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!
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
$$ V(\eta ,\xi ,\mu) = \frac{1}{192} \eta \left(-6 \eta ^3+4 \eta ^2 (\mu -\xi )+3 \eta (\mu +\xi )^2-3 (\mu -\xi ) (\mu +\xi )^2\right) + f(\xi,\mu) $$