Simple Example of Method of Characteristics

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To find the general solution of an equation such as $u_{x} - yu_{y} = 0$, it is clear by the method of characteristics that the characteristic curves satisfy $dx = -\frac{dy}{y}$ and so we get the relation that $u$ depends only on the constant $k = ye^{x}$ on these curves. It seems to me that it is more useful (for nonhomogeneous problems) to approach this by parametrizing the curve with respect to a new variable. I want to get comfortable with such an approach, but I'm not sure how to proceed. Here is what I tried:

By parametrizing the curve with respect to a new variable $t$, I want to solve this problem through the system of equations:

\begin{cases} \frac{dx}{dt} = 1 \\ \frac{dy}{dt} = -y \\ \frac{dz}{dt} = 0 \end{cases}

giving me a curve $(x(t), y(t), z(t))$ with $x(t) = t + c_{1}$, $y(t) = c_{2}e^{-t}$, and $z(t) = c_{3}$ for some constants. By redefining new constants, I can identify $x$ with $t$ and substitute it into the expression for $y(t)$.

At this point, I'm wondering what I do next to infer that $u$ is constant on these curves and depends only on the constant $k = ye^{x}$?

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=y_0e^{-x}$

$\dfrac{du}{dt}=0$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=f(y_0)=f(ye^x)$