Sections 1 to 6: structure of the general solution, undetermined coefficients (polynomial, exponential, trig), the modification rule, superposition, variation of parameters, IVP synthesis, and forced-response applications第1至6节：通解结构、待定系数法（多项式、指数、三角函数）、修正规则、叠加原理、参数变易法、初值问题综合，以及强迫响应应用CALC IV
Name:姓名：Date:日期：
PART I · CORE TECHNIQUES第一部分 · 核心技术Undetermined coefficients · 28 marks待定系数法 · 28分
Structure and Undetermined Coefficients通解结构与待定系数法
For every question in this part: (i) solve the associated homogeneous equation to find $y_c$; (ii) identify the correct trial form for $y_p$, checking whether it overlaps with $y_c$; (iii) substitute and match coefficients; (iv) write the complete general solution $y = y_c + y_p$. Show all algebra.本部分每道题均需：(i) 求解对应齐次方程，得到余函数 $y_c$；(ii) 确定 $y_p$ 的正确试探形式，检验其是否与 $y_c$ 重叠；(iii) 代入并比较系数；(iv) 写出完整通解 $y = y_c + y_p$。需展示全部代数运算过程。
(a)Find the complementary solution $y_c$ by solving the associated homogeneous equation.通过求解对应齐次方程，求余函数 $y_c$。[2]
(b)Explain why the trial $y_p = At^{2} + Bt + C$ (a full degree-2 polynomial) is the correct form, and not merely $y_p = At^{2}$. Substitute and determine $A$, $B$, and $C$.解释为什么试探解 $y_p = At^{2} + Bt + C$（完整的二次多项式）是正确形式，而非仅取 $y_p = At^{2}$。代入并确定 $A$、$B$、$C$ 的值。[4]
(c)Write the general solution and verify that your $y_p$ satisfies the original ODE.写出通解，并验证所求 $y_p$ 满足原方程。[2]
Q2MEDIUMCOREexponential and trig forcing, superposition指数与三角函数强迫项，叠加原理[8 marks]
Find the general solution of $y'' + 4y = 3e^{t} + 8\sin t$ using superposition.用叠加原理求 $y'' + 4y = 3e^{t} + 8\sin t$ 的通解。
(a)State the complementary solution $y_c$.写出余函数 $y_c$。[2]
(b)Use superposition: find $y_{p,1}$ for the forcing $3e^{t}$ and $y_{p,2}$ for the forcing $8\sin t$ separately. For $y_{p,2}$ you must include both $\sin t$ and $\cos t$ in the trial and explain why.运用叠加原理：分别求强迫项 $3e^{t}$ 对应的 $y_{p,1}$ 和强迫项 $8\sin t$ 对应的 $y_{p,2}$。对于 $y_{p,2}$，试探解中必须同时包含 $\sin t$ 和 $\cos t$，并解释原因。[4]
Q3HARDCOREresonance and the modification rule, simple-root overlap共振与修正规则，单根重叠情形[6 marks]
Find a particular solution of $y'' - 4y' + 4y = e^{2t}$.求 $y'' - 4y' + 4y = e^{2t}$ 的一个特解。
(a)Find $y_c$. Show that the standard trial $y_p = Ae^{2t}$ fails, and explain precisely why it must fail by reference to the characteristic equation. State the correct modified trial.求 $y_c$。证明标准试探解 $y_p = Ae^{2t}$ 失效，并通过特征方程精确解释其必然失效的原因。写出正确的修正试探解。[3]
(b)Use the correct modified trial to find $y_p$, showing the full substitution. Verify by substituting back into the ODE.用正确的修正试探解求 $y_p$，展示完整代入过程。将结果代回原方程进行验证。[3]
(a)Find $y_c$. Show that the standard trial $y_p = e^{-t}(A\cos 2t + B\sin 2t)$ fails because of resonance, and state the correct modified trial.求 $y_c$。证明标准试探解 $y_p = e^{-t}(A\cos 2t + B\sin 2t)$ 因共振而失效，并写出正确的修正试探解。[2]
