Case Study in Action: Advanced Problem-Solving for CCEA Year 13 Further Mathematics | 案例分析实战：CCEA Year 13 进阶数学综合解题演练

📚 Case Study in Action: Advanced Problem-Solving for CCEA Year 13 Further Mathematics | 案例分析实战：CCEA Year 13 进阶数学综合解题演练

In Year 13 CCEA Further Mathematics, students face challenging problems that integrate multiple topics from the pure and applied syllabus. This case study will guide you through a realistic, multi-step modelling problem involving differential equations, complex numbers, and graphical analysis – typical of high-mark exam questions. By working through this example, you’ll sharpen your problem-solving skills and see how seemingly separate topics link together.

在 CCEA Year 13 进阶数学中，学生会遇到融合纯数学和应用数学多个主题的综合性难题。本案例分析将带你逐步完成一个真实的多步骤建模问题，涉及微分方程、复数、图形分析等——这些都是高分值考题的典型特征。通过演练这个例子，你能锤炼解题能力，并理解看似独立的知识点之间的联系。

1. The Scenario: Modelling an RLC Circuit | 场景：RLC 电路建模

Consider a series RLC circuit connected to an alternating voltage source V(t) = V₀ cos(ωt). Using Kirchhoff’s voltage law, the differential equation for the charge q(t) on the capacitor is L d²q/dt² + R dq/dt + (1/C) q = V₀ cos(ωt). We are given L = 0.5 H, R = 4 Ω, C = 0.02 F, V₀ = 10 V, and ω = 5 rad/s. The initial conditions are q(0) = 0 and q'(0) = 0. Our goal is to find q(t) and analyse its behaviour.

考虑一个串联 RLC 电路，连接到交流电压源 V(t) = V₀ cos(ωt)。利用基尔霍夫电压定律，电容器电荷 q(t) 的微分方程为 L d²q/dt² + R dq/dt + (1/C) q = V₀ cos(ωt)。给定参数 L = 0.5 H，R = 4 Ω，C = 0.02 F，V₀ = 10 V，ω = 5 rad/s。初始条件为 q(0) = 0 且 q'(0) = 0。我们的目标是求出 q(t) 并分析其行为。

2. Formulating the Differential Equation | 构造微分方程

Substituting the numerical values gives 0.5 d²q/dt² + 4 dq/dt + (1/0.02) q = 10 cos(5t), which simplifies to 0.5 d²q/dt² + 4 dq/dt + 50 q = 10 cos(5t). Multiply through by 2 to obtain a standard form: d²q/dt² + 8 dq/dt + 100 q = 20 cos(5t). This is a second-order linear non-homogeneous ordinary differential equation.

代入数值得到 0.5 d²q/dt² + 4 dq/dt + (1/0.02) q = 10 cos(5t)，化简为 0.5 d²q/dt² + 4 dq/dt + 50 q = 10 cos(5t)。两边乘以 2 得到标准形式：d²q/dt² + 8 dq/dt + 100 q = 20 cos(5t)。这是一个二阶线性非齐次常微分方程。

3. Solving the Homogeneous Equation | 求解齐次方程

The associated homogeneous equation is d²q/dt² + 8 dq/dt + 100 q = 0. Its characteristic equation is r² + 8r + 100 = 0. The discriminant Δ = 8² – 4·100 = 64 – 400 = –336. Hence the roots are complex: r = –4 ± i√84 = –4 ± 2i√21. The homogeneous solution is therefore q_h(t) = e⁻⁴ᵗ (A cos(2√21 t) + B sin(2√21 t)), where A and B are arbitrary constants.

