Companion to the University-Style Practice Set大学风格练习题配套解答
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
Consider $y'' - 3y' + 2y = 4t^{2} - 2$. (a) Find $y_c$. (b) Determine the correct trial and find $A$, $B$, $C$. (c) Write the general solution and verify $y_p$.考虑常微分方程 $y'' - 3y' + 2y = 4t^{2} - 2$。(a) 求余函数 $y_c$。(b) 确定正确的试探形式并求 $A$、$B$、$C$。(c) 写出通解并验证 $y_p$。
The characteristic equation is $r^{2} - 3r + 2 = (r - 1)(r - 2) = 0$, giving roots $r = 1$ and $r = 2$. (M1)特征方程为 $r^{2} - 3r + 2 = (r - 1)(r - 2) = 0$，根为 $r = 1$ 和 $r = 2$。(M1)
$$ y_c = C_1 e^{t} + C_2 e^{2t}. $$(A1)
The forcing $4t^{2} - 2$ is a degree-2 polynomial. The trial must be a complete degree-2 polynomial $y_p = At^{2} + Bt + C$, not merely $At^{2}$, because differentiating twice mixes all three powers: $y_p' = 2At + B$ contains a degree-1 term, and $-3y_p' = -6At - 3B$ introduces a constant that is needed to match the constant $-2$ in the forcing. Omitting $Bt + C$ would leave those terms with no free parameter to absorb them. (M1 for reasoning)强迫项 $4t^{2} - 2$ 是一个二次多项式。试探解必须取完整的二次多项式 $y_p = At^{2} + Bt + C$，而非仅取 $At^{2}$，原因在于求导后各次项相互混合：$y_p' = 2At + B$ 包含一次项，而 $-3y_p' = -6At - 3B$ 引入了常数项，这对匹配强迫项中的常数 $-2$ 是必要的。若省略 $Bt + C$，这些项便没有自由参数可供吸收。(M1 推理得分)
Compute derivatives: $y_p' = 2At + B$ and $y_p'' = 2A$. Substitute: (M1)计算导数：$y_p' = 2At + B$，$y_p'' = 2A$。代入：(M1)
$$ y_p'' - 3y_p' + 2y_p = 2A - 3(2At + B) + 2(At^{2} + Bt + C) = 2At^{2} + (-6A + 2B)t + (2A - 3B + 2C). $$Match coefficients with $4t^{2} - 2$:与 $4t^{2} - 2$ 比较系数：
So $y_p = 2t^{2} + 6t + 6$.因此 $y_p = 2t^{2} + 6t + 6$。
(A1) Verification: $y_p'' = 4$, $y_p' = 4t + 6$. Then $y_p'' - 3y_p' + 2y_p = 4 - 3(4t + 6) + 2(2t^{2} + 6t + 6) = 4 - 12t - 18 + 4t^{2} + 12t + 12 = 4t^{2} - 2$. This matches the forcing exactly. (B1)(A1) 验证：$y_p'' = 4$，$y_p' = 4t + 6$。则 $y_p'' - 3y_p' + 2y_p = 4 - 3(4t + 6) + 2(2t^{2} + 6t + 6) = 4 - 12t - 18 + 4t^{2} + 12t + 12 = 4t^{2} - 2$。与强迫项完全吻合。(B1)
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。
Characteristic equation: $r^{2} + 4 = 0$, so $r = \pm 2i$. (M1)特征方程：$r^{2} + 4 = 0$，故 $r = \pm 2i$。(M1)
$$ y_c = C_1\cos 2t + C_2\sin 2t. $$(A1)
By superposition, find $y_{p,1}$ for $y'' + 4y = 3e^{t}$ and $y_{p,2}$ for $y'' + 4y = 8\sin t$ independently, then add.运用叠加原理，分别求 $y'' + 4y = 3e^{t}$ 的 $y_{p,1}$ 和 $y'' + 4y = 8\sin t$ 的 $y_{p,2}$，再相加。
For $3e^{t}$:对于 $3e^{t}$： Since $r = 1$ is not a root of $r^{2} + 4 = 0$, try $y_{p,1} = Ae^{t}$. Then $y_{p,1}'' + 4y_{p,1} = Ae^{t} + 4Ae^{t} = 5Ae^{t} = 3e^{t}$, so $A = \tfrac{3}{5}$. (M1·A1)因为 $r = 1$ 不是 $r^{2} + 4 = 0$ 的根，取 $y_{p,1} = Ae^{t}$。则 $y_{p,1}'' + 4y_{p,1} = Ae^{t} + 4Ae^{t} = 5Ae^{t} = 3e^{t}$，故 $A = \tfrac{3}{5}$。(M1·A1)
