Conceptual help, Why are we allowed to fix variables to solve PDE's? I am struggling to understand conceptually why we are allowed to fix a variable to solve PDE's and still get out a general solution, take the following example for instance. 
Say we have a partial differential equation of the form 
$$ U_{x} + 2U_{y} + (2x-y)U = 2x^2 + 3xy - 2y^2$$
we can use the following change of coordinates
$$\widetilde{x}=x+2y \\ \widetilde{y}=2x-y$$
and we are left with 
$$5U_\widetilde{x} + \widetilde{y}U=\widetilde{x}\widetilde{y}$$
$\textbf{We can now solve this by fixing $\widetilde{y}$ to get}$
$$U(\widetilde{x},\widetilde{y})= (\widetilde{x} - \frac{5}{\widetilde{y}})+\exp(\frac{-\widetilde{x} \widetilde{y}}{5})f(\widetilde{y})$$
Whice when reverted back to original coordinates gives
$$U(x,y) = (x+2y - \frac{5}{2x-y})+e^{-x-2y}g(2x-y)$$
Where I struggle conceptually is trying to understand why we are allowed to fix $\widetilde{y}$? Further why is the solution then applicable to all $\widetilde{y}$  how comes the solution isn't limited to only fixed $\widetilde{y}$ values,since that's what we used to get the solution? 
 A: Strictly speaking you're not fixing $y$, but you're using an integrating factor to integrate a partial derivative with respect to $x$. 
If you multiply both sides of the equation by $\frac{e^{\tilde{x} \tilde{y}}}{5}$, the PDE becomes: 
\begin{equation}
\dfrac{\partial }{\partial \tilde x} \left(U e^{\frac{\tilde x \tilde y}{5}} \right) = \frac{1}{5} \tilde x \tilde y e^{\frac{\tilde x \tilde y}{5}} =  \dfrac{\partial }{\partial \tilde x} \left( \left( \tilde x -\frac{5}{\tilde y} \right)e^{\frac{\tilde x \tilde y}{5}} \right),
\end{equation}
that is 
\begin{equation}
\dfrac{\partial }{\partial \tilde x} \left(\left(U-\tilde x + \frac{5}{\tilde y} \right) e^{\frac{\tilde x \tilde y}{5}} \right) = 0.
\end{equation}
So you have a function whose partial derivative WRT $\tilde x$ is zero, so no matter how much you change $\tilde x$, the function will remain  the same  if  $\tilde y$ is unchanged. This means that the argument inside the differential operator can only be a function of $\tilde y$. A priori this is unknown and  you may call it $f(\tilde y)$ - it is then determined by initial/boundary data.
