📚 Solving First-Order Linear Unforced Equations | 求解一阶线性无外力方程
In IB Mathematics, differential equations model how quantities change. A first-order linear unforced equation is one where the rate of change of a variable y with respect to x is proportional to y itself, possibly modified by a function of x, but with no external driving term. Mastering its solution provides a foundation for tackling forced equations and real-world applications such as population growth, radioactive decay, and cooling processes.
在IB数学中，微分方程用于描述量的变化。一阶线性无外力方程是指变量 y 关于 x 的变化率与 y 本身成正比，可能还乘以一个关于 x 的系数函数，但没有外部驱动项。掌握其求解方法为处理有外力方程以及人口增长、放射性衰变和冷却过程等实际应用打下坚实基础。
1. Introduction to First-Order Linear Equations | 一阶线性方程介绍
A first-order linear differential equation can be written in the standard form:
dy/dx + P(x) y = Q(x), where P and Q are functions of x. The term 'first-order' indicates that only the first derivative dy/dx appears, and 'linear' means that y and its derivative are raised to the first power and not multiplied together. The function Q(x) represents an external forcing term or driving function.
一阶线性微分方程可以写成标准形式：dy/dx + P(x) y = Q(x)，其中 P 和 Q 是 x 的函数。“一阶”表示只出现一阶导数 dy/dx，“线性”意味着 y 及其导数都是一次幂且不相乘。函数 Q(x) 代表外部强迫项或驱动函数。
n P is constant (dy/dx + k y = 0) or variable-coefficient models like dy/dx + (1/x) y = 0.
当 Q(x) = 0 时，方程称为无外力或齐次方程。这是最简单的情形，也是我们关注的重点。例子包括 P 为常数时的指数衰减（dy/dx + k y = 0），或变系数模型如 dy/dx + (1/x) y = 0。
2. The Unforced (Homogeneous) Equation | 无外力（齐次）方程
For unforced equations, the standard form reduces to:
dy/dx + P(x) y = 0.
This equation tells us that the rate of change of y is directly proportional to y itself, with a coefficient –P(x) that may vary with x. Solutions will always involve an exponential function whose exponent is an integral of –P.
对于无外力方程，标准形式简化为：
dy/dx + P(x) y = 0。
这个方程告诉我们 y 的变化率与 y 本身成正比，比例系数为 –P(x)，可能随 x 变化。解总是包含一个指数函数，其指数是 –P 的积分。
The equation is separable because we can rewrite it as dy/dx = – e unforced case an excellent starting point for understanding first-order linear dynamics.
方程是可分离的，因为我们可以将其改写为 dy/dx = –P(x) y。若 y ≠ 0，我们可以两边同除以 y 并乘以 dx，直接分离变量。这种简单性使得无外力情形成为理解一阶线性动力学的绝佳起点。
3. Standard Form and Identification | 标准形式与识别
). Identify P(x) carefully; mistakes often occur when equations are presented as dy/dx = P̃ are rearranged. For example, x dy/dx + 2 y = 0 must be divided by x to reveal P(x) = 2/x.
在求解之前，你必须始终将给定方程改写为标准形式 dy/dx + P(x) y = Q(x)。仔细识别 P(x)；当方程以 dy/dx = P̃(x) y 的形式给出或项需要移项时，容易出错。例如，x dy/dx + 2 y = 0 必须除以 x 才能得到 P(x) = 2/x。
ower, and no products like y·(dy/dx) are present. Confirm Q(x) = 0 for unforced equations.
确认 dy/dx 的系数为 1。若不是，将整个方程除以该系数。写成标准形式后，验证方程确实是线性的：y 和 dy/dx 只能以一次幂出现，且不存在 y·(dy/dx) 等乘积。确认无外力时有 Q(x) = 0。
4. Separation of Variables | 分离变量法
Starting from dy/dx + P(x) y = 0, isolate dy/dx = –P(x) y. If y ≠ 0, divide by y and multiply by dx:
(1/y) dy = –P(x) dx.
We have now separated the variables: all y-dependent terms are on the left, and all x-dependent terms are on the right.
从 dy/dx + P(x) y = 0 出发，分离出 dy/dx = –P(x) y。若 y ≠ 0，两边除以 y 再乘以 dx：
(1/y) dy = –P(x) dx。
现在变量已经分离：所有含 y 的项在左，所有含 x 的项在右。
ion to the unforced equation; it will later be included in the general solution by allowing the arbitrary constant to be zero. In IB exams, it is safe to proceed assuming y ≠ 0 and then verify the constant case.
