I have to solve the problem
$U_{xx}-U_{xy}+2U_y+Uyy-3U_{yx}+4U=0$
using diagonal matrix as described in this article page 44 section 3.2
But my problem is there the matrix A is symmetric matrix assuming derivatives commute. But in the given problem coefficient of $U_{xy}=-1$ and coefficient of $U_{yx}=-3$.
Then the corresponding matrix
A=$$
\begin{bmatrix}
1/2 & -1 \\
\\
-3 & 0 \\
\end{bmatrix}
$$
Then the matrix A is not symmetric
As posted in the commentaries depends on the hypothesis about the solutions. First I'm going to assume $U_{xy}=U_{yx}$. On your article the derivatives of grade $2$ are writed as the next form: $$\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}\frac{\partial^2u}{\partial x_i\, \partial x_j} $$ What can be writed as: $$\sum_{i=1}^{n}a_{ii}+\sum_{i=1}^{n}\sum_{j=1,i\neq j}^{n}a_{ij}\frac{\partial^2u}{\partial x_i\, \partial x_j} $$ Assuming that your derivatives commute: $$\sum_{i=1}^{n}a_{ii}+\sum_{i=1}^{n}\sum_{j<i}^{n}(a_{ij}+a_{ji})\frac{\partial^2u}{\partial x_i\, \partial x_j} $$ Here is the keypoint exist more than one matrix $A$ that match with this pattern for example: $$ A= \begin{bmatrix} 1 & -1 \\ \\ -3 & 1 \\ \end{bmatrix} $$ But if you assume that the matrix is symmetric: $$ A= \begin{bmatrix} 1 & -2 \\ \\ -2 & 1 \\ \end{bmatrix} $$ What is the result of putting your equation in the next way: $$U_{xx}-U_{xy}+2U_y+Uyy-U_{yx}-2U_{yx}+4U=0$$ $$U_{xx}-U_{xy}+2U_y+Uyy-U_{xy}-2U_{yx}+4U=0$$ $$U_{xx}-2U_{xy}+2U_y+Uyy-2U_{yx}+4U=0$$ Using the fact that $U_{xy}=U_{yx}$. The advantage of assuming $A$ symmetric allows you to use the diagonalization tool as you can see in rest of the article.