对应的齐次方程为 d²q/dt² + 8 dq/dt + 100 q = 0。其特征方程为 r² + 8r + 100 = 0。判别式 Δ = 8² – 4·100 = 64 – 400 = –336。因此特征根为复数：r = –4 ± i√84 = –4 ± 2i√21。于是齐次解为 q_h(t) = e⁻⁴ᵗ (A cos(2√21 t) + B sin(2√21 t))，其中 A 和 B 为任意常数。

4. Particular Solution Using Complex Impedance | 利用复数阻抗求特解

Since the forcing term is 20 cos(5t), we use a complex exponential approach. Seek a particular solution of the form q_p(t) = Re{ Q eⁱ⁵ᵗ }. Substituting into the equation with the forcing term replaced by 20 eⁱ⁵ᵗ yields ( (i5)² + 8(i5) + 100 ) Q eⁱ⁵ᵗ = 20 eⁱ⁵ᵗ. Simplify: (–25 + 40i + 100) Q = 20 → (75 + 40i) Q = 20. Thus Q = 20 / (75 + 40i). Rationalise: Q = 20(75 – 40i) / (75² + 40²) = (1500 – 800i) / 7225 = (60 – 32i)/289 after simplifying by 25. The modulus |Q| = √(60² + 32²)/289 = √(3600 + 1024)/289 = √4624/289 = 68/289 = 4/17. The argument φ = arctan(32/60) = arctan(8/15). Therefore the particular solution is q_p(t) = (4/17) cos(5t – φ), where φ = arctan(8/15) ≈ 0.490 rad (or 28.1°).

由于驱动项为 20 cos(5t)，我们采用复指数方法。寻找形式为 q_p(t) = Re{ Q eⁱ⁵ᵗ } 的特解。将驱动项替换为 20 eⁱ⁵ᵗ 代入方程，得到 ((i5)² + 8(i5) + 100) Q eⁱ⁵ᵗ = 20 eⁱ⁵ᵗ。化简：( –25 + 40i + 100 ) Q = 20 → (75 + 40i) Q = 20。因此 Q = 20 / (75 + 40i)。分母有理化：Q = 20(75 – 40i) / (75² + 40²) = (1500 – 800i) / 7225 = (60 – 32i)/289（约分 25 后）。模长 |Q| = √(60² + 32²)/289 = √(3600+1024)/289 = √4624/289 = 68/289 = 4/17。辐角 φ = arctan(32/60) = arctan(8/15)。因此特解为 q_p(t) = (4/17) cos(5t – φ)，其中 φ = arctan(8/15) ≈ 0.490 rad（或 28.1°）。

5. General Solution and Applying Initial Conditions | 通解与施加初始条件

The general solution is q(t) = q_h(t) + q_p(t) = e⁻⁴ᵗ (A cos(α t) + B sin(α t)) + (4/17) cos(5t – φ), with α = 2√21. Using q(0) = 0 gives A + (4/17) cos φ = 0. Since cos φ = 15/17 (from the triangle with tan φ = 8/15), we get A = –(4/17)·(15/17) = –60/289. Next, differentiate q(t): q'(t) = e⁻⁴ᵗ [(–4A + Bα) cos(α t) + (–4B – Aα) sin(α t)] – (20/17) sin(5t – φ). At t = 0, q'(0) = –4A + Bα – (20/17) sin(–φ) = 0. Using sin φ = 8/17 and –sin(–φ) = sin φ, we solve for B. After substitution, B = (4A – (20/17) sin φ)/α = ( –240/289 – (20/17)·(8/17) ) / (2√21) = ( –240/289 – 160/289 ) / (2√21) = ( –400/289 ) / (2√21) = –200/(289√21). Thus the full solution is q(t) = e⁻⁴ᵗ [ (–60/289) cos(α t) – (200/(289√21)) sin(α t) ] + (4/17) cos(5t – φ).

