Solving an ODE with parametrized variables My question is related to another I asked some days ago about the method of characteristics. Although I now understand how the method works, I still don't understand how, for example given $x(s)$ and $y(s)$, we can solve 
$$\frac{dy}{-y}=\frac{dx}{x}$$
like that:
$$\int^y_{y(s)}\frac{dy}{-y}=\int^x_{x(s)}\frac{dx}{x}$$
I don't understand why we use the parameter like that while integrating, could somebody show me more detailed steps ?
 A: I looked at your previous post and this is my interpretation:
1.) $(x(s),y(s))$ is a point on the curve with parameter $s$.
2.) $(x,y)$ is just a fixed point on the solution curve
Let me elaborate further on some notational issues that might be vexing.
I liked the previous post where the integration variables were primed:
$$ \int_{y(s)}^{y} \frac{dy'}{-y'} = \int_{x(s)}^{x} \frac{dx'}{x'} $$
However, I would rather use $(x_o,y_o)$ for the fixed point on the solution curve, then we could write:
$$ \int_{y(s)}^{y_o} \frac{dy'}{-y'} = \int_{x(s)}^{x_o} \frac{dx'}{x'} $$
or, if you'd rather,
$$ \int_{y(s)}^{y_o} \frac{dy}{-y} = \int_{x(s)}^{x_o} \frac{dx}{x}. $$
Now, why do the bounds appear as they do? I think it's just the u-substitution theorem. If we change variables we must change bounds. It's really quite natural. In applications, if two physical variables share a common parameter (usually time) then to integrate we simply integrate each over the set of variables which corresponds to a common range of the parameter. For example, $a = \frac{dv}{dt}$ for one-dimensional motion. It is convenient to eliminate time by noting $a = \frac{dx}{dt}\frac{dv}{dx} = v\frac{dv}{dx}$. Consequently,
$$ \int _{x_o}^{x_1} a \, dx = \int _{v_o}^{v_1} vdv $$
where $x(t_o)=x_o$ and $x(t_1) = x_1$ and $y(t_o)=y_o$ and $y(t_1) = y_1$.
In the question you are asking, perhaps the use of $x$ instead of $x_o$ is confusing? It's a matter of style in my opinion.
