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For a first order ODE in the form of

$$ \frac{dy}{dx}= f(x)g(y) $$

it can be solved by

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

I feel quite uncomfortable with this method. It seems to me that $\frac{dy}{dx}$ is a function which happens to have a (sensible) notation $\frac{dy}{dx}$. It feels unjustified to be able to multiply part of a notation as if it is a first class mathematical object. But it seems that a lot of methods to solve differential equations rely upon these sort of free form $dx$, $dy$ manipulation.

I searched a bit on this question. It seems that $dx$ and $dy$ can have meaning in the theory of manifolds? I would like to know more about it. In particular, I want some illumination on the connection between manifold theory and ordinary calculus. So I guess I'm looking for something like, $\frac{dy}{dx}$ can actually be reinterpreted in the framework of differential geometry, so that all of the manipulations are justified. But I'm not really sure.

Rui Liu
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  • You may like these: https://math.stackexchange.com/questions/1577421/the-connection-between-differential-forms-and-ode and https://math.stackexchange.com/questions/27425/what-am-i-doing-when-i-separate-the-variables-of-a-differential-equation – cmk Jun 22 '19 at 20:35

1 Answers1

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Let $H$ be an anti-derivative of $1/g$. Then by the chain rule $$ \frac{d}{dx}H(y(x))=\frac{dH}{dy}(y(x))\cdot \frac{dy}{dx}(x)=\frac1{g(y(x))}\cdot f(x)g(y(x))=f(x) $$ Which implies that $H(y(x))=F(x)+C$ where $f$ is an anti-derivative of $f$.

The integral notation is just a shorter way to write this method down, and as usual the Leibniz notation works intuitively well with the chain rule.

Lutz Lehmann
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