I have what may help as a perspective as I often find myself trying to explain what differential equations is to people that took calculus but stopped after that. You seem to be in that position. We'll stick with the traditional definition (something along the lines of what) Astronuc gave you: a derivative is the instantaneous slope of a function.
Let's generalize this definition a bit, and let just make clear that when i say "n-th order derivative" of y(x) , I'm referring to the 0-th order of y(x) as just y(x), i.e. y^{(n)}(x), such that n=0. Notice that the equation you provided for y^{'}(x) = 2x is just an equation involving one "n-th order derivative" of y(x).
A differential equation involves more than one of these n-th order derivatives. A differential equation can be expressed as a function, we'll call it g, set to 0 (if required by simply moving all terms over to one side). g(x,y^{(n)}(x),y^{(k)}(x),...)=0. It's still not a differential equation until I tell you that n=/=k. and keep in mind it can involve even more derivatives.
A differential equation is a generalized relationship between different orders of derivatives. "Solving the equation" means finding a closed form expression for y(x).
In most calculus problems, you're usually given f(x) or y(x) , or you've found it from the physics of a given problem, and you're asked to find the derivative - you follow rules and patterns you learn from the limit definition. Usually, however (at least in more realistic scenarios), we can't directly come up with an expression for f(x) or y(x), just the relationship between various derivatives of y(x) or f(x) and "solve the differential equation".