When we write:

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

to me the intuition is that the $dx$s cancel (I am aware this is not a fully mathematically correct way to think about it). This arises every time we solve a separable first order ordinary differential equation.

But in changing variables of the integration, there must be a substitution. If there were limits on the first integral they would need to change for the second. So what substitution are we technically making?

  • $\begingroup$ I am not sure if I get the question correctly but the famous transformation theorem is behind this: Let Ω be an open set and $\phi$ a diffeomorphism on Ω. Then you get the nice formula $$\int_{\phi(Ω)} f(y) \ \text{d}y = \int_Ω f(\phi(x)) \ |\text{det D}\phi(x) | \ \text{d} x$$ In your case it seems $\phi(x)=y$ and therefore $\text{D}\phi(x)=\text{D}_x y$. $\endgroup$
    – Cahn
    Commented Jan 21, 2017 at 15:12
  • $\begingroup$ Is there not a simple substitution we use to change the variable of integration from $x$ to $y$? In the same way that we make basic changes of variables to solve integrals. $\endgroup$ Commented Jan 21, 2017 at 15:29

1 Answer 1


I've been pondering this, and I have two different answers. I can't decide which one is more to your point.

First, if this really were from a separable DE and there was an initial condition $y(x_0) = y_0,$ then I think we'd write:

$$\int_{s=x_0}^{s=x} f(y(s))\frac{dy}{ds} \; ds = \int_{t=y_0}^{t=y} f(t) \; dt.$$

So the substitution is $y = y(x)$, even though we don't know what $y(x)$ is yet.

Second thought: Maybe there isn't a substitution. Suppose $y=h(x).$ If $h'(x) = \frac{dy}{dx}$ then the definition of $dy$ and $dx$ is that they are related by the relation $dy = h'(x)\; dx$. So when we separate a DE and write

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

we exploit this relation ship to write

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

$$f(y) \; h'(x) = g(x)$$

$$f(y) \; h'(x) \; dx = g(x) \; dx$$

$$f(y) \; dy = g(x) \; dx$$

and (indefinite) integrate both sides. By this we've done no substitution, but have only observed that the differentials on each side have the same set of antiderivatives.

I hope that sheds a little light. If not, gather up all your friends and downvote me into oblivion.

  • $\begingroup$ I think the first part is more or less exactly what I was looking for. It makes a lot of sense that $y = y(x)$ is the substitution. However, I do have a question - if we start with the separable DE: $$ \frac {dy}{dx} = \frac {f(x)}{g(y)} $$ with initial conditions $y(x_0)=y_0$ (indeed this is roughly where my question comes from), then how do we rearrange that into the sort of thing you have written above? Sorry - I am aware that my initial question is probably not phrased particularly well. $\endgroup$ Commented Jan 22, 2017 at 15:29
  • $\begingroup$ And, how does the LHS of the equation you wrote transform to the RHS with the substitution $y=y(x)$? $\endgroup$ Commented Jan 22, 2017 at 15:35
  • 1
    $\begingroup$ @bnosnehpets I used $f$ for two different things, which was probably confusing. I'll edit a bit. $\endgroup$
    – B. Goddard
    Commented Jan 22, 2017 at 22:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .