My PDEs are a bit rusty, and this seems all too simple, but something isn't quite right. I have the following equation of Thomson: \begin{equation} \frac{\partial}{\partial t}v(x,t)=\gamma\frac{\partial^2}{\partial x^2}v(x,t)-\alpha v(x,t)\,, \end{equation} with $\alpha$, $\gamma>0$. Now I am to show that \begin{equation} v(x,t)=e^{-\alpha t}u(x,t) \end{equation}is its general solution, where $u(x,t)$ solves $\partial u(x,t)/\partial t$=$\gamma\partial^2 u(x,t)/\partial x^2$ (the heat equation).
My normal approach to such a thing would be to differentiate the assumed solution with $t$, set the result equal to the RHS of the original equation, rearrange to isolate $u(x,t)$, then respectively differentiate that once with $t$ and twice with $x$. The latter two differentiations should produce expressions that are off by a factor of $\gamma$, verifying the needed properties. But some cursory inspection discounts that possibility: there'd be a fourth derivative in one of the expressions, and a mixed partial in the other.
What am I missing, and how can I start on the correct foot?
I think you're overthinking this. Observe that
\begin{align*} v_t &= -\alpha v + e^{-\alpha t} u_t \\ &= -\alpha v + \gamma e^{-\alpha t}u_{xx}\\ &= -\alpha v + \gamma v_{xx}. \end{align*}
The first equality comes from differentiating $v$, the second equality comes from the heat equation, and the third equality comes again from differentiating $v$.