Form of Reaction-Advection-Diffusion Equation My textbook (Essential Partial Differential Equations by Griffiths, Dold, and Silvester) says the following:

$$u'' = a(x)u' + b(x)u - f(x)$$
This is called a reaction-advection-diffusion equation since the second derivative term represents "diffusion", the term $a(x)u''$ represents "advection" (if $a > 0$ then there is a "wind" blowing from left to right with strength $a$) and the term $b(x)u$ represents "reaction". The term $f$ is often called the "source" term.

I'm guessing that should be $a(x)u'$ instead of $a(x)u''$?
I tried researching this, but every time I found an equation that was described in a similar way to the above one, it was in a wildly different form.
Thanks for the clarification.
 A: I believe that the book should speak of the stationnary version of such an equation.
Indeed, if we consider the following reaction-advection-diffusion equation
$$
\partial_t u + a(x) \partial_x u = \partial_{xx} u - b(x) u + f(x)
$$
and seek for its steady-state solutions ($\partial_t u=0$), we obtain the above differential equation. 
Now, let us consider the convection-diffusion equation from the Wikipedia article:
$$
\partial_t c + \nabla\cdot (\vec v c) = \nabla\cdot(D\nabla c) + R \, .
$$
The use of the words convection and advection is the same in this context, so that the previous equation could be called an advection-diffusion equation. Under the assumption of incompressible flow ($\nabla\cdot \vec v = 0$), the convection term becomes
\begin{aligned}
\nabla\cdot (\vec v c) &= \vec v\cdot\nabla c + c\nabla\cdot \vec v\\
&= \vec v\cdot\nabla c \, .
\end{aligned}
For a constant diffusion coefficient $D=1$, the diffusion term becomes $\nabla\cdot(D\nabla c) = \nabla^2 c$. In the case of one-dimensional flow with velocity $\vec v = v \vec x$, we therefore have
$$
\partial_t c + v \partial_x c = \partial_{xx} c + R \, ,
$$
which corresponds to the previous reaction-advection-diffusion equation, where the reaction coefficient is $b(x) = 0$.