For $8\sin t$:对于 $8\sin t$： The trial must include both $A\cos t$ and $B\sin t$, even though the forcing contains only $\sin t$, because $L$ mixes sine and cosine through its derivative terms: $y'' + 4y$ applied to $A\cos t$ alone gives $(4 - 1)A\cos t$ with no $\sin t$ produced, so if $B = 0$ the $\sin t$ equation is satisfied here. However, for a general operator $L[y] = ay'' + by' + cy$ with $b \ne 0$ the derivative terms do couple them, making both necessary. Here $b = 0$ and $L[\sin t] = -\sin t + 4\sin t = 3\sin t$ while $L[\cos t] = -\cos t + 4\cos t = 3\cos t$, so try $y_{p,2} = A\cos t + B\sin t$. (M1 for including both)试探解必须同时包含 $A\cos t$ 和 $B\sin t$，尽管强迫项仅含 $\sin t$，原因是算子 $L$ 通过导数项将正弦与余弦混合：对 $A\cos t$ 单独施加 $y'' + 4y$ 只得 $(4-1)A\cos t$，不产生 $\sin t$ 项，故此处令 $B = 0$ 时 $\sin t$ 方程确实满足。但对于一般含 $b \ne 0$ 的算子 $L[y] = ay'' + by' + cy$，导数项确实会耦合正弦与余弦，使两者均不可缺。此处 $b = 0$，$L[\sin t] = -\sin t + 4\sin t = 3\sin t$，$L[\cos t] = -\cos t + 4\cos t = 3\cos t$，故取 $y_{p,2} = A\cos t + B\sin t$。(M1，含两项得分)
$$ y_{p,2}'' + 4y_{p,2} = (-A\cos t - B\sin t) + 4(A\cos t + B\sin t) = 3A\cos t + 3B\sin t = 8\sin t. $$So $3A = 0 \Rightarrow A = 0$ and $3B = 8 \Rightarrow B = \tfrac{8}{3}$. Thus $y_{p,2} = \tfrac{8}{3}\sin t$. (A1)故 $3A = 0 \Rightarrow A = 0$，$3B = 8 \Rightarrow B = \tfrac{8}{3}$。因此 $y_{p,2} = \tfrac{8}{3}\sin t$。(A1)
(A1) Verification of $y_{p,2}$: $y_{p,2}'' = -\tfrac{8}{3}\sin t$; so $y_{p,2}'' + 4y_{p,2} = -\tfrac{8}{3}\sin t + \tfrac{32}{3}\sin t = \tfrac{24}{3}\sin t = 8\sin t$. Correct. (B1)(A1) 验证 $y_{p,2}$：$y_{p,2}'' = -\tfrac{8}{3}\sin t$；则 $y_{p,2}'' + 4y_{p,2} = -\tfrac{8}{3}\sin t + \tfrac{32}{3}\sin t = \tfrac{24}{3}\sin t = 8\sin t$。正确。(B1)
Find a particular solution of $y'' - 4y' + 4y = e^{2t}$. Show the standard trial fails and use the correct modified trial.求 $y'' - 4y' + 4y = e^{2t}$ 的一个特解。证明标准试探解失效，并使用正确的修正试探解。
Characteristic equation: $r^{2} - 4r + 4 = (r - 2)^{2} = 0$. The root $r = 2$ is a double root. Thus $y_c = (C_1 + C_2 t)e^{2t}$. (M1)特征方程：$r^{2} - 4r + 4 = (r - 2)^{2} = 0$。根 $r = 2$ 是二重根。因此 $y_c = (C_1 + C_2 t)e^{2t}$。(M1)
Substitute $y_p = Ae^{2t}$ into $L[y_p]$: $y_p'' - 4y_p' + 4y_p = 4Ae^{2t} - 8Ae^{2t} + 4Ae^{2t} = 0$. The result is identically zero for any $A$ because $e^{2t}$ lies in $\ker L$ (it is a solution of the homogeneous equation). No value of $A$ can make $0 = e^{2t}$, so the standard trial is useless. (R1)将 $y_p = Ae^{2t}$ 代入 $L[y_p]$：$y_p'' - 4y_p' + 4y_p = 4Ae^{2t} - 8Ae^{2t} + 4Ae^{2t} = 0$。结果对任意 $A$ 恒等于零，因为 $e^{2t}$ 属于 $\ker L$（即它是齐次方程的解）。没有任何 $A$ 值能使 $0 = e^{2t}$，故标准试探解无效。(R1)
