Here there are Maxwell equations https://en.wikipedia.org/wiki/Maxwell%27s_equations . Let's take the first in particular, Gauss's law. In its integral form, the left side is equal to the integral over the volume $\Omega$ of $\nabla \cdot E$ (because of Stokes's theorem). So I have a triple integral of $\nabla \cdot E$ on the left side, and a triple integral (over the same volume) of $\frac{\rho}{\epsilon_0}$ on the other side. I can't see what they mean by "differential equations". This equation in particoular becomes $\nabla \cdot E= \frac{\rho}{\epsilon_0}$. I just don't understand what this means. It seems like that if two integrals are equal, then their integrands are also equal, which is obviously false.
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Let's break this down into pieces.
A differential equation just means an equation that uses differentiation. ∇⋅E=ρ/ϵ0 is a differential equation because it involves a gradient.
You're mostly right. Both sides become a triple integral over the same volume (for any volume) which implies that the integrands of both are the same thing. You'll notice that the differential equation is simply stating this.
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$\begingroup$ To make things even more clear. The main point of the answer is the "for any volume" part. $\endgroup$ Commented Oct 4, 2017 at 18:57
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$\begingroup$ ^^ Thanks, I probably should have stressed this more $\endgroup$– GeorgeCommented Oct 4, 2017 at 19:15
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$\begingroup$ So, you are saying that those are equal because the equality holds for every volume, also "infinitesimal" ones. I think I got it now. The thing that puzzled me was that it looked like "integrals are the same, so integrands are the same too" $\endgroup$ Commented Oct 4, 2017 at 19:50
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$\begingroup$ Yes, exactly. If your integral equation works for all volumes, it works for differential volumes and hence the integrand may form a differential equation. $\endgroup$– GeorgeCommented Oct 4, 2017 at 20:52