这一步假设 y ≠ 0。常数解 y = 0 总是无外力方程的一个平凡解；稍后通过允许任意常数为零可将它包含在通解中。在IB考试中，可以安全地假设 y ≠ 0 继续求解，然后验证常数情况。
5. General Solution Derivation | 通解的推导
Integrate both sides of the separated equation:
∫ (1/y) dy = –∫ P(x) dx + C, where C is the constant of integration. This yields:
ln|y| = –∫ P(x) dx + C.
The absolute value is necessary because the domain of the logarithm is positive real numbers.
对分离后的方程两边积分：
∫ (1/y) dy = –∫ P(x) dx + C，其中 C 是积分常数。得到：
ln|y| = –∫ P(x) dx + C。
绝对值是必要的，因为对数函数的定义域是正实数。
Exponentiating both sides gives |y| = eC · e–∫ P(x) dx. Removing the absolute value introduces a ± sign: y = ± eC e–∫ P(x) dx. We absorb the constant ± eC into a new arbitrary constant A (which can be any real number except possibly zero, but including zero gives the trivial solution). Thus the general solution is:
y = A e–∫ P(x) dx, where A ∈ ℝ.
两边同时取指数，得 |y| = eC · e–∫ P(x) dx。去掉绝对值会引入正负号：y = ± eC e–∫ P(x) dx。我们将常数 ± eC 吸收为一个新的任意常数 A（A 可以是除零外的任何实数，但包含零即得平凡解）。因此通解为：
y = A e–∫ P(x) dx，其中 A ∈ ℝ。
6. Handling the Absolute Value in Integration | 积分中绝对值的处理
Many students wonder why we can drop the absolute value and simply write A e–∫ P dx. The reason is that the constant A = ± eC bsolute value. When A = 0, we recover the trivial solution y = 0. Thus the formula y = A e–∫ P dx accounts for all possible real solutions without explicitly writing |y|.
许多学生会疑问为什么可以去掉绝对值而直接写成 A e–∫ P dx。原因在于常数 A = ± eC 可正可负，涵盖了绝对值的两种情况。当 A = 0 时，我们得到平凡解 y = 0。因此公式 y = A e–∫ P dx 已经包含了所有可能的实数解，无需显式写出 |y|。
In rigorous terms, the step from ln|y| erpret A appropriately. IB examiners expect you to present the general solution as y = A e–∫ P dx without absolute value signs.
严格来说，从 ln|y| 到 y = ... 的步骤假设 y 在某个区间上连续且非零。最终含 A 的表达式在 y 可能为零的点处同样有效，只要我们恰当地解释 A。IB考官期望你将通解表示为 y = A e–∫ P dx，而不带绝对值符号。
7. The Role of the Integrating Factor | 积分因子的作用
proach for all first-order linear equations. The integrating factor is defined as μ(x) = e∫ P(x) dx. Multiply the standard form dy/dx + P y = 0 by μ:
μ dy/dx + μ P y = 0.
尽管分离变量法对于无外力方程已足够，但积分因子法统一了所有一阶线性方程的解法。积分因子定义为 μ(x) = e∫ P(x) dx。将标准形式 dy/dx + P y = 0 乘以 μ：
μ dy/dx + μ P y = 0。
Thus we have d(μ y)/dx = 0, which integrates to μ y = C. Solving for y gives y = C/μ = C e–∫ P dx, identical to the earlier result. This method is elegant and avoids separating variables, making it powerful when Q(x) ≠ 0.
左侧恰好是乘积 μ y 的导数，因为 d(μ y)/dx = μ dy/dx + (dμ/dx) y，且 dμ/dx = μ P。因此我们有 d(μ y)/dx = 0，积分得 μ y = C。解出 y 得到 y = C/μ = C e–∫ P dx，与之前的结果完全相同。该方法十分优雅，且无需分离变量，在处理 Q(x) ≠ 0 时威力巨大。
8. Worked Example 1: Constant Coefficient | 例题1：常系数
Problem: Find the general solution of dy/dx + 3y = 0.
问题：求 dy/dx + 3y = 0 的通解。
Solution: : The equation is already in standard form with P(x) = 3 and Q(x) = 0. Using the formula y = A e–∫ 3 dx, we compute ∫ 3 dx = 3x, so the general solution is y = A e–3x xponential decay; all solution curves approach zero as x → ∞ and grow without bound as x → –∞ for A ≠ 0.
解：方程已是标准形式，P(x) = 3，Q(x) = 0。使用公式 y = A e–∫ 3 dx，计算 ∫ 3 dx = 3x，因此通解为 y = A e–3x。这表示指数衰减；所有解曲线当 x → ∞ 时趋近于零，当 A ≠ 0 且 x → –∞ 时无限增长。
Alternatively, by separation: dy/dx = –3y ⇒ (1/y) dy = –3 dx ⇒ ln|y| = –3x + C ⇒ y = A e–3x ondition is provided, e.g., y(0) = 2 gives A = 2, yielding the particular solution y = 2 e–3x.
