# 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 ?

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did my answer help? Do you have further questions? – James S. Cook Sep 6 '13 at 1:16

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.

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