Modified version of heat equation Consider the following equation:
$$u_t(t,x)=\frac12u_{xx}(t,x)-x\cdot u(t,x),\qquad u(0,x)=x\exp(-x^2/2).$$
After searching in Wikipedia, I can see that this is similar to the heat equation, with an extra $-x\cdot u(t,x)$ term.
Given that I know very little PDE theory,
I usually use mathematica to analyse them. However, mathematica doesn't seem to do anything with the formula, which I interpret as it saying that it can't find a solution.
After looking in the wikipedia page for the Heat equation, I've tried writing $u(t,x)=a(t)b(x)$ and then reasoning about this, but it leads to a contradiction. Are there elementary tricks that would be expected to work in such a case, or is this equation well-known and have a name?
 A: Separation of variables works.  You get $u_t = a^{\prime}b$ and $u_{xx}=ab^{\prime\prime}$, so that your equation becomes $$a^{\prime}b ={1\over 2}ab^{\prime\prime}-xab = \left({1\over 2} b^{\prime\prime}-xb\right)a.$$
Divide by $ab$ to get $${a^{\prime}\over a} =\frac{\left({1\over 2} b^{\prime\prime}-xb\right)}{b}.$$
Since the left side is a function of $t$ alone and the right side is a function of $x$ alone, the two sides must be equal to a constant $\lambda$.  So we have two ODE's: 
$$a^{\prime}(t) = \lambda a(t) \mbox{ and } {1\over 2} b^{\prime\prime}(x)-(x+\lambda)b(x) = 0.$$
Without some boundary conditions (like $u(t,0)=u(t,\alpha)=0$), it's hard to continue, but the solutions for the second equation will be in terms of Airy functions.  The choice of $\alpha$ might determine values of $\lambda$, in which case the first ODE is easy to solve.  You may get a sequence of solutions $u_{\lambda} = a_{\lambda}b_{\lambda}$ for whatever $\lambda$'s work.  Then some linear combination of the $u_\lambda$ could be forced to make $u(0,x) = x\exp({-x^2/2}).$  
