General Solution for Partial Differential Equation To be honest I'm a bit lost on this, and I would like to get a hint or something that can help me, thanks.
I need to find the general solution $U(x,y,z)$ of the next equation: 
$$U_{xx}+U_{yy}+4U_{zz}-2U_{xy}+4U_{xz}-4U_{yz}=xyz$$
I know that most sure exists a change of variable that would help to solve the equation, but I don't know how to find it, our teacher of Multivariable Calculus asked to solve this.
Thanks!
 A: This PDE is linear so it's solution can be obtained as
$$
U = U_h + U_p
$$
We have
$$
\left(\partial_{xx}^2+\partial_{yy}^2+4\partial_{zz}^2-2\partial_x\partial_y + 4\partial_x\partial_z-4\partial_y\partial_z\right)U = \left(\partial_x-\partial_y+2\partial_z\right)^2U = x y z
$$
so calling
$$
V = \left(\partial_x-\partial_y+2\partial_z\right)U
$$
we have
$$
\left(\partial_x-\partial_y+2\partial_z\right)V = x y z
$$
using the characteristic method, we have
$$
\frac{dx}{1}=\frac{dy}{-1} = \frac{dz}{2}
$$
giving the characteristics
$$
x+y = \eta\\
2y+z = \xi\\
2x-z = \mu
$$
Choosing instead
$$
x-y = \eta\\
2y+z = \xi\\
2x-z = \mu
$$
because $\xi + \mu = 2(x+y)$ and introducing now this change of variables into the full PDE we will obtain
$$
V_{\eta}(\eta ,\xi ,\mu )=\frac{1}{64} (2 \eta -\mu +\xi ) (-2 \eta +\mu +\xi ) (2 \eta +\mu +\xi )
$$
now considering $V = V_h + V_p$ this PDE can be easily solved. 
$$
V_h(\eta ,\xi ,\mu) = C_1+f(\xi,\mu) 
$$
In this case $V_p(\eta,\xi,\mu)$ is a polynomial form.
Finally after solving $V$ we will solve
$$
\left(\partial_x-\partial_y+2\partial_z\right)U = V
$$
with the same process. 
NOTE
Here 
$$
V_p(\eta ,\xi ,\mu) = \frac{1}{192} \eta  \left(-6 \eta ^3+4 \eta ^2 (\mu -\xi )+3 \eta  (\mu +\xi )^2-3 (\mu -\xi ) (\mu +\xi )^2\right)
$$
then
$$
V(\eta ,\xi ,\mu) = \frac{1}{192} \eta  \left(-6 \eta ^3+4 \eta ^2 (\mu -\xi )+3 \eta  (\mu +\xi )^2-3 (\mu -\xi ) (\mu +\xi )^2\right) + f(\xi,\mu)
$$