(b)Using the correct modified trial, compute $y_p'$ and $y_p''$, substitute into the ODE, and solve for $A$ and $B$.利用正确的修正试探解，计算 $y_p'$ 和 $y_p''$，代入方程，求解 $A$ 和 $B$。[4]
PART II · DEFINITIONS AND PROOF第二部分 · 定义与证明Rigorous arguments · 26 marks严格论证 · 26分
Proofs and Derivations证明与推导
These items are graded on the logic and completeness of the argument. State every hypothesis before invoking a conclusion. In proofs about the modification rule or variation of parameters, show the algebra that forces the specific form of the trial or the specific system of equations.本部分按论证的逻辑性和完整性评分。在引用结论前须先陈述所有假设。在证明修正规则或参数变易法时，须展示迫使试探解取特定形式或形成特定方程组的代数推导过程。
Q5HARDPROOFaffine structure of the solution set解集的仿射结构[6 marks]
Let $L[y] = y'' + p(t)y' + q(t)y$ where $p$ and $q$ are continuous on an interval $I$. Let $y_p$ be any particular solution of $L[y] = g(t)$, and let $y_1$, $y_2$ be a fundamental pair for $L[y] = 0$.设 $L[y] = y'' + p(t)y' + q(t)y$，其中 $p$ 和 $q$ 在区间 $I$ 上连续。设 $y_p$ 是 $L[y] = g(t)$ 的任意一个特解，$y_1$、$y_2$ 是 $L[y] = 0$ 的一组基本解对。
(a)Prove that $y = C_1 y_1 + C_2 y_2 + y_p$ satisfies $L[y] = g(t)$ for every choice of constants $C_1, C_2$. State the linearity property you use.证明对任意常数 $C_1, C_2$，$y = C_1 y_1 + C_2 y_2 + y_p$ 都满足 $L[y] = g(t)$。指出所用的线性性质。[3]
(b)Prove that every solution of $L[y] = g(t)$ on $I$ has this form. That is, if $\tilde{y}$ is any solution of $L[\tilde{y}] = g(t)$, show there exist constants $C_1, C_2$ such that $\tilde{y} = C_1 y_1 + C_2 y_2 + y_p$.证明 $L[y] = g(t)$ 在 $I$ 上的每个解都具有此形式，即若 $\tilde{y}$ 是 $L[\tilde{y}] = g(t)$ 的任意解，则存在常数 $C_1, C_2$ 使得 $\tilde{y} = C_1 y_1 + C_2 y_2 + y_p$。[3]
Q6HARDPROOFwhy the modification rule works: resonance kills the trial修正规则的原理：共振使标准试探解失效[8 marks]
Consider $L[y] = y'' + by' + cy = e^{\alpha t}$ where $\alpha$ is a simple root of the characteristic polynomial $r^{2} + br + c = (r - \alpha)(r - \beta)$ with $\alpha \ne \beta$.考虑 $L[y] = y'' + by' + cy = e^{\alpha t}$，其中 $\alpha$ 是特征多项式 $r^{2} + br + c = (r - \alpha)(r - \beta)$（$\alpha \ne \beta$）的一个单根。
(a)Substitute $y_p = Ae^{\alpha t}$ (the standard trial) into $L[y_p]$ and show that $L[Ae^{\alpha t}] = 0$ regardless of the value of $A$. Explain why this means the standard trial cannot produce a particular solution.将标准试探解 $y_p = Ae^{\alpha t}$ 代入 $L[y_p]$，证明无论 $A$ 取何值，$L[Ae^{\alpha t}] = 0$ 恒成立。解释这意味着标准试探解无法给出特解的原因。[3]
(b)Now try $y_p = Ate^{\alpha t}$. Compute $y_p'$ and $y_p''$, substitute into $L[y_p]$, and show that the result simplifies to $A(\alpha - \beta)e^{\alpha t}$. Hence determine $A$.改取 $y_p = Ate^{\alpha t}$。计算 $y_p'$ 和 $y_p''$，代入 $L[y_p]$，证明结果化简为 $A(\alpha - \beta)e^{\alpha t}$，进而确定 $A$。[4]