Since $r = 2$ is a double root (multiplicity 2), the modification rule requires multiplying the base trial by $t^{s}$ with $s = 2$. The correct trial is $y_p = At^{2}e^{2t}$. (A1)由于 $r = 2$ 是二重根（重数为2），修正规则要求将基础试探解乘以 $t^{s}$，其中 $s = 2$。正确的试探解为 $y_p = At^{2}e^{2t}$。(A1)
Compute derivatives. Let $y_p = At^{2}e^{2t}$. Then (M1):计算导数。设 $y_p = At^{2}e^{2t}$，则 (M1)：
$$ y_p' = Ae^{2t}(2t^{2} + 2t) = 2Ate^{2t}(t + 1), \qquad y_p'' = Ae^{2t}(4t^{2} + 8t + 2). $$Substitute into $y_p'' - 4y_p' + 4y_p$:代入 $y_p'' - 4y_p' + 4y_p$：
$$ Ae^{2t}(4t^{2} + 8t + 2) - 4Ae^{2t}(2t^{2} + 2t) + 4At^{2}e^{2t} $$ $$ = Ae^{2t}\bigl[(4t^{2} + 8t + 2) - (8t^{2} + 8t) + 4t^{2}\bigr] = Ae^{2t} \cdot 2 = 2Ae^{2t}. $$Set $2Ae^{2t} = e^{2t}$, so $A = \tfrac{1}{2}$. (A1) Thus $y_p = \tfrac{1}{2}t^{2}e^{2t}$.令 $2Ae^{2t} = e^{2t}$，故 $A = \tfrac{1}{2}$。(A1) 因此 $y_p = \tfrac{1}{2}t^{2}e^{2t}$。
Verification: $2 \cdot \tfrac{1}{2} \cdot e^{2t} = e^{2t}$. Confirmed. (B1)验证：$2 \cdot \tfrac{1}{2} \cdot e^{2t} = e^{2t}$。得证。(B1)
Find a particular solution of $y'' + 2y' + 5y = e^{-t}\cos 2t$.求 $y'' + 2y' + 5y = e^{-t}\cos 2t$ 的一个特解。
Characteristic equation: $r^{2} + 2r + 5 = 0$, so $r = \dfrac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$. (M1)特征方程：$r^{2} + 2r + 5 = 0$，故 $r = \dfrac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$。(M1)
$$ y_c = e^{-t}(C_1\cos 2t + C_2\sin 2t). $$The forcing $e^{-t}\cos 2t$ corresponds exactly to the complex root $r = -1 + 2i$, which is already a root of the characteristic equation. Thus there is resonance: the trial $e^{-t}(A\cos 2t + B\sin 2t)$ lies inside $y_c$ and must be modified by multiplying by $t$. The correct trial is $y_p = te^{-t}(A\cos 2t + B\sin 2t)$. (A1)强迫项 $e^{-t}\cos 2t$ 恰好对应复数根 $r = -1 + 2i$，它已经是特征方程的根。因此存在共振：试探解 $e^{-t}(A\cos 2t + B\sin 2t)$ 已包含在 $y_c$ 中，必须乘以 $t$ 进行修正。正确的试探解为 $y_p = te^{-t}(A\cos 2t + B\sin 2t)$。(A1)
Let $y_p = te^{-t}(A\cos 2t + B\sin 2t)$. Write $u = A\cos 2t + B\sin 2t$ so $y_p = te^{-t}u$. (M1)设 $y_p = te^{-t}(A\cos 2t + B\sin 2t)$，令 $u = A\cos 2t + B\sin 2t$，则 $y_p = te^{-t}u$。(M1)
Using the product rule twice:两次使用乘积法则：
$$ y_p' = e^{-t}u + te^{-t}u' - te^{-t}u = e^{-t}u + te^{-t}(u' - u). $$ $$ y_p'' = e^{-t}(u' - u) + e^{-t}(u' - u) + te^{-t}(u'' - 2u' + u) = e^{-t}(2u' - 2u) + te^{-t}(u'' - 2u' + u). $$Substitute $y_p'' + 2y_p' + 5y_p$ and collect terms at $te^{-t}$ and $e^{-t}$ (M1):代入 $y_p'' + 2y_p' + 5y_p$，整理 $te^{-t}$ 和 $e^{-t}$ 各项 (M1)：
The $te^{-t}$ group: $u'' - 2u' + u + 2(u' - u) + 5u = u'' + 4u$. With $u = A\cos 2t + B\sin 2t$, we have $u'' = -4A\cos 2t - 4B\sin 2t$, so $u'' + 4u = 0$. These terms vanish, as expected from the resonance.$te^{-t}$ 组：$u'' - 2u' + u + 2(u' - u) + 5u = u'' + 4u$。由 $u = A\cos 2t + B\sin 2t$，有 $u'' = -4A\cos 2t - 4B\sin 2t$，故 $u'' + 4u = 0$。这些项消失，与共振的预期一致。
