Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
did my answer help? Do you have further questions? –  James S. Cook Sep 6 '13 at 1:16
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.