(e) ~~I will work on this part in a little while. So far I think I have gotten all of the same solutions as Rong Wei.~~ Added my solution below.

For the case of the half-line and Dirichlet boundary condition, we will have the solution: \begin{equation}

u(x,t) = \frac{e^{\alpha{}x + \beta{}t}}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/4kt} - e^{-(x+y)^2/4kt}]g(y)dy \end{equation}

In the case of Neumann boundary conditions, we cannot use a similar method.

I don't think the solution is right here, unless we assume $v(x,0) = g(x)$. But usually we use $u(x,0) = g(x)$, then in this case \begin{equation}u(x,0) = v(x,0)e^{\alpha x} = g(x) \rightarrow v(x,0) = g(x)e^{-\alpha x} \end{equation}

Dirichlet condition transforms to:

\begin{equation} u(0,t) = v(0,t)e^{\beta t} = 0 \rightarrow v(0,t) = 0 \end{equation}

Thus we need to solve \begin{equation}v_t = kv_{xx} \end{equation}

\begin{equation} v(x,0) = g(x)e^{-\alpha x} \end{equation}

\begin{equation} v(0,t) = 0 \end{equation}

The answer is then the general result:

\begin{equation}

v(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/4kt} - e^{-(x+y)^2/4kt}]g(y)e^{-\alpha y}dy \end{equation}

Then

\begin{equation}

u(x,t) = \frac{e^{\frac{c}{2k}x -\frac{c^2}{4k}t}}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/4kt} - e^{-(x+y)^2/4kt}]g(y)e^{-\frac{c}{2k} y}dy \end{equation}