> But then came the physics class, why we could go from dy/dx = f(x) to dy = f(x).dx and integrate both sides remain a mystery. That is just integrating both s

> But then came the physics class, why we could go from dy/dx = f(x) to dy = f(x).dx and integrate both sides remain a mystery. That is just integrating both sides with respect to x, but you're skipping the step that makes that clear. Start with dy/dx = f(x). Integrate both sides with respect to x: ∫(dy/dx)dx = ∫f(x)dx. Doing the integrals produces: y = F(x) + C, where F(x) is the integral of f(x) and C is an arbitrary constant of integration. Above, the integral on the left doesn't do anything except cancel the derivative. A similar but more substantive method is solving a DE by separation of variables, where the "mystery" is integration by substitution (a.k.a. the chain rule backwards). Start with dy/dx = f(x)g(y). This is a "separable" DE. Rearrange to (1/g(y)) dy/dx = f(x). Integrate both sides with respect to x: ∫(1/g(y))(dy/dx)dx = ∫f(x)dx. We are integrating with respect to x, but where is the x in 1/g(y)? It is hiding inside the y: y = y(x). Define a substitution: Let u = y(x). Then du/dx = dy/dx ⇒ du = (dy/dx)dx, and our equation becomes ∫(1/g(u))du = ∫f(x)dx. Do we really need to introduce another variable? No. We can just use y: ∫(1/g(y))dy = ∫f(x)dx. We often jump straight to this point which makes it pretty confusing, as it looks like we're doing different things to each side of the equation.