Finding the characteristic ODE from a nonlinear PDE I am studying for a PDE exam on Tuesday, and I am getting pretty confused about one specific type of problem and I am thinking that perhaps I am misinterpreting the correct procedure to follow. The problem is as follows: solve the equation $$y(\frac{\partial{u}}{\partial{x}})+x( \frac{\partial{u}}{\partial{y}}) = 0 $$ $$u(0,y) = e^{-y^2} $$
The way I wanted to do it is as follows:
$$
\frac{\partial{u}}{\partial{x}} + \frac{x}{y} (\frac{\partial{u}}{\partial{y}} )= 0
$$
$$
\frac{\partial{y}}{\partial{x}} + \frac{x}{y} = 0
$$
$$
\frac{\partial{y}}{\partial{x}} = - \frac{x}{y}
$$
$$
y \space \partial{y} = -x \space \partial{x}
$$
Integrating both sides yields
$$
\frac{y^2}{2} + \frac{x^2}{2} = C
$$
However, on the following solution (pictured below), there are two things I don't understand: how can one possibly arrive at $\frac{\partial{y}}{\partial{x}} = \frac{x}{y}$??? I have seen "slightly wrong" steps like this on several problems of this nature, and so I suspect I am simply misunderstanding the step that is being taken. Of course if $\frac{\partial{y}}{\partial{x}} = \frac{x}{y}$ is correct then it is indeed possible to arrive at the equation $x^2 - y^2 = C$ but I am truly at a loss as to how you can just introduce a minus sign to only one term in an equation. Is the author not actually solving the PDE but doing something else? I see that the exercise asks to find the characteristics; is this something related? 
Also, is this PDE nonlinear? I thought that nonlinear meant that there were terms where partial derivatives of a function with respect to one of its independent variables were multiplied by other order derivatives of the same function (e.g. $u(x,y)(\frac{\partial{u(x,y)}}{\partial{y}})$) or when partial derivatives of a function were multiplied by the function itself, or independent variables of the function (e.g. $y\frac{\partial{u(x,y)}}{\partial{y}}$).

 A: Denote the PDE as
$$a(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} = 0, \tag{1}$$
where 
$$a(x, y) = y \text{ and } b(x, y) = x. \tag{2}$$

Is this PDE nonlinear?
A PDE is nonlinear if the solution space of the homogeneous equation (with RHS = 0) is not a vector space. Suppose that $u$ and $v$ are solutions of the PDE (1-2), and that $k \in \mathbb{R}$ is an arbitrary constant. It can be shown that $k u$ and $u + v$ are solutions of the PDE (which is a homogeneous equation). Therefore the solution space of (1-2) is a vector space; therefore (1-2) is a linear PDE.

How can one possibly arrive at $\frac{d y}{d x} = \frac{x}{y}$?
Using the method of characteristics (MoC). I will give a simplified description of the MoC that is applicable to your current problem. As I recall, Strauss discusses the MoC in more generality.
Consider a curve $\Gamma$ given by the parametrization $\Gamma(s) = \begin{bmatrix} x(s), & y(s) \end{bmatrix}$, $s \in I$ ($I$ is an arbitrary interval). Moreover, suppose that $x(s)$ and $y(s)$ satisfy the following system of ODEs:
$$\frac{d x}{d s} = a(x, y) \text{ and } \frac{d y}{d s} = b(x, y). \tag{3}$$.
Then (1) becomes
\begin{align}
0 = a(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} &= \frac{d x}{d s} \frac{\partial u}{\partial x} + \frac{d y}{d s} \frac{\partial u}{\partial y}. \tag{4}
\end{align}
You may recognize the last expression in (4) as the derivative $\frac{d u(x, y)}{d s}$ due to the chain rule:
$$\frac{d u(x, y)}{d s} = \frac{\partial u}{\partial x} \frac{d x}{d s} + \frac{\partial u}{\partial y} \frac{d y}{d s}.$$
Therefore the PDE (1) has been reduced to the following ODE along the curve $\Gamma(s)$:
$$\frac{d u}{d s} = 0. \tag{5}$$
From Eqn (5), we know something about the solution of the PDE (1). Along the curve $\Gamma(s)$ (defined by the system of ODEs (3)), the solution $u$ of the PDE (1) is constant.
The system of ODEs (3) for the particular PDE (1-2) are
$$\frac{d x}{d s} = y \text{ and } \frac{d y}{d s} = x,$$
alternatively
\begin{align}
\frac{\frac{d y}{d s}}{\frac{d x}{d s}} &= \frac{x}{y} \\
\frac{d y}{d s} \frac{d s}{d x} &= \frac{x}{y} \\
\frac{d y}{d x} &= \frac{x}{y}.
\end{align}
By separation of variables we find that $x^2 - y^2 = \text{constant}$.
A: $\newcommand{\+}{^{\dagger}}%
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$\dot{x} = y$ and $\dot{y} = x\quad$ yields $\quad\dd y/\dd x = x/y\quad$ which means $x^{2} - y^{2} = $ constant. Then ${\rm u}\pars{x, y} = {\rm f}\pars{x^{2} - y^{2}}$:
$$
{\rm u}\pars{0,y} = \expo{-y^{2}}\quad\imp\quad
{\rm f}\pars{-y^{2}} = \expo{-y^{2}}.\quad
\mbox{So,}\quad {\rm f}\pars{x} = \expo{x}
$$ 
and
$${\large
{\rm u}\pars{x,y} = \expo{x^{2} - y^{2}}}
$$
Let's checked
$$
y\,\partiald{u}{x} = 2xy\expo{x^{2} - y^{2}}\,,
\qquad
x\,\partiald{u}{y} = -2xy\expo{x^{2} - y^{2}}
$$
