**Part (a):**

$$\begin{equation*}

u(x,t) = U(x - vt, t)\\

Let \xi = x - vt, \eta = t \\

u_x = U_x = \frac{\partial U}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial x} = U_x\\

u_{xx} = U_{\xi\xi}\\

u_t = U_t = \frac{\partial U}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial t} = -vU_{\xi} + U_{\eta}\\

-vU_{\xi} + U_{\eta} + vU_{\xi} = kU_{\xi\xi} \\

U_{\eta} = kU_{\xi\xi}\\

U(\xi,\eta) = \int_{-\infty}^{\infty}G(\xi,y,\eta)g(y)dy\\

u(x,t) = \int_{-\infty}^{\infty}G(x - vt,y,t)g(y)dy \\

= \int_{-\infty}^{\infty} \frac{1}{2 \sqrt{k \pi t}} e^{\frac{-(x-vt-y)^2}{4kt}}g(y)dy

\end{equation*}$$

**Part (b):**

Let $$

\begin{equation*}

u(x,t) = e^{\alpha t + \beta x}V(x,t)\\

u_t = \alpha e^{\alpha t + \beta x}V + e^{\alpha t + \beta x}V_t\\

u_x = \beta e^{\alpha t + \beta x} + e^{\alpha t + \beta x}V_t\\

u_{xx} = \beta^2 e^{\alpha t + \beta x}V + 2\beta e^{\alpha t + \beta x}V_x + e^{\alpha t + \beta x}V_{xx}\\

\end{equation*}$$

Now plug these into the given heat equation.

$$

\begin{equation*}

\alpha V + V_t + v\beta V + vV_x = k\beta^2 V + 2k\beta V_x + kV_{xx} \end{equation*}$$

set $ \alpha + v\beta = k\beta^2$ and $ v = 2k\beta$ then,

$$\begin{equation*}

\beta = \frac{v}{2k}, \alpha = \frac{-v^2}{4k} \\

u(x,t) = e^{\alpha t + \beta x} \int_{0}^{\infty} G_D(x,y,t)g(y)dy\end{equation*} \\$$

This works because if $u(0,t) = 0,$ then $ V(0,t) = 0.$