How do i solve this simple PDE? $ \partial_x u + \partial_y u =0$ Can you please give me a hint on how to start this differential equation? $u$ is a two variable real function.
$$ \partial_x u + \partial_y u =0$$
Thank you in advance!
 A: One method for solving first order pde equations is as follows. 
\begin{align}
\frac{dx}{1} = \frac{dy}{1} = \frac{du}{0}.
\end{align}
Use the set $dx = dy$ and upon integration yields $x-y= c$, where $c$ is a constant. Now for $du = (0) dx = 0$ yields $u(x,y) = c_{1}$, where $c_{1}$ is a constant. Comparison of the possible characteristic curves, ie the constant curves, then the general solution of the first order pde is
\begin{align}
u(x,y) = F(x-y),
\end{align}
where $F(z)$ is a general function. 
This may also be found in the presentation pde
A: This PDE can be solved by the method of separation of variables, to obtain solutions of a particular form. That is, we assume that there is a solution of the for $u(x, y) = f(x)g(y)$, where $f(x)$ is purely a function of $x$, and $g(y)$, purely a function of $y$. Then we test this by substitution in the given equation.
Let $u(x, y) = f(x)g(y)$.
Then $\partial_x u = f'(x)g(y)$, and $\partial_y f(x)g(y) = f(x)g'(y)$.
Substitution these in the equation $\partial_x u + \partial_y u = 0$:
$
f'(x)g(y) + f(x)g'(y) = 0 \Rightarrow\\
f'(x)g(y) = -f(x)g'(y) \Rightarrow\\
\dfrac{f'(x)}{f(x)} = -\dfrac{g'(y)}{g(y)}
$
But the LHS is purely a function of $x$ and does not contain any terms of $y$, and the RHS is purely a function of $y$, and does not contain any terms of $x$. Therefore, if these are equal, neither contains any terms of $x$ or $y$, so they equal a constant, say $k$.
$\dfrac{f'(x)}{f(x)} = -\dfrac{g'(y)}{g(y)} = k \Rightarrow\\
f'(x) = kf(x), \quad -g'(y) = kg(y) \Rightarrow
$
These are well-known first order linear ODE, and the solutions are:
$f(x) = c_1 e^{kx}, \quad g(y) = c_2 e^{-ky}$
Thus, $u(x, y) = f(x)g(y) = c_1 e^{kx} \times c_2 e^{-kx}$.
$$\boxed{u(x, y) = c\,e^{k(x - y)}}$$
This is a solution for any values of the constants $c$ and $k$, and since the PDE is homogeneous, linear combinations of different solutions are also solutions. For example, when initial conditions are given, an appropriate linear combination is formed based on the functions in the conditions. The more general solution is
$$\boxed{u(x, y) = c_1 e^{k_1 (x - y)} + c_2 e^{k_2 (x - y)} + \cdots}$$
Note: This is not the most general solution possible. For example, $u = x - y$ is an obvious solution that cannot be written in the above form.
