Since all terms of the PDE are in same order and constant coefficient, you can use the "factorization method" to solve it:
The auxiliary polynomial of this PDE can be $A\lambda^2+B\lambda+C$ or $C\lambda^2+B\lambda+A$
By factorizing them in $\mathbb{C}$ :
$A\lambda^2+B\lambda+C=A\left(\lambda+\dfrac{B+\sqrt{B^2-4AC}}{2A}\right)\left(\lambda+\dfrac{B-\sqrt{B^2-4AC}}{2A}\right)$ or $C\lambda^2+B\lambda+A=C\left(\lambda+\dfrac{B+\sqrt{B^2-4AC}}{2C}\right)\left(\lambda+\dfrac{B-\sqrt{B^2-4AC}}{2C}\right)$
$\therefore$ the general solution of $Au_{xx}+Bu_{xy}+Cu_{yy}=0$ is $u(x,y)=f_1\left(2Ay+\left(B+\sqrt{B^2-4AC}\right)x\right)+f_2\left(2Ay-\left(B+\sqrt{B^2-4AC}\right)x\right)$ or $u(x,y)=f_1\left(2Cx+\left(B+\sqrt{B^2-4AC}\right)y\right)+f_2\left(2Cx-\left(B+\sqrt{B^2-4AC}\right)y\right)$