Solving a heat equation with time dependent boundary conditions Problem statement
Solve the following PDE for $ u(x,t) (0<x<\ell, t>0)$ $$ u_t=ku_{xx} +bu_x+cu $$ 
With initial and boundary conditions as follows, $$u(x,0)=f(x), u(0,t)=g_0(t), u(\ell,t)=g_1(t)$$.
My attempt
Taking a Fourier transform,
        \begin{align*}
   \mathcal{F}(u_t)&=\mathcal{F}(ku_{xx}+b_u{x}+cu)=k\mathcal F(u_{xx})+b\mathcal F(u_x)+c\mathcal F(u)\\
   \frac{d}{dt}\hat{u}(\omega,t )&=-k\omega^2 \hat{u}(\omega, t)+ib\omega \hat{u}(\omega,t)+\hat{u}(\omega,t)
\end{align*}
This is an ODE with the following general solution
\begin{align*}
   \hat{u}(\omega, t)&=a(\omega)e^{(-k\omega^2+ib\omega+1)t}
\end{align*}
            We know $ \hat{u}(\omega, 0)=a(\omega)=\mathcal F(f(x))=\hat{f}(\omega)$. Now we use this and take an inverse Fourier transform.
\begin{align*}
   \hat{u}(\omega,t)&=\hat{f}(\omega)e^{(-k\omega^2+ib\omega+1)t}\\
   u(x,t)&=\mathcal{F}^{-1}\left[\hat{f}(\omega)e^{(-k\omega^2+ib\omega+1)t} \right](x,t)\\
   &=\mathcal{F}^{-1}\left[\hat{f}(\omega)e^{(-k\omega^2+1)t}\right](x+bt,t)\\
   &=\frac{1}{2 \pi} \left(f(x) * \sqrt{\frac{\pi }{k t}}e^{1-\frac{x^2}{4kt}} \right)(x+bt,t)\\
   u(x,t)&=\frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^\infty f(y)e^{1-\frac{(x+bt-y)^2}{4kt}}\, dy
  \end{align*}
Question
I am unsure how to proceed from the last step above. Mainly, I don't know how to implement the boundary conditions.
 A: Unfortunately I don't have much time so the answer is short. By the linearity of the heat equation you can set $u(x,t)=v(x,t)+w(x,t)$. Then consider the following problems:
\begin{equation}
v_t=kv_{xx} +bv_x+cv, \qquad 0<x<\ell, t>0, \\
\qquad v(0,t)=g_{0}(t), v(x,0)=f(x)
\end{equation}
and
\begin{equation}
w_t=kw_{xx} +bw_x+cw, \qquad 0<x<\ell, t>0, \\
\qquad w(l,t)=g_{1}(t), w(x,0)=0
\end{equation}
Once defined the Gauss's kernel
\begin{equation}
A(x,t)=\frac{1}{\sqrt{4\pi kt}} e^{1-\frac{(x+bt)^2}{4kt}}
\end{equation}
You looked for the solution of the heat equation with initial temperature distribution $f(x)$
\begin{equation}
u(x,t)=\int_{-\infty}^\infty A(x-y)f(y) dy 
\end{equation}
It should be possible to prove that the solution to the equation with variable $v(x,t)$ is given by
\begin{equation}
v(x,t)=\int_{0}^\infty F(x,y,t)f(y) dy 
\end{equation}
with $F(x,y,t)=A(x-y,t)-A(x+y,t)$. This can be done by making an odd extension of $f(x)$ to $-\infty<x<0$, thus solving the initial value problem for the extended $f_{odd}(x)$ given by
\begin{equation}
f_{odd}(x)=
\begin{cases}
 f(x) & \quad 0 < x <\ell \\
 -f(-x) & \quad -\ell <x < 0 \\
  0 & \quad  x=0
\end{cases}
\end{equation}
The general method here is called the method of images.
