Linear homogeneous equation $u_x + u_y + u = 0$ Equation: $u_x + u_y + u = 0$
I'm stuck on how to solve this linear homogeneous equation. As I study from the book, it only mentioned the general solution is $u(x,y) = f(ay - bx)$ without the additional $u$ term. How should I start a good approach?
 A: $$u_x+u_y+u=0$$
Use Lagrange characteristic method to solve the first order PDE:
$$\frac {dx}{1}=\frac {dy}{1}=-\frac {du}{u}$$
First differential equation is easy to solve:
$$ {dx}={dy}$$
$$\implies x =  y +c_1$$ $$ \implies c_1=x-y$$
The second DE gives us:
$$\frac {dy}{1}=-\frac {du}{u}$$
$$ \implies y = -\ln u +c $$
$$\implies e^yu=c_2$$
Now the solution opf the PDE is
$$f(c_1)=c_2 \implies e^yu=f \left (x-y\right )$$
$$\boxed {u(x,y)=e^{-y}f \left (x-y \right )}$$
A: Setting $v=e^{x}u$, we have the partial derivatives $v_x=e^{x}(u_x+u)$ and $v_y=e^{x}u_y$, so that
$$
v_x + v_y = e^{x}(u_x+u_y+u) = 0 .
$$
Now the result of the book $v(x,y) = f(y-x)$ can be used. Finally, $$u(x,y) = e^{-x}f(y-x)\, ,$$ where $f$ is a function to be determined by using the boundary conditions.
NB. The coordinates $x$ and $y$ play symmetric roles, and thus, we may write  $u(x,y) = e^{-y}g(x-y)$.
A: Separation of variables provides some traction here, to wit:
Set
$u(x, y) = v(x)w(y); \tag 1$
then
$u_x = v_x w, \tag 2$
and
$u_y = v w_y; \tag 3$
then the given equation
$u_x + u_y + u  = 0 \tag 4$
becomes
$v_x w + v w_y + vw = 0; \tag 5$
in any region of the $xy$-plane where
$v w \ne 0, \tag 6$
we may write this as
$\dfrac{v_x}{v} + \dfrac{w_y}{w} + 1 = 0.  \tag 7$
Since $v$ depends only on $x$ and $w$ only on $y$, it follows that each of the functions $v_x/v$ and $w_y/w$ is constant; this may be more rigorously seen by differentiating (7) with respect to $x$, obtaining
$\left ( \dfrac{v_x}{v} \right )_x = 0, \tag 8$
implying
$\dfrac{v_x}{v} = \lambda, \; \text{a constant}, \tag 9$
with a similar result holding for $w_y/w$; in light of (7) we thus have
$\dfrac{w_y}{w} = -1 - \lambda.  \tag{10}$
(9) and (10) are easily solved for $v$ and $w$, provided we supply initial conditions of the form
$v(x_0) = v_0, \; w(y_0) = w_0 \tag{11}$
at some point $(x_0, y_0)$; of course we must ensure that
$v_0 w_0 = u_0 = u(x_0, y_0) \tag{12}$
if we are given $u$ at this point.  The solutions to (9) and (10) are then
$v(x) = v_0 e^{\lambda(x - x_0)} \tag{13}$
and
$w(y) = w_0 e^{(-1 - \lambda)(y - y_0)}; \tag{14}$
thus
$u(x, y) = v_0 w_0 e^{\lambda(x - x_0) + (-1 - \lambda)(y - y_0)}$
$= u_0  e^{\lambda(x - x_0) + (-1 - \lambda)(y - y_0)}. \tag{15}$
