Why are we allowed to move over "dt" terms when solving separable differential equations?

Question

When solving the differential equation:

$\frac{dx}{dt}=x(t)'=f(x)g(t)$,

we manipulate it so that

$\frac{1}{f(x)}dx=g(t)dt \Rightarrow \int \frac{1}{f(x)}dx= \int g(t)dt$

But from what I understand, it's not strictly allowed to split the $dx$ and $dt$ terms in this manner, since you can't simply multiply the $dt$ term to both sides.

So then my instinct is to look at

$\frac{1}{f(x)}\frac{dx}{dt}=g(t) \Rightarrow \int\frac{1}{f(x)}\frac{dx}{dt}dt=\int g(t)dt$

But even in this case, are we allowed to just cancel out the $dt$ terms on the left hand side? My intuition tells me that $dt$ isn't a number, so you shouldn't be allowed to do this? Can someone explain what's happening here, and if you are allowed to just cancel them... why?

I guess the concept of being able to split up and cancel out $dt$ terms is relevant for the chain rule and U-sub as well, so even in these 2 cases, why are we allowed to cancel/split up these differential terms?

Thank you in advance!

Answer 1

Yes, it's precisely the chain rule that says you can do this. Integration by substitution is justified by the chain rule. If $x=g(t)$, then $$\int_a^b f(g(t))g'(t)\,dt = \int_{g(a)}^{g(b)} f(x)\,dx.$$ Why? Let $F$ be an antiderivative of $f$; i.e., $F'(x) = f(x)$. Then \begin{align*} \int_a^b F'(g(t))g'(t)\,dt &= \int_a^b (F\circ g)'(t)\,dt = (F\circ g)(t)\Big|_a^b \\ &= F(g(b))-F(g(a)). \end{align*} That's precisely what $\displaystyle\int_{g(a)}^{g(b)} f(x)\,dx$ is.