📚 Complex Roots of the Characteristic Equation and General Solutions of Differential Equations | 特征方程复数根与微分方程通解
In IB Mathematics Analysis and Approaches HL, solving second-order linear differential equations with constant coefficients is a key skill. When the characteristic equation has complex roots, the general solution takes a beautiful form involving exponential and trigonometric functions. This article explains the derivation, the role of Euler’s formula, and how to apply the method to initial value problems.
在IB数学分析与方法HL中，求解常系数二阶线性微分方程是一项核心技能。当特征方程具有复数根时，通解呈现出指数函数与三角函数结合的优美形式。本文将解释其推导过程、欧拉公式的作用，以及如何将方法应用于初值问题。
1. The Characteristic Equation | 特征方程
Consider a homogeneous second-order linear differential equation with constant coefficients:
考虑一个常系数齐次二阶线性微分方程：
a y” + b y’ + c y = 0
We look for solutions of the form y = erx. Substituting this into the differential equation gives the characteristic equation:
我们寻找形如 y = erx 的解。将其代入微分方程，得到特征方程：
a r2 + b r + c = 0
This quadratic equation in r determines the behaviour of the general solution. The discriminant b2 − 4ac decides whether the roots are real and distinct, real and repeated, or complex conjugates.
这个关于 r 的二次方程决定了通解的行为。判别式 b2 − 4ac 决定了根是相异实根、重根还是共轭复根。
2. Why Complex Roots Appear | 为什么会出现复数根
When the discriminant is negative, b2 − 4ac < 0, the characteristic equation has two complex conjugate roots. They are usually written as r = α ± iβ, where α = −b/(2a) and β = √(4ac − b2)/(2a).
当判别式为负，即 b2 − 4ac < 0 时，特征方程具有一对共轭复根。通常写作 r = α ± iβ，其中 α = −b/(2a)，β = √(4ac − b2)/(2a)。
These roots still correspond to valid solutions y1 = e(α+iβ)x and y2 = e(α−iβ)x. However, these solutions are complex-valued, whereas the original differential equation is real. To obtain real-valued general solutions, we combine them using Euler’s formula.
这些根仍然对应有效的解 y1 = e(α+iβ)x 和 y2 = e(α−iβ)x。然而，这些解是复值函数，而原微分方程是实系数的。为了得到实值通解，我们需要借助欧拉公式将它们组合。
3. Euler’s Formula | 欧拉公式
Euler’s formula states that for any real number θ:
欧拉公式指出，对于任意实数 θ：
eiθ = cos θ + i sin θ
It also gives the conjugate relation e−iθ = cos θ − i sin θ. Adding and subtracting these two identities yields:
同时也给出共轭关系 e−iθ = cos θ − i sin θ。将两式相加和相减可得：
cos θ = (eiθ + e−iθ)/2, sin θ = (eiθ − e−iθ)/(2i)
These identities allow us to replace complex exponentials with real sinusoidal functions, which is the key step in constructing real solutions.
这些恒等式使我们能够用实正弦函数替换复指数函数，这是构造实值解的关键步骤。
4. Deriving the General Solution | 推导通解
Assume the roots are r = α ± iβ. The two independent complex solutions are y = e(α+iβ)x and y = e(α−iβ)x. Because the differential equation is linear, any linear combination is also a solution. Define:
假设根为 r = α ± iβ。两个线性无关的复解为 y = e(α+iβ)x 和 y = e(α−iβ)x。由于微分方程是线性的，任意线性组合仍是解。定义：
u(x) = (y1 + y2)/2 = eαx cos(βx)
v(x) = (y1 − y2)/(2i) = eαx sin(βx)
Using Euler’s formula, both u and v are real-valued and satisfy the differential equation. Moreover, they are linearly independent provided β ≠ 0. Therefore the general solution is a linear combination of u and v.
利用欧拉公式，u 和 v 都是实值且满足微分方程。此外，当 β ≠ 0 时它们线性无关。因此通解是 u 和 v 的线性组合。
5. The Form of the General Solution | 通解的形式
For a second-order homogeneous linear differential equation with characteristic roots α ± iβ, the general solution is:
对于特征根为 α ± iβ 的二阶齐次线性微分方程，通解为：
y = eαx(C1 cos(βx) + C2 sin(βx))
Here C1 and C2 are arbitrary constants determined by initial or boundary conditions. The term eαx controls the amplitude growth or decay, while the sine and cosine terms produce oscillation with angular frequency β.
其中 C1 和 C2 是由初始条件或边界条件确定的任意常数。因子 eαx 控制振幅的增大或衰减，而正弦和余弦项产生角频率为 β 的振荡。
This form is valid for all real x. If α = 0, the solution reduces to pure oscillation C1 cos(βx) + C2 sin(βx), which is common in undamped mechanical and electrical systems.
该形式对所有实数 x 均成立。若 α = 0，解退化为纯振荡 C1 cos(βx) + C2 sin(βx)，这在无阻尼机械和电路系统中很常见。
6. Worked Example 1 | 例题1
Solve the differential equation y” − 4y’ + 13y = 0.
求解微分方程 y” − 4y’ + 13y = 0。
The characteristic equation is r2 − 4r + 13 = 0. Using the quadratic formula:
特征方程为 r2 − 4r + 13 = 0。使用求根公式：
r = [4 ± √(16 − 52)]/2 = [4 ± √(−36)]/2 = 2 ± 3i
Thus α = 2 and β = 3. The general solution is:
因此 α = 2，β = 3。通解为：
y = e2x(C1 cos(3x) + C2 sin(3x))
You can verify this by differentiating twice and substituting back into the original equation.