通解为 q(t) = q_h(t) + q_p(t) = e⁻⁴ᵗ (A cos(α t) + B sin(α t)) + (4/17) cos(5t – φ)，其中 α = 2√21。利用 q(0) = 0 得 A + (4/17) cos φ = 0。由于 cos φ = 15/17（由 tan φ = 8/15 的三角形可得），求得 A = –(4/17)·(15/17) = –60/289。接着对 q(t) 求导：q'(t) = e⁻⁴ᵗ [(–4A + Bα) cos(α t) + (–4B – Aα) sin(α t)] – (20/17) sin(5t – φ)。在 t = 0 处，q'(0) = –4A + Bα – (20/17) sin(–φ) = 0。利用 sin φ = 8/17 及 –sin(–φ) = sin φ，解出 B。代入后得 B = (4A – (20/17) sin φ)/α = ( –240/289 – (20/17)·(8/17) ) / (2√21) = ( –240/289 – 160/289 ) / (2√21) = ( –400/289 ) / (2√21) = –200/(289√21)。于是最终解为 q(t) = e⁻⁴ᵗ [ (–60/289) cos(α t) – (200/(289√21)) sin(α t) ] + (4/17) cos(5t – φ)。

6. Analysing Transient and Steady-State Behaviour | 分析暂态与稳态行为

The term e⁻⁴ᵗ multiplied by the trigonometric functions represents the transient response. Its amplitude decays exponentially with a time constant of 1/4 = 0.25 s. After approximately 4τ = 1 s, the transient becomes negligible. The steady-state solution is q_ss(t) = (4/17) cos(5t – φ), with amplitude ≈ 0.235 C and a phase lag of about 28.1°. In an exam, you would be expected to sketch the charge as a function of time, showing an initial irregular oscillation that quickly settles into a pure cosine wave at the driving frequency.

e⁻⁴ᵗ 乘以三角函数的部分表示暂态响应。其振幅以时间常数 τ = 1/4 = 0.25 s 指数衰减。大约经过 4τ = 1 s 后，暂态项可忽略不计。稳态解为 q_ss(t) = (4/17) cos(5t – φ)，振幅约为 0.235 C，相位滞后约 28.1°。在考试中，你需要画出电荷随时间变化的草图，展示初始不规则振荡迅速过渡到以驱动频率振动的纯余弦波。

7. Resonance and Frequency Response | 共振与频率响应

Now examine how the steady-state amplitude depends on the driving frequency ω. Returning to the complex impedance, the amplitude is A(ω) = 20 / √( (100 – ω²)² + 64 ω² ). To find the resonant frequency ω_r that maximises A(ω), minimise the denominator squared D(ω) = (100 – ω²)² + 64 ω². Setting D'(ω) = 0 gives 2(100 – ω²)(–2ω) + 128ω = 0 → –4ω(100 – ω²) + 128ω = 0 → 4ω(ω² – 100 + 32) = 0, hence ω_r = √68 ≈ 8.25 rad/s. The peak amplitude is A(ω_r) = 20 / √( (100 – 68)² + 64·68 ) = 20 / √(32² + 4352) = 20 / √(1024 + 4352) = 20 / √5376 ≈ 20/73.32 ≈ 0.273 C. This is larger than the amplitude at ω = 5 rad/s (0.235 C), illustrating resonance.

现在考察稳态振幅如何随驱动频率 ω 变化。回到复数阻抗，振幅为 A(ω) = 20 / √( (100 – ω²)² + 64 ω² )。为求使 A(ω) 最大的共振频率 ω_r，最小化分母平方 D(ω) = (100 – ω²)² + 64 ω²。令 D'(ω) = 0 得 2(100 – ω²)(–2ω) + 128ω = 0 → –4ω(100 – ω²) + 128ω = 0 → 4ω(ω² – 100 + 32) = 0，因此 ω_r = √68 ≈ 8.25 rad/s。峰值振幅为 A(ω_r) = 20 / √( (100 – 68)² + 64·68 ) = 20 / √(32² + 4352) = 20 / √(1024 + 4352) = 20 / √5376 ≈ 20/73.32 ≈ 0.273 C。这比 ω = 5 rad/s 时的振幅（0.235 C）更大，体现了共振现象。

8. Critical Damping and Variation of Parameters | 临界阻尼与参数变化

Back in the homogeneous equation, the damping is governed by R. The characteristic equation derived from L d²q

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