Solve the 1-D Heat Equation with the given boundary values and initial conditions Solve:
$\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2}+4\frac{\partial u}{\partial x}+2u$ for $0<x<\pi$, $t>0$ with boundary conditions $u(0,t)=u(\pi,t)=0$ for $t>0$ and initial condition $u(x,0)=x(\pi-x)$ for $0<x<\pi$. 
The way I am solving it is by reducing it to one involving the standard heat equation by setting $u(x,t)=(e^{\alpha x+\beta t})U(x,t)$. Then substituting this into the problem and choosing $\alpha$ and $\beta$ to obtain a standard problem for $U(x,t)$ for a bar with ends kept at zero temperature. I have gotten to the point where I differentiated the equation and plugged it back into the initial PDE and took out $e^{\alpha x+\beta t}$. It's after here that I am unsure of what to do. What I dont understand is choosing alpha and beta or the following steps to finish the overall problem. Also, I apologize for the syntax, I don't have Latex or anything on my computer and the derivative notation is supposed to be partial derivatives. 
 A: Note that when I made the derivations, I considered ${{\rm e}^{\alpha\,t+\beta\,x}}$ not $ {{\rm e}^{\alpha\,x+\beta\,t}} $ which does not change the result.  
The heat equation has the form
$$ \frac{\partial u}{\partial t}=C\frac{\partial^2u}{\partial x^2}\,, $$
where $C$ is a constant.
As you mentioned above, you have reached with the problem at the following stage
$$  \left( \alpha-4\,\beta-{\beta}^{2}-2 \right) U \left( x,t \right) +
 \left( -4-2\,\beta \right) {\frac {\partial }{\partial x}}U \left( x,
t \right) -{\frac {\partial ^{2}}{\partial {x}^{2}}}U \left( x,t
 \right) +{\frac {\partial }{\partial t}}U \left( x,t \right) 
 =0$$
So, you can see from the above equation that you have to get rid of the first two terms on the left of the above equation which implies that
$$  \left( \alpha-4\,\beta-{\beta}^{2}-2 \right)=0 \,,$$
$$ \left( -4-2\,\beta \right) = 0 $$
Solving the above system for $\alpha$ and $\beta$, you should have $\alpha=-2$ and $\beta = -2$.
You solve the resulting heat equation in $U(x,t)$ and then your final solution should be
$$ u(x,t) = {{\rm e}^{-2\,(t+x)}}U(x,t) \,.$$
A: Setting $u(x,t)=e^{\alpha x+\beta t}U(x,t)$ we have for $(x,t) \in (0,\pi)\times(0,\infty)$:
$$
\frac{\partial U}{\partial t}(x,t)=\frac{\partial^2U}{\partial x^2}(x,t)+2(\alpha+2)\frac{\partial U}{\partial x}(x,t)+[(\alpha+2)^2-\beta-2]U(x,t).
$$
Choosing
$$
\alpha=\beta=-2,
$$
the problem reduces to the Heat Equation
$$
\begin{cases}
\frac{\partial U}{\partial t}(x,t)=\frac{\partial^2U}{\partial x^2}(x,t) &\text{ for } (x,t) \in (0,\pi)\times(0,\infty),\\
U(0,t)=U(\pi,t)=0 &\text{ for } t \in (0,\infty),\\
U(x,0)=x(\pi-x)e^{2x} &\text{ for } x \in (0,\pi),
\end{cases}
$$
whose solution is given by
$$\tag{1}
U(x,t)=\sum_{n=1}^\infty D_n\sin\left(\frac{n\pi x}{\pi}\right)\exp\left(-\frac{n^2\pi^2t}{\pi^2}\right)=\sum_{n=1}^\infty D_n\sin(nx)e^{-n^2t},
$$
where
$$\tag{2}
D_n
:=\frac{2}{\pi}\int_0^\pi U(x,0)\sin\left(\frac{n\pi x}{\pi}\right)dx
=\frac{2}{\pi}\Im\int_0^\pi x(\pi-x)e^{(2+in)x}dx.
$$
Your solution is given by $u(x,t)=e^{-2(x+t)}U(x,t)$, with $U$ defined by (1) and (2).