你可以通过两次求导并代回原方程来验证该解。
7. Worked Example 2 | 例题2
Solve y” + 9y = 0 subject to y(0) = 2 and y'(0) = 6.
求解 y” + 9y = 0，满足初始条件 y(0) = 2，y'(0) = 6。
The characteristic equation is r2 + 9 = 0, so r = ±3i. Here α = 0 and β = 3. The general solution is:
特征方程为 r2 + 9 = 0，所以 r = ±3i。这里 α = 0，β = 3。通解为：
y = C1 cos(3x) + C2 sin(3x)
Apply y(0) = 2: C1 = 2. Differentiate: y’ = −3C1 sin(3x) + 3C2 cos(3x). Then y'(0) = 3C2 = 6, so C2 = 2. The particular solution is y = 2 cos(3x) + 2 sin(3x).
应用 y(0) = 2：C1 = 2。求导：y’ = −3C1 sin(3x) + 3C2 cos(3x)。于是 y'(0) = 3C2 = 6，所以 C2 = 2。特解为 y = 2 cos(3x) + 2 sin(3x)。
8. Initial Value Problems | 初值问题
To find C1 and C2, you need two conditions. Typically these are y(x0) and y'(x0). Substitute x0 into y and y’, then solve the resulting linear system.
要确定 C1 和 C2，需要两个条件。通常是 y(x0) 和 y'(x0)。将 x0 代入 y 和 y’，然后求解得到的线性方程组。
Sometimes the solution can be rewritten as a single shifted sinusoid: y = R eαx cos(βx − φ), where R = √(C1² + C2²) and φ is the phase shift. This form is useful for describing amplitude and phase in physics contexts.
有时解可以重写为单个平移正弦形式：y = R eαx cos(βx − φ)，其中 R = √(C1² + C2²)，φ 为相位差。这种形式在物理背景下描述振幅和相位非常有用。
9. Connection to Oscillations and Damping | 与振荡和阻尼的联系
In mechanics, the equation m x” + c x’ + k x = 0 describes a damped oscillator. Here α = −c/(2m) and β = √(k/m − c²/(4m²)). If the discriminant is negative, the system is underdamped: it oscillates while the amplitude decays exponentially.
在力学中，方程 m x” + c x’ + k x = 0 描述阻尼振荡器。这里 α = −c/(2m)，β = √(k/m − c²/(4m²))。若判别式为负，系统为欠阻尼：它在振荡的同时振幅按指数衰减。
In electrical circuits, the same mathematics models the charge in an RLC circuit. The complex roots indicate the natural response oscillates at resonant frequency with a decaying envelope.
在电路中，同样的数学描述 RLC 电路的电荷变化。复数根表明自然响应以共振频率振荡并伴随衰减包络。
Understanding this connection helps you interpret the constants α and β physically rather than memorising formulas blindly.
理解这一联系有助于从物理意义上解释常数 α 和 β，而不是盲目记忆公式。
10. Common Mistakes and Tips | 常见错误与提示
A common mistake is writing the solution as eαx(C1 eiβx + C2 e−iβx) and stopping there. This is correct but not in real form, which the exam usually expects. Always convert to sine and cosine.
一个常见错误是写出 eαx(C1 eiβx + C2 e−iβx) 后就此止步。这虽然正确，但不是考试通常要求的实值形式。务必转换为正弦和余弦。
Another mistake is using the wrong sign for α or β when reading roots from a characteristic equation. Write the roots clearly as α ± iβ before substituting into the formula.
另一个错误是读取特征根时弄错 α 或 β 的符号。在代入公式前，先将根明确写成 α ± iβ。
Also, remember that the general solution has exactly two arbitrary constants for a second-order equation. If you find yourself with more, you have likely introduced extra parameters inadvertently.
此外，记住二阶方程的通解恰好包含两个任意常数。如果你发现常数多于两个，很可能无意中引入了额外参数。
11. Practice Problems | 练习
Try solving these on your own:
请自行尝试求解以下题目：
- y” + 2y’ + 5y = 0
- y” − 6y’ + 25y = 0, with y(0) = 1 and y'(0) = 0
- y” + 4y = 0, with y(π/4) = 0 and y'(π/4) = 2
For the first equation, the roots are −1 ± 2i, giving y = e−x(C1 cos 2x + C2 sin 2x). Check your answers by differentiation.
对于第一个方程，根为 −1 ± 2i，所以 y = e−x(C1 cos 2x + C2 sin 2x)。通过求导检验你的答案。
12. Conclusion | 结论
Complex roots of the characteristic equation lead to general solutions that combine exponential and trigonometric functions. The key is to extract α and β from the roots, write down y = eαx(C1 cos βx + C2 sin βx), and then use initial conditions to find the constants. Master this process and you will handle any second-order linear homogeneous equation with complex roots confidently in the IB exam.
特征方程的复数根导致通解由指数函数和三角函数结合而成。关键在于从根中提取 α 和 β，写出 y = eαx(C1 cos βx + C2 sin βx)，然后利用初始条件确定常数。掌握这一过程，你就能在IB考试中自信应对任何具有复数根的二阶线性齐次方程。
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