(c)State, without proof, the corresponding modification when $\alpha$ is a double root of the characteristic polynomial.无需证明，直接写出当 $\alpha$ 是特征多项式的二重根时对应的修正形式。[1]
Q7HARDPROOFderivation of variation of parameters参数变易法的推导[12 marks]
Let $y_1$, $y_2$ be a fundamental pair for $y'' + p(t)y' + q(t)y = 0$ with Wronskian $W = y_1 y_2' - y_2 y_1'$. Seek a particular solution of $y'' + p(t)y' + q(t)y = g(t)$ in the form $y_p = u_1 y_1 + u_2 y_2$.设 $y_1$、$y_2$ 是 $y'' + p(t)y' + q(t)y = 0$ 的基本解对，朗斯基行列式为 $W = y_1 y_2' - y_2 y_1'$。设 $y'' + p(t)y' + q(t)y = g(t)$ 的特解形如 $y_p = u_1 y_1 + u_2 y_2$。
(a)Write $y_p' = u_1' y_1 + u_1 y_1' + u_2' y_2 + u_2 y_2'$. Impose the simplifying constraint $u_1' y_1 + u_2' y_2 = 0$ and explain why this is a free choice that reduces the order of the differentiation.写出 $y_p' = u_1' y_1 + u_1 y_1' + u_2' y_2 + u_2 y_2'$。施加化简约束 $u_1' y_1 + u_2' y_2 = 0$，并解释为何这是一个自由选择，且能降低微分阶数。[3]
(b)Under this constraint, compute $y_p''$ and substitute $y_p$, $y_p'$, $y_p''$ into $y_p'' + p y_p' + q y_p = g$. Use the fact that $y_1$ and $y_2$ satisfy the homogeneous equation to reduce the result to the second equation $u_1' y_1' + u_2' y_2' = g$.在此约束下，计算 $y_p''$，将 $y_p$、$y_p'$、$y_p''$ 代入 $y_p'' + p y_p' + q y_p = g$。利用 $y_1$ 和 $y_2$ 满足齐次方程的条件，将结果化简为第二个方程 $u_1' y_1' + u_2' y_2' = g$。[4]
(c)Solve the $2 \times 2$ linear system from (a) and (b) for $u_1'$ and $u_2'$ using Cramer's rule (expressing the answer in terms of $W$, $g$, $y_1$, $y_2$), then write the formulas for $u_1$ and $u_2$ as integrals.用克拉默法则求解由(a)和(b)得到的 $2 \times 2$ 线性方程组（答案用 $W$、$g$、$y_1$、$y_2$ 表示），然后将 $u_1$ 和 $u_2$ 写成积分形式。[3]
(d)State why $W \ne 0$ is guaranteed on $I$ and why this guarantees the system is always solvable.说明为何 $W \ne 0$ 在 $I$ 上有保证，以及这为何能保证方程组始终可解。[2]
PART III · APPLICATIONS AND SYNTHESIS第三部分 · 应用与综合Extended problems · 28 marks综合题 · 28分
Variation of Parameters, IVPs, and Forced Response参数变易法、初值问题与强迫响应
Set up each problem from first principles. For IVPs, apply initial conditions to the complete solution $y = y_c + y_p$ after the particular solution is determined. Show all integration and algebraic simplification steps.每道题均需从基本原理出发建立方程。对于初值问题，在确定特解后，将初始条件代入完整解 $y = y_c + y_p$。需展示全部积分和代数化简步骤。
Q8HARDAPPLIEDvariation of parameters with a non-elementary forcing非初等函数强迫项的参数变易法[10 marks]
Find a particular solution of $y'' + y = \sec t$, valid on $-\tfrac{\pi}{2} < t < \tfrac{\pi}{2}$, using variation of parameters. (Undetermined coefficients cannot handle this forcing.)用参数变易法求 $y'' + y = \sec t$ 在 $-\tfrac{\pi}{2} < t < \tfrac{\pi}{2}$ 上的一个特解。（待定系数法无法处理此强迫项。）
