关于常微分方程及“常”字字面含义的技术咨询

Great question—it’s easy to gloss over naming conventions, but they actually carry useful context! Let’s break this down since you already understand the mathematical definition but want to dig into what "ordinary" really means here.

First, let’s clarify ordinary derivative: The word "ordinary" doesn’t refer to "basic" or "common" (though these derivatives are indeed ubiquitous). Instead, it’s used in direct contrast to "partial" derivatives, and its literal meaning centers on being unrestricted to a single part of a multi-variable system.

Think of it this way: If you have a function that depends on only one independent variable—say, y(t) where t represents time—then dy/dt is an ordinary derivative. There’s no ambiguity here: you’re measuring the full rate of change of y with respect to its only input. You never have to specify "which variable" you’re differentiating against, because there’s just one option.

Now, extend that to Ordinary Differential Equations (ODEs): These are equations that include ordinary derivatives. Every function in an ODE relies on exactly one independent variable, so all derivatives in the equation are these "unqualified" full derivatives. For example, the equation y'' + 3y' + 2y = e^(-t) is an ODE—y only depends on t, and every derivative is taken with respect to t.

To make the contrast crystal clear, compare this to Partial Differential Equations (PDEs). In a PDE, the function depends on multiple independent variables (like u(x,t), which depends on both position x and time t). Here, you’ll see partial derivatives like ∂u/∂t or ∂²u/∂x²—these are "partial" because you’re only measuring the rate of change with respect to one part of the input (holding all other variables constant).

So, to wrap it up: "Ordinary" in this context means relating to a single independent variable, with derivatives that measure the full, unpartitioned rate of change of the function—a direct counterpoint to the "partial" derivatives used in equations with multiple variables.

备注：内容来源于stack exchange，提问作者Θάνος Κ.