PDE (similar to Heat equation) tranformation, how to solve. I suddenly ran into this equation: Let $u:[a,b]\times \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying:
$$\partial_t u = u' + \frac{1}{2}u'', \quad (1)$$
with some boundary conditions that are not relevant for the question.
I solved this equation by choosing $u(t,x):= e^{ax+bt}v(t,x)$ where $v$ solves the well-known Heat equation: $\partial_t v = \frac{1}{2} v''$ with corresponding modified boundary conditions.
The question is: What about considering the more general version of equation $(1)$?
$$\partial_t u = p(x)u' + \frac{1}{2}u''$$
where $p(x) = \sum_{k=0}^n a_k x^k$ is a polynomial of degree $n$? Can one use a similar "tranformation trick" like "$u(t,x)= \mbox{something}\times v(t,x)$" where $v$ solves the Heat equation or even $(1)$ or any other solvable PDE?
If this trick is not possible, are there other tricks, is there a theory on how to solve such PDEs?
If this is too demanding, would it be at least possible for the case $p(x)=x$ or $p(x)=x^2$?
Thanks a lot! :)
 A: My remark is too long for a comment so I post it as an answer. Let
\begin{equation*}
L=p(x)\partial _{x}+\frac{1}{2}\partial _{x}^{2}
\end{equation*}
and suppose we can find its eigenvalues and eigenfunctions.Then we can find
the solution of
\begin{equation*}
Lu=p(x)\partial _{x}u+\frac{1}{2}\partial _{x}^{2}u
\end{equation*}
by expanding $u$ in terms of the latter. However $L$ is not self-adjoint in $%
L^{2}(\mathbb{R})$ so this complicates matters. In case $p(x)\partial _{x}$
is replaced by $\frac{1}{2}\{p(x)\partial _{x}-\partial _{x}p(x)\}$ the
situation improves. Now
\begin{equation*}
Lv_{r}=\lambda _{r}v_{r},\;\lambda _{r}\in \mathbb{R}
\end{equation*}
and we can expand
\begin{equation*}
u=\int drc_{r}(t)v_{r}
\end{equation*}
(in view of the form of $L$ we expect a continuum of $r$'s). Now, using
orthogonality,
\begin{eqnarray*}
\partial _{t}c_{r}(t) &=&\lambda _{r}c_{r}(t)\Rightarrow
c_{r}(t)=c_{r}(0)\exp [\lambda _{r}t] \\
u(x,t) &=&\int drc_{r}(0)\exp [\lambda _{r}t]v_{r}(x)
\end{eqnarray*}
But with your original expression for $L$ the problem seems much more
complicated.
