How to solve a convection-diffusion equation analytically? For a 1-d convection equation $u_t + c u_x = 0$ on $\mathbb{R}_+ \times \mathbb{R}$ with a constant $c$, we know the solution is simply $u(x,t)=u_0(x-ct)$ for an initial condition $u_0 := u(x,0)$. However, if we add some diffusion to the RHS of the equation, with a constant $b$, to have
$$
u_t + c u_x = b u_{xx}
$$
then how can we solve this type of convection-diffusion equation analytically?
In my point of view, we may solve the heat equation $u_t = b u_{xx}$ first, by separation of variables, and then add the "transport" features to our solution. It seems requiring some knowledge from Fourier transform, which I am not familiar with, to get the solution of the heat equation, and the form is not fundamental but in a form of convolution. Anyway, I guess we may expect to have a final solution to the convection-diffusion equation in the form like:
$$
u(x,t) = \frac{1}{\sqrt{4 \pi b t}} \int_{-\infty}^{\infty} e^{-\frac{(x-ct-s)^2}{4 b t}} u(s,0) ds
$$
based on Prof. Gilbert Strang's lecture note.
However, I am questioning about the proof in details that the form above is exactly the solution to the convection-diffusion equation given.
 A: If you consider $v(t,x)=u(t,x+ct)$, then $v_t=u_t+cu_x$ and $v_{xx}=u_{xx}$ so that the equation in $u$ is equivalent to the usual heat equation in $v$. Translating the standard solution back should result in the given solution formula.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Write $\ds{\on{u}\pars{x,t}}$ as
$$
\on{u}\pars{x,t} \equiv
\exp\pars{\alpha x - \beta t}\on{U}\pars{x,t}
$$
where $\ds{\alpha}$ and $\ds{\beta}$ are constants.

Then,
\begin{align}
0 & =
-\beta\exp\pars{\alpha x - \beta t}\on{U}\ +
\exp\pars{\alpha x - \beta t}\on{U}_{t}
\\[2mm] &
+ \alpha c\exp\pars{\alpha x - \beta t}\on{U} +
c\exp\pars{\alpha x - \beta t}\on{U}_{x}
\\[2mm] &
-b\alpha^{2}\exp\pars{\alpha x - \beta t}\on{U} -
2b\alpha\exp\pars{\alpha x - \beta t}\on{U}_{x} \\[2mm] &
-b\exp\pars{\alpha x - \beta t}\on{U}_{xx}
\end{align}
Therefore,
\begin{align}
0 & = \pars{\color{red}{-\beta + \alpha c - b\alpha^{2}}}
\on{U} +
\pars{\color{red}{c - 2b\alpha}}\on{U}_{x} +
\on{U}_{t} - b\on{U}_{xx}
\end{align}
Now, I choose $\ds{\color{red}{-\beta + \alpha c - b\alpha^{2}} = 0}$ and
$\ds{\color{red}{c - 2b\alpha} = 0}$
$\ds{\implies \bbx{\alpha = {c \over 2b}}}$ and
$\ds{\bbx{\beta = {c^{2} \over 4b}}}$

Then,
\begin{align}
\on{u}\pars{x,t} & =
\exp\pars{{c \over 2b}\,x -
{c^{2} \over 4b}\,t}\on{U}\pars{x,t}
\\[2mm] & \mbox{where}\ \on{U}\pars{x,t}\
\mbox{satisfies}\
\bbx{\on{U}_{t} = b\on{U}_{xx}} \\ & 
\end{align}