或者用分离变量法：dy/dx = –3y ⇒ (1/y) dy = –3 dx ⇒ ln|y| = –3x + C ⇒ y = A e–3x。若给出初始条件，如 y(0) = 2，可得 A = 2，得到特解 y = 2 e–3x。
9. Worked Example 2: Variable Coefficient | 例题2：变系数
Problem: Solve x dy/dx = 2y, for x > 0.
问题：求解 x dy/dx = 2y，其中 x > 0。
Solution: First rewrite in standard form: divide by x to get dy/dx – (2/x) y = 0. Here P(x) = –2/x. Compute ∫ P(x) dx = ∫ –2/x dx = –2 ln|x| = –2 ln x (since x > 0). Plug into the general solution formula: y = A e–∫ P dx = A e–(–2 ln x) = A e2 ln x = A x2. Thus y = A x2 is the general solution. This is a family of parabolas through the origin, demonstrating that unforced first-order linear equations can yield non-exponential solutions when P(x) is non-constant.
解：首先改写为标准形式：除以 x 得 dy/dx – (2/x) y = 0。这里 P(x) = –2/x。计算 ∫ P(x) dx = ∫ –2/x dx = –2 ln|x| = –2 ln x（因为 x > 0）。代入通解公式：y = A e–∫ P dx = A e–(–2 ln x) = A e2 ln x = A x2。因此通解为 y = A x2。这是通过原点的抛物线族，说明当 P(x) 不是常数时，无外力一阶线性方程亦可得到非指数函数解。
Check by differentiation: y = A x2 ⇒ dy/dx = 2A x. Then dy/dx – (2/x) y = 2A x – (2/x)(A x2) = 2A x – 2A x = 0, confirming the solution.
验证：y = A x2 ⇒ dy/dx = 2A x。则 dy/dx – (2/x) y = 2A x – (2/x)(A x2) = 2A x – 2A x = 0，解成立。
10. Applying Initial Conditions | 应用初始条件
ne particular solution from the infinite family. Substitute x = x₀ and y = y₀ into y = A e–∫ P dx ntegral, evaluate the antiderivative at x₀, often using a definite integral form: y = y₀ e–∫x₀x P(t) dt.
形如 y(x₀) = y₀ 的初始条件可以从无穷解族中选出一个特解。将 x = x₀ 和 y = y₀ 代入 y = A e–∫ P dx，解出 A。对于通解积分，通常在 x₀ 处计算不定积分，也可使用定积分形式：y = y₀ e–∫x₀x P(t) dt。
For example, given dy/dx + 2x y = 0 and y(1) = 3. Here P(x) = 2x, ∫ P dx = x2, so y = A e–x2. Plug in x=1, y=3: 3 = A e–1 ⇒ A = 3 e. The particular solution is y = 3 e1 – x2. This form highlights how the initial condition fixes the amplitude of the exponential.
例如，给定 dy/dx + 2x y = 0 且 y(1) = 3。这里 P(x) = 2x，∫ P dx = x2，因此 y = A e–x2。代入 x=1, y=3：3 = A e–1 ⇒ A = 3 e。特解为 y = 3 e1 – x2。该形式凸显了初始条件如何确定指数的幅度。
11. Graphical Interpretation and Solution Curves | 几何解释与解曲线
The general solution y = A e–∫ P dx represents a family of curves. If P(x) > 0 on an interval, the exponent – ay or grow depending on the sign of A. When P(x) is positive and constant, we see classic exponential decay (for A>0) or growth (if the equation had a plus sign, but here decay). The curves never cross except possibly at points where the exponential factor becomes singular, though for smooth P they form a non-intersecting foliation.
通解 y = A e–∫ P dx 代表一族曲线。若在区间上 P(x) > 0，指数 –∫P dx 递减，因此解根据 A 的符号衰减或增长。当 P(x) 为正常数时，我们见到经典的指数衰减（A>0 时）或增长（若方程符号不同，但此处为衰减）。曲线永不相交，除非指数因子出现奇异，但对于光滑的 P，它们构成互不相交的叶状结构。
In the variable-coefficient example y = A x2 ss. These curves can intersect at x=0, where the differential equation is singular (P(x) = –2/x is undefined), illustrating a breakdown of the uniqueness theorem.
在变系数例子 y = A x2 中，解曲线为全部通过原点的抛物线。对 x>0，不同的 A 值控制陡峭程度。这些曲线在 x=0 处可以相交，此处微分方程具有奇性（P(x) = –2
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