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Suppose there is an ode: $$ \frac{dx}{dt} = a(t)x $$ Then we can solve it by following steps: $$ \frac{dx}{x} = a(t)dt $$ integrate on both side and we get: $$ \int \frac{dx}{x} = \int a(t)dt + C $$ Why can I integrate differentials, what does it exactly mean by integrate a differential? Is there an elementary explanation?

Florian Huo
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Suppose there is an equation of differentials: $$ F(x)dx = G(t)dt $$ The equation only makes sense when $x$ is exactly a function of $t$, we denote it by $x = x(t)$, then the equation becomes $$ F(x(t))dx(t) = G(t)dt $$ what the equation exactly says is that $$ F(x(t))x'(t)dt = G(t)dt $$ namely $$ F(x(t))x'(t) = G(t) $$ We can integrate on both side since they are functions and there is no ambiguity, and we get $$ \int F(x(t))x'(t)\, dt = \int G(t) \, dt + C $$ And $$ \int F(x(t))x'(t)\, dt = \int F(x) \, dx + C $$ is what we already know. So we get $$ \int F(x) \, dx = \int G(t) \, dt + C $$

Florian Huo
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    I will not say much about the most popular question on MSE why it is that we can do algebra with differentials as if they were ordinary numbers. Just this: do it and check if it solves the ODE. Regarding your way of "proving" it. It is simpler: From $dx/dt=ax$ we get $$\frac{1}{x}\frac{dx}{dt}=a$$ (fully rigorous!). Now integrate both sides w.r.t. $t$ and use the fully rigorous change of variables formula to convert the $dt$ integral into a $dx$ integral. You should also mention that the solution you get is $$\textstyle x(t)=x(0)\exp(\int_0^ta(s),ds),.$$ – Kurt G. Aug 13 '23 at 05:57
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    Even if we arrived here unrigorously we could check now that it solves the ODE. – Kurt G. Aug 13 '23 at 05:57