(a)State $y_c$. Identify $y_1 = \cos t$ and $y_2 = \sin t$, compute the Wronskian $W$, and confirm $W \ne 0$ on the interval.写出 $y_c$。取 $y_1 = \cos t$，$y_2 = \sin t$，计算朗斯基行列式 $W$，并确认 $W \ne 0$ 在该区间上成立。[2]
(b)Write down the system for $u_1'$ and $u_2'$, then solve to find $u_1'$ and $u_2'$ explicitly.写出关于 $u_1'$ 和 $u_2'$ 的方程组，然后求解得到 $u_1'$ 和 $u_2'$ 的显式表达式。[3]
(c)Integrate to find $u_1$ and $u_2$. You may use $\int \tan t\, dt = -\ln|\cos t| + C$ and $\int 1\, dt = t + C$.积分求 $u_1$ 和 $u_2$。可使用 $\int \tan t\, dt = -\ln|\cos t| + C$ 和 $\int 1\, dt = t + C$。[3]
(d)Assemble $y_p = u_1 \cos t + u_2 \sin t$ and verify by substituting back into $y'' + y = \sec t$.组合得 $y_p = u_1 \cos t + u_2 \sin t$，并代回 $y'' + y = \sec t$ 进行验证。[2]
Q9HARDAPPLIEDfull IVP: find $y_c$, find $y_p$, fit constants to the complete solution完整初值问题：求 $y_c$，求 $y_p$，对完整解确定常数[10 marks]
Solve the initial value problem $y'' - y' - 6y = 12e^{-t}$, $y(0) = 2$, $y'(0) = -1$.求初值问题 $y'' - y' - 6y = 12e^{-t}$，$y(0) = 2$，$y'(0) = -1$ 的解。
(b)Confirm that $e^{-t}$ is not a root of the characteristic equation, choose the trial $y_p = Ae^{-t}$, and determine $A$. Verify $y_p$ satisfies the ODE.确认 $e^{-t}$ 不是特征方程的根，选取试探解 $y_p = Ae^{-t}$，确定 $A$，并验证 $y_p$ 满足方程。[3]
(c)Write the complete solution $y = y_c + y_p$. Apply the initial conditions $y(0) = 2$ and $y'(0) = -1$ to the complete solution (not to $y_c$ alone) to find $C_1$ and $C_2$.写出完整解 $y = y_c + y_p$。将初始条件 $y(0) = 2$ 和 $y'(0) = -1$ 代入完整解（而非仅代入 $y_c$），求 $C_1$ 和 $C_2$。[4]
(d)State the final solution and verify both initial conditions are satisfied.写出最终解，并验证两个初始条件均满足。[1]
Q10HARDAPPLIEDforced mechanical oscillator: steady-state and transient response受迫机械振子：稳态响应与瞬态响应[8 marks]
A damped mass-spring system satisfies $x'' + 4x' + 13x = 10\cos 3t$, where $x(t)$ is displacement and the right side is a periodic driving force.一个阻尼弹簧质量系统满足 $x'' + 4x' + 13x = 10\cos 3t$，其中 $x(t)$ 为位移，右侧为周期性驱动力。
(a)Find the complementary solution $x_c(t)$ (the transient response). Show that $x_c(t) \to 0$ as $t \to \infty$ because of the damping, and identify the natural frequency $\omega_0$ of the free oscillations.求余函数 $x_c(t)$（即瞬态响应）。证明由于阻尼的存在，$x_c(t) \to 0$（当 $t \to \infty$ 时），并确定自由振荡的固有频率 $\omega_0$。[3]
(b)Find the particular solution $x_p(t)$ (the steady-state response) using undetermined coefficients. Check whether resonance is present.用待定系数法求特解 $x_p(t)$（即稳态响应）。检验是否存在共振。[4]
(c)The driving frequency is $\omega = 3$ rad/s. Explain in one or two sentences why the system does not exhibit resonance here, even though the driving frequency is close to $\omega_0$, and state what condition on the driving frequency would cause pure resonance in an undamped version of this system.驱动频率为 $\omega = 3$ rad/s。用一两句话解释为何该系统在此条件下不发生共振（尽管驱动频率接近 $\omega_0$），并说明在无阻尼版本中，驱动频率满足何种条件才会引发纯共振。[1]