The $e^{-t}$ group: $2u' - 2u + 2u = 2u'$. Now $u' = -2A\sin 2t + 2B\cos 2t$, so $2u' = -4A\sin 2t + 4B\cos 2t$.$e^{-t}$ 组：$2u' - 2u + 2u = 2u'$。而 $u' = -2A\sin 2t + 2B\cos 2t$，故 $2u' = -4A\sin 2t + 4B\cos 2t$。
Set equal to the forcing coefficient $\cos 2t$ (the $e^{-t}$ has been factored out): (M1)令其等于强迫项系数 $\cos 2t$（$e^{-t}$ 已提出）：(M1)
$$ 4B\cos 2t - 4A\sin 2t = \cos 2t. $$Matching: $4B = 1 \Rightarrow B = \tfrac{1}{4}$ and $-4A = 0 \Rightarrow A = 0$. (A1)比较系数：$4B = 1 \Rightarrow B = \tfrac{1}{4}$，$-4A = 0 \Rightarrow A = 0$。(A1)
Thus $y_p = \tfrac{1}{4}te^{-t}\sin 2t$. (A1)因此 $y_p = \tfrac{1}{4}te^{-t}\sin 2t$。(A1)
Let $L[y] = y'' + p(t)y' + q(t)y$, $y_p$ a particular solution of $L[y] = g$, and $y_1, y_2$ a fundamental pair for $L[y] = 0$. (a) Prove $C_1 y_1 + C_2 y_2 + y_p$ satisfies $L[y] = g$. (b) Prove every solution has this form.设 $L[y] = y'' + p(t)y' + q(t)y$，$y_p$ 是 $L[y] = g$ 的特解，$y_1, y_2$ 是 $L[y] = 0$ 的基本解对。(a) 证明 $C_1 y_1 + C_2 y_2 + y_p$ 满足 $L[y] = g$。(b) 证明每个解都具有此形式。
Let $y = C_1 y_1 + C_2 y_2 + y_p$. Apply $L$, using the fact that $L$ is a linear operator: (M1)设 $y = C_1 y_1 + C_2 y_2 + y_p$。利用 $L$ 是线性算子的性质施加 $L$：(M1)
$$ L[y] = L[C_1 y_1 + C_2 y_2 + y_p] = C_1 L[y_1] + C_2 L[y_2] + L[y_p]. $$Since $y_1$ and $y_2$ are solutions of the homogeneous equation, $L[y_1] = 0$ and $L[y_2] = 0$. Since $y_p$ is a particular solution, $L[y_p] = g(t)$. (A1) Therefore $L[y] = 0 + 0 + g(t) = g(t)$. (R1)由于 $y_1$ 和 $y_2$ 是齐次方程的解，故 $L[y_1] = 0$，$L[y_2] = 0$。由于 $y_p$ 是特解，$L[y_p] = g(t)$。(A1) 因此 $L[y] = 0 + 0 + g(t) = g(t)$。(R1)
Let $\tilde{y}$ be any solution of $L[\tilde{y}] = g(t)$. Define $w = \tilde{y} - y_p$. Then (M1):设 $\tilde{y}$ 是 $L[\tilde{y}] = g(t)$ 的任意解。定义 $w = \tilde{y} - y_p$，则 (M1)：
$$ L[w] = L[\tilde{y}] - L[y_p] = g(t) - g(t) = 0. $$So $w$ satisfies the homogeneous equation $L[w] = 0$. Since $y_1, y_2$ is a fundamental pair (their Wronskian is non-zero on $I$, by Abel's theorem), every solution of the homogeneous equation is a linear combination: $w = C_1 y_1 + C_2 y_2$ for some constants $C_1, C_2$. (A1)故 $w$ 满足齐次方程 $L[w] = 0$。由于 $y_1, y_2$ 是基本解对（根据阿贝尔定理，其朗斯基行列式在 $I$ 上非零），齐次方程的每个解都是线性组合：存在常数 $C_1, C_2$ 使 $w = C_1 y_1 + C_2 y_2$。(A1)
Therefore $\tilde{y} = w + y_p = C_1 y_1 + C_2 y_2 + y_p$. Since $\tilde{y}$ was arbitrary, every solution has this form. (R1)因此 $\tilde{y} = w + y_p = C_1 y_1 + C_2 y_2 + y_p$。由于 $\tilde{y}$ 是任意的，每个解都具有此形式。(R1)
