# dy/dx when x and y are functions

Let $x$ and $y$ be continuously differentiable real-valued functions defined on $\mathbf{R}$ and suppose that $x'(t) = f\,(x(t),y(t))$ and $y'(t) = g(x(t),y(t))$ where $f$ and $g$ are continuously differentiable on $\mathbf{R}^2$. According to the ODE book I am reading, if $f$ is never 0, then $\frac{dy}{dx} = \frac{g(x,y)}{f(x,y)}$ $y(x_0) = y_0$ where $x_0,y_0$ are constants, is an initial value problem.

Now what does $dy/dx$ mean here? Is it just the function $y'(x)$? Or is the author being sloppy and $x$ is actually a dummy variable? If the former is true, how is this an initial value problem? Normally, an initial value problem is given as $y'(t) = h(y(t),t)$, $y(t_0) = y_0$ where $t$ is variable.

PS. Why is it that ODE books are sloppy with notation and unshamefully nonrigorous?

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Possible duplicate: math.stackexchange.com/questions/8040/… –  Hans Lundmark Feb 14 '11 at 6:29

If $x$ and $y$ are functions of $t$, $\frac{\mathrm dy}{\mathrm dx}$ means, in all likelihood, the following: where possible by the inverse function theorem, write $t$ in terms of $x$; then $y=y(t)=y(t(x))$ is a function of $x$: differentiate that.

If you think about it a bit, you'll see that $\frac{\mathrm dy}{\mathrm dx}$ is in fact the ratio of change of $y$ with respect to changes in $x$, as it should be!

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I did not think about the inverse function theorem. In any case, when you write $y(t(x))$, are you treating $x$ as a dummy variable? Your last statement gave me an idea: Perhaps $dy/dx$ is the function whose value at $t$ is just $y'(t)/x'(t)$? –  echoone Feb 14 '11 at 15:29
@echoone, when I write $y(t(x))$ I mean the function resulting from composing $y$ with the inverse function of $t$. As for the last sentence of your comment: it is true that the function $dy/dx$ is equal to $y'(t)/x'(t)$, but not by definition: you actually have to prove that. –  Mariano Suárez-Alvarez Feb 14 '11 at 16:15

The condition that f is never 0 is important .by the continuity of f,f must be always positive or negative ,which means x(t) is monotonous,so it has inverse function t(x),which is also smooth.so y(t)=y(t(x)).

by

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Just wanted to comment ( but I don't have enough points), that , most likely, when you see a reference to "f never being 0" in problems of this sort; other than the obvious not wanting to divide by 0, this is a reference to the implicit/inverse function theorem, which guarantees a(n) (at least) local differentiable inverse. It then often ends up being a case of playing around with the chain rule until you get the right expressions.

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I asked my ODE professor and he said that $dy/dx$ is the function whose value at $t$ gives the slope of the graph of $(x,y)$ at the point $(x(t), y(t))$. This means that $dy/dx$ is the function defined by $\frac{dy}{dx}(t) = \lim_{s \to t} \frac{y(t+s) - y(t)}{x(t+s) - x(t)}$ and this is why $(dy/dx)(t) = y'(t)/x'(t) = g(x(t),y(t))/f(x(t),y(t))$.

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