# How $\frac{dx}{dy}=f(x)g(y) \Leftrightarrow \int \frac{dx}{f(x)} = \int g(y)dy$?

In my intro differential equations class we have often used the "equivalence" stated in title. It seems to me that somehow, the intermediate step $$\frac{dx}{f(x)} = g(y)dy$$ is being used, in which case $dx$ and $dy$ are being used as numbers from a fraction! The equivalence is intuitively clear to me, but I would like to know what the justification for this informal process is. How would one give rigor to this equivalence?

To be clear, I am looking for a formal treatment as well as some motivation as to why we can work with $dx$'s in this manner.

• I don't truly understand hyperreal numbers and how to manipulate them formally, but I think this might be a start Wiki- hyperreal numbers Apr 9, 2013 at 2:21
• @KevinDriscoll : very interesting, I will look into that... Apr 9, 2013 at 2:29
• I think try interpret this into $\delta-\epsilon$ language will work.
– Easy
Apr 9, 2013 at 2:53
• Hmm. Well, such a quantity has meaning in differential geometry. For the intro class, it's just a technique to solve. By uniqueness (sometimes), it doesn't matter $\textit{how}$ you got the solution. Apr 9, 2013 at 2:58
• @Euler....IS_ALIVE : yes, but I personally would like to know. Apr 9, 2013 at 3:14

Take $$\frac{dy}{dx} = f(y)g(x) \implies \frac{1}{f(y)} \frac{dy}{dx} = g(x)$$ then we integrate over $x$

$$\int \frac{1}{f(y)} \frac{dy}{dx} dx= \int g(x)dx.$$ Since we're thinking of $y$ as a function of $x$, we 'integrate by substitution' or perform '$u$-substitution'.

$$\int \frac{1}{f(y(x))} \frac{dy}{dx} dx= \int\frac{dy}{f(y)}.$$

This is simply the integral equivalent of the chain rule.

• very clear, but its still funny how using the $dy/dx$ as a fraction somehow works out! Apr 9, 2013 at 18:06
• It's one of those amazing things in mathematics that seem so intuitive but hint at a deeper structure. It's the sort of thing that spurred me to learn more. Apr 10, 2013 at 2:20

$\frac{dx}{dy}=f(x)g(y)\Leftrightarrow\Delta x=f(x)g(y)\Delta y+o(\Delta y)\Leftrightarrow\frac{\Delta x}{f(x)}=g(y)\Delta y+\frac{o(\Delta y)}{f(x)}\Leftrightarrow\lim_{\Delta x\rightarrow0}\sum\frac{\Delta x}{f(x)}=\lim_{\Delta y\rightarrow0}\sum(g(y)\Delta y+\frac{o(\Delta y)}{f(x)})=\lim_{\Delta y\rightarrow0}\sum g(y)\Delta y\Leftrightarrow\int\frac{dx}{f(x)}=\int g(y)dx$

Note: $\Delta x=(x+\Delta x)-x$ is the infinitesimal variation at the point $x$; $o(\Delta x)$ is a function which satisfies $\lim_{\Delta x\rightarrow0}\frac{o(\Delta x)}{\Delta x}=0$.

• could you please define $\Delta x$ and $o(\Delta x)$ ? Apr 9, 2013 at 3:13
• @Coffee_Table, check out my definition. Note that the first equivalence comes from the definition of limit, and $\Delta x$ and $\Delta y$ simultaneously goes to $0$ is due to the continuity of the function.
– Easy
Apr 9, 2013 at 3:26
• "$\Delta x=(x+\Delta x)-x$" - the quantity is defined in terms of itself? perhaps you means $\Delta x=(x+h)-x$ where $h$ goes to $0$ ? Apr 9, 2013 at 3:42
• @Coffee_Table, it is ok if you feel comfortable about the second way of definition. In fact they are the same as $\Delta x=h$.
– Easy
Apr 9, 2013 at 6:07

Since one of the tags attached to this question is "infinitesimals", I thought I would elaborate on MrSlunk's comment above: "It's one of those amazing things in mathematics that seem so intuitive but hint at a deeper structure". To clarify the structure involved, note that in the context of a hyperreal extension of the reals one can indeed represent the derivative as the ratio of infinitesimals $dy$ over $dx$. Then the relation $\frac{dy}{dx}=f(x)g(y)$ can literally be rewritten as $\frac{dx}{f(x)}=g(y)dy$, and then integrated to produce the result. This treatment appears in Keisler's "Elementary calculus" on pages 464-465, see http://www.math.wisc.edu/~keisler/calc.html, and retroactively justifies the expression "separation of variables".