📚 Introduction to Differential Equations | 微分方程入门
A differential equation is an equation that connects an unknown function with one or more of its derivatives. At AQA A-Level, the focus is usually on first-order equations, where the highest derivative present is the first derivative dy/dx or dP/dt. Solving a differential equation means finding the original function that satisfies the equation, not just a number.
微分方程是将未知函数与其一个或多个导数联系起来的方程。在 AQA A-Level 考试中，重点通常是一阶微分方程，即出现的最高阶导数是 dy/dx 或 dP/dt。解微分方程意味着找到满足该方程的原始函数，而不仅仅是一个数值。
1. What is a Differential Equation? | 什么是微分方程？
3x² is a first-order differential equation because it contains the derivative dy/dx. The right-hand side can depend on x, on y, or on both.
微分方程包含未知函数及其至少一个导数。例如，dy/dx = 3x² 是一个一阶微分方程，因为它含有导数 dy/dx。等式右边可以依赖于 x、y，或者同时依赖于两者。
A solution is any function y = y(x) that makes the equation true for all x in its domain.
在 AQA 考题中，你经常会看到形如 dy/dx = f(x) 或 dy/dx = f(x)g(y) 的方程。第一种类型可以通过直接积分求解，而第二种类型需要使用分离变量法。解是任何在其定义域内使等式对所有 x 都成立的函数 y = y(x)。
2. Direct Integration: The Simplest Case | 直接积分：最简单的情形
3, the general solution is found by writing y = ∫(2x + 3) dx, which gives y = x² + 3x + C.
如果导数只表示为 x 的函数，我们可以立即对方程两边积分。对于 dy/dx = 2x + 3，通解可以通过 y = ∫(2x + 3) dx 求得，结果为 y = x² + 3x + C。
dy/dx = 2x + 3 ⇒ y = x² + 3x + C
The constant C appears because integration is the reverse of differentiation and any vertical shift of the curve has the same derivative. Therefore, a differential equation by itself usually gives infinitely many solutions, unless extra information is provided.
积分常数 C 的出现是因为积分是微分的逆运算，曲线的任意竖直平移都具有相同的导数。因此，除非给出额外信息，否则微分方程本身通常会给出无穷多个解。
3. Order of a Differential Equation | 微分方程的阶数
² + 5 dy/dx − 6y = 0 is a second-order equation because the highest derivative is d²y/dx².
微分方程的阶数是方程中出现的最高阶导数。例如，dy/dx = 2x + 1 是一阶方程，因为它只包含一阶导数。方程 d²y/dx² + 5 dy/dx − 6y = 0 是二阶方程，因为最高阶导数是 d²y/dx²。
At this stage, AQA mainly requires you to solve first-order equations. You should still be able to identify the order of higher-order equations and recognise that their solutions involve more arbitrary constants.
在此阶段，AQA 主要要求你会求解一阶方程。你仍然应该能够识别高阶方程的阶数，并认识到高阶方程的解包含更多的任意常数。
4. Verifying a Given Solution | 验证给定解
If a question gives a candidate solution, you can check it by differentiating the function and substituting it back into the original differential equation. The left-hand side should match the right-hand side exactly.
如果题目给出了一个可能的解，你可以通过对该函数求导并将其代回原微分方程来检验它。左边应与右边完全相等。
For example, show that y = e²ˣ satisfies dy/dx − 2y = 0. Differentiating gives dy/dx = 2e²ˣ. Substituting into the left-hand side gives 2e²ˣ − 2e²ˣ = 0, so the equation is satisfied.
例如，证明 y = e²ˣ 满足 dy/dx − 2y = 0。求导得 dy/dx = 2e²ˣ。代入左边得到 2e²ˣ − 2e²ˣ = 0，所以方程成立。
