How to solve this PDE? I was wondering how to solve $$a(x-1)\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} =0 ?$$
$a$ is a constant. 
(1) I am still having trouble to understand the method of characteristics recommended in comments. May I have some more explanation? Usually, which type of PDE can be solved by the method of characteristics? 
(2) Can separation of variables apply here? Usually, which type of PDE can be solved by separation of variables?
Thanks and regards!!
 A: The method of characteristics yields the equations
$$
\begin{eqnarray}
\dot x &=& a(x-1)\;,\\
\dot y &=& 1\;,\\
\dot z &=& 0\;,
\end{eqnarray}
$$
with solutions
$$
\begin{eqnarray}
x - 1 &=& c_1\mathrm e^{at}\;,\\
y &=& t + c_2\;,\\
z &=& c_3\;.
\end{eqnarray}
$$
So the function value $z=f(x,y)$ is constant along these curves. We can combine the solutions for $x(t)$ and $y(t)$ into an equation
$$(x-1)\mathrm e^{-ay}=c_4\;.$$
This tells us that $f$ only depends on $x$ and $y$ through $(x-1)\mathrm e^{-at}$, which is Didier's solution in the comments.
Yes, you can also solve this by separation of variables. The ansatz $f(x,y)=X(x)Y(y)$ leads to
$$\frac{X'}{X}a(x-1)=-\frac{Y'}{Y}\;.$$
Setting both sides equal to a constant $c$ gives you two ordinary differential equations, with solutions
$$
\begin{eqnarray}
X(x)&=&c_1(x-1)^{\frac{c}{a}}\;,\\
Y(y)&=&c_2\mathrm e^{-cy}\;.
\end{eqnarray}
$$
That gives you solutions
$$
\begin{eqnarray}
f(x,y)&=&c_3(x-1)^{\frac{c}{a}}\mathrm e^{-cy}\\
&=&c_3\left((x-1)\mathrm e^{-ay}\right)^{\frac{c}{a}}\;.
\end{eqnarray}
$$
Since you can choose $c$ freely, you can combine these into arbitrary functions of $(x-1)\mathrm e^{-ay}$.
A: I would recommend you to read the book Partial Differential Equations: An Introduction by Strauss. Your PDE appears in the very beginning of the book.
