mathbf{x_N}-\mathbf{x_1}||)_2  & \cdots & \phi(||\mathbf{x_N}-\mathbf{x_N}||)_2  \\
    \end{matrix}\right]
$$
$$ \mathbf{\alpha}=\left[\begin{matrix}
\mathbf{\alpha_1} \\
\vdots \\
\mathbf{\alpha_N} \\
\end{matrix}\right]$$
For ATPS (Augmented Thin Plate Splines) the matrix of radial basis functions becomes:
$$ \mathbf{x_i}=(x_i,y_i) $$
$$ r=\sqrt{ (x_i-x_j)^2 + (y_i+y_j)^2 }=||\mathbf{x_i}-\mathbf{x_j}|| $$
$$f(\mathbf{x_i})=\Sigma_{j=1}^{N}\alpha_j r^2 log(r) + \beta_1+\beta_2x_j+\beta_3y_j $$
$$\Sigma_{j=1}^N \alpha_j=\Sigma_{j=1}^N \alpha_j x_j=\Sigma_{j=1}^N \alpha_j y_j=0
 $$
$$ \mathbf{P}= \left[
    \begin{matrix}
    1 & 1 & \cdots & 1 \\
    x_1 & x_2 & \cdots & x_N \\
    y_1 & y_2 & \cdots & y_N \\
    \end{matrix} \right]
$$
$$ \mathbf{F^*}= \left[
    \begin{matrix}
    \mathbf{F} && \mathbf{P^T} \\
    \mathbf{P} && \mathbf{0} \\
    \end{matrix} \right]
$$
$$ \left[
    \begin{matrix}
    \mathbf{u} \\
    \mathbf{0} \\
    \end{matrix} \right]= \left[
    \begin{matrix}
    \mathbf{F} && \mathbf{P^T} \\
    \mathbf{P} && \mathbf{0} \\
    \end{matrix} \right]\left[
    \begin{matrix}
    \mathbf{\alpha} \\
    \mathbf{\beta} \\
    \end{matrix} \right]=\mathbf{F^*\alpha^*}
$$
In the following reference, Golberg suggests that the particular solution to this radial basis function approximation is as below:
Golberg, Michael A. "The method of fundamental solutions for Poisson's equation." Engineering Analysis with Boundary Elements 16, no. 3 (1995): 205-213. 
$$ \hat{u}= \Sigma_{j=1}^{N}\alpha_j\Phi_j+\Sigma_{j=1}^{N}\beta_j\Psi_j$$
where $ \nabla^2\Phi_j=\Sigma_{j=1}^{N}\alpha_j r^2 log(r)$ and $\nabla^2\Psi_j= \beta_1+\beta_2x_j+\beta_3y_j $ 
Golberg finds (incorrectly) that $$\Phi=\frac{r^4 log(r)}{16} - \frac{r^3}{32}$$ When the actual result is $$\Phi=\frac{r^4 log(r)}{16} - \frac{r^4}{32}$$
The error is probably a simple typo.
Golberg goes on to say that $$\Psi= \frac{x^2+y^2}{4} + \frac{x^3}{6} + \frac{y^3}{6}$$
How does he obtain this result?
Why would this be the desired particular solution instead of something that is symmetric like with the boundary element solution kernel and the thin plate spline particular solution?
The other references I can find for ATPS in the literature related to DRBEM either suggest that obtaining the particular solutions is trivial or that they can be found in the literature. Some suggest that the polynomial particular solutions are obtained by the method of unknown coefficients.

Your particular solution isn't correct. Once you add in angular dependency, the Laplacian becomes
$$ \nabla^2 \Psi = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial \Psi}{\partial \rho}\right) + \frac{1}{\rho^2}\frac{\partial^2 \Psi}{\partial \theta^2} $$
So $\theta$ can't be treated as a constant
Instead, you look for a solution of the form
$$ \Psi = \beta_1 f(\rho) + \beta_2 g(x) + \beta_3 h(y) $$
where each term satisfies
\begin{align}
\nabla^2 f &= \frac{1}{\rho}\frac{d}{d \rho}\left(\rho\frac{df}{d\rho}\right) = 1 \\
\nabla^2 g &= \frac{d^2 g}{dx^2} = x \\
\nabla^2 h &= \frac{d^2 h}{dy^2} = y
\end{align}
Solving each equation individually gives $f = \dfrac{\rho^2}{4} = \dfrac{x^2+y^2}{4}$, $g = \dfrac{x^3}{6}$ and $h = \dfrac{y^3}{6}$
Note that $\Psi$ can't be radially symmetric, because $\nabla^2\Psi$ itself isn't radially symmetric.
