Suppose we have a homogeneous linear PDE of order $n$ with the form $$\frac{\partial^nf(x,y)}{\partial x^n}+p_{n-1}(x)\frac{\partial^{n-1}f(x,y)}{\partial x^{n-1}}+\cdots+p_0(x)f(x,y)=0$$ where the $p_i$'s are smooth. Must it be true that $f(x,y)=X_1(x)Y_1(y)+\cdots+X_n(x)Y_n(y)$ for some functions $X_i$ and $Y_i$?
I can prove that this is true for $n=1$:
Let $q(x)$ be such that $q'(x)=p_0(x)$, then $$e^{q(x)}f_x(x,y)+p_0(x)e^{q(x)}f(x,y)=0$$ $$\frac{\partial}{\partial x}(e^{q(x)}f(x,y))=0$$ $$f(x,y)=Y(y)e^{-q(x)}$$ For $n=2$, assuming that $f(x,y_0)$ is non-zero everywhere, we can set $f(x,y)=f(x,y_0)g(x,y)$ and use reduction of order to get a similar conclusion. But I don't know if that assumption can be dropped.
This question arose from solving an equation in spherical coordinates, so the domain of the equation is $(0,\infty)\times[0,\pi]\times[0,2\pi]$. (I believe the number of absent variables should not matter, so I used an equation with two variables only.) But more general answers are also welcomed.
Let $\{X_1,X_2,\dots,X_n\}$ be a fundamental set of solutions for the $n$-th order linear ordinary differential equation \begin{equation} X^{(n)} + p_{n-1} X^{(n-1)} + \cdots + p_0 X = 0. \end{equation} Then the function \begin{equation} f(x,y) := \left( C_1 X_1(x) + C_2 X_2(x) + \cdots + C_n X_n(x) \right) Y(y) \end{equation} is a solution of your linear partial differential equation, for any numbers $C_1, C_2, \dots, C_n \in \mathbb{R}$ and for any function $Y$. Define functions $Y_i(y) := C_i Y(y)$, $i =1,2,\dots,n$, to obtain the desired form \begin{equation} f = X_1 Y_1 + X_2 Y_2 + \cdots + X_n Y_n. \end{equation}