How did Einstein integrate $\frac{\partial \tau}{\partial x'}+\frac{v}{c^2-v^2}\frac{\partial \tau}{\partial t}=0$? 
In his paper "On the Electrodynamics of Moving Bodies", Einstein writes the equation
$$\dfrac{\partial \tau}{\partial x'}+\dfrac{v}{c^2-v^2}\dfrac{\partial \tau}{\partial t}=0$$
where
  
  
*
  
*$\tau=\tau(x',y,z,t)$ is a linear function (i.e. $\tau=Ax'+By+Cz+Dt$)
  
*$x'=x-vt$
  
*$\dfrac{\partial \tau}{\partial y}=0$ (i.e. $B=0$)
  
*$\dfrac{\partial \tau}{\partial z}=0$ (i.e. $C=0$)
  
*$c$ is a constant
  
*$x,x',y,z,t,v$ are variables
  
  
  and he derives that
$$\tau=a\left(t-\dfrac{v}{c^2-v^2}x'\right)$$
where $a=a(v)$

Could someone please walk me through step-by-step how he derived this? I am not very familiar with integrals invovling partial derivatives, so I would be appreciative if any answers are quite explicit.
Additionally, if I modified the question to say that $\tau$ was an affine function (i.e. $\tau=Ax'+By+Cz+Dt+E$), would it make any difference to the result (I suspect it wouldn't)?
 A: From the definitions given:
$$\partial_{x'}\tau = A, \quad \partial_t \tau = D$$
Also:
$$\partial_{y}\tau = B = 0, \quad \partial_z \tau = C = 0$$
From the differential equation we get:
$$A + \frac{v}{c^2 - v^2}D = 0 \\
\implies A = - \frac{v}{c^2 - v^2}D$$
Now using the definition given for $\tau$:
$$\tau = - \frac{v}{c^2 - v^2}Dx' + Dt = D\bigg(t - \frac{v}{c^2 - v^2}x'\bigg)$$
So if you define $D=a$ you get the final expression for $\tau$. As you have noticed, requiring $\tau$ to be affine doesn't change the results at all - your $\tau$ would be:
$$ \tau = a\bigg(t - \frac{v}{c^2 - v^2}x'\bigg) + E$$
The reason $a=a(v)$ is because $D$, who is actually $a$ in disguise, does not depend on $x', y, z, t$ but is assumed to depend on $v$. Otherwise, the relation wouldn't be linear. Without further information, $a$ is a function of potentially anything except those four variables. 
A: There is really no need to assume that $\tau$ is linear or affine to derive the general form of $\tau$. Write the equation as 
$$
\frac{\partial\tau}{\partial x'}+k\frac{\partial\tau}{\partial t} = 0 ,
$$
where $k={v}/({c^2-v^2})$. Change coordinates from $(x',t)$ to $(\xi,u)$ by 
$$
\begin{cases}\xi= x'\\
u=t-k x'
\end{cases}
\qquad
\text{or equivalently,}\qquad
\begin{cases}x'=\xi\\
t=u+k\xi.
\end{cases}
$$
This gives
$$
\frac{\partial\tau}{\partial\xi} = \frac{\partial\tau}{\partial x'} \frac{\partial x'}{\partial\xi}+ \frac{\partial\tau}{\partial t}\frac{\partial t}{\partial\xi} = \frac{\partial\tau}{\partial x'}+k\frac{\partial\tau}{\partial t} = 0 ,
$$
meaning that in the new coordinates, $\tau$ is a function of only $u$. Hence
$$
\tau=a(u)=a(t-k x'),
$$
for some function $a$. At this point, you can use the assumption that $\tau$ is linear (or affine) to deduce that $a$ is linear (or affine).
A: Making the change of variables
$$
x'=x- vt\\
t'=\alpha t
$$
we have
$$
(v^2+\alpha(c^2-v^2))\frac{\partial\tau}{\partial x'}+v\frac{\partial \tau}{\partial t'}=0
$$
so choosing 
$$
\alpha = \frac{v^2}{v^2-c^2}
$$
we get
$$
\frac{\partial \tau}{\partial t'} = 0\Rightarrow \tau(x',t') = f(x') = f(x-v t)
$$