Consider $L[y] = y'' + by' + cy = e^{\alpha t}$ where $\alpha$ is a simple root of $r^{2} + br + c = (r - \alpha)(r - \beta)$, $\alpha \ne \beta$. (a) Show the standard trial $Ae^{\alpha t}$ fails. (b) Show the modified trial $Ate^{\alpha t}$ gives $A(\alpha - \beta)e^{\alpha t}$ and hence find $A$. (c) State the double-root modification.考虑 $L[y] = y'' + by' + cy = e^{\alpha t}$，其中 $\alpha$ 是 $r^{2} + br + c = (r - \alpha)(r - \beta)$（$\alpha \ne \beta$）的单根。(a) 证明标准试探解 $Ae^{\alpha t}$ 失效。(b) 证明修正试探解 $Ate^{\alpha t}$ 给出 $A(\alpha - \beta)e^{\alpha t}$，进而求 $A$。(c) 写出二重根对应的修正形式。
Let $y_p = Ae^{\alpha t}$. Then $y_p' = A\alpha e^{\alpha t}$ and $y_p'' = A\alpha^{2} e^{\alpha t}$. Substitute: (M1)设 $y_p = Ae^{\alpha t}$，则 $y_p' = A\alpha e^{\alpha t}$，$y_p'' = A\alpha^{2} e^{\alpha t}$。代入：(M1)
$$ L[Ae^{\alpha t}] = A(\alpha^{2} + b\alpha + c)e^{\alpha t}. $$Since $\alpha$ is a root of $r^{2} + br + c$, we have $\alpha^{2} + b\alpha + c = 0$. Therefore $L[Ae^{\alpha t}] = 0$ for every $A$. (A1) No choice of $A$ can produce $e^{\alpha t} \ne 0$ on the right side, so the standard trial is incapable of being a particular solution. (R1)由于 $\alpha$ 是 $r^{2} + br + c$ 的根，故 $\alpha^{2} + b\alpha + c = 0$。因此对任意 $A$，$L[Ae^{\alpha t}] = 0$。(A1) 没有任何 $A$ 的选取能使右端为非零的 $e^{\alpha t}$，故标准试探解无法成为特解。(R1)
Let $y_p = Ate^{\alpha t}$. Differentiate: (M1)设 $y_p = Ate^{\alpha t}$，求导：(M1)
$$ y_p' = Ae^{\alpha t}(1 + \alpha t), \qquad y_p'' = Ae^{\alpha t}(\alpha^{2}t + 2\alpha). $$Substitute into $L[y_p] = y_p'' + by_p' + cy_p$: (M1)代入 $L[y_p] = y_p'' + by_p' + cy_p$：(M1)
$$ Ae^{\alpha t}\bigl[(\alpha^{2}t + 2\alpha) + b(1 + \alpha t) + ct\bigr] = Ae^{\alpha t}\bigl[(\alpha^{2} + b\alpha + c)t + (2\alpha + b)\bigr]. $$Since $\alpha^{2} + b\alpha + c = 0$, the $t$-coefficient vanishes. Using the factored form: $r^{2} + br + c = (r - \alpha)(r - \beta)$, so $b = -(\alpha + \beta)$ and thus $2\alpha + b = 2\alpha - \alpha - \beta = \alpha - \beta$. (A1)由于 $\alpha^{2} + b\alpha + c = 0$，$t$ 系数消失。利用因式分解形式：$r^{2} + br + c = (r - \alpha)(r - \beta)$，故 $b = -(\alpha + \beta)$，从而 $2\alpha + b = 2\alpha - \alpha - \beta = \alpha - \beta$。(A1)
$$ L[Ate^{\alpha t}] = A(\alpha - \beta)e^{\alpha t}. $$Setting this equal to $e^{\alpha t}$: $A(\alpha - \beta) = 1$. Since $\alpha \ne \beta$, we get $A = \dfrac{1}{\alpha - \beta}$. (A1)令其等于 $e^{\alpha t}$：$A(\alpha - \beta) = 1$。由于 $\alpha \ne \beta$，得 $A = \dfrac{1}{\alpha - \beta}$。(A1)
If $\alpha$ is a double root, then $y_c = (C_1 + C_2 t)e^{\alpha t}$ and both $e^{\alpha t}$ and $te^{\alpha t}$ lie in $\ker L$. The correct trial is $y_p = At^{2}e^{\alpha t}$. (B1)若 $\alpha$ 是二重根，则 $y_c = (C_1 + C_2 t)e^{\alpha t}$，且 $e^{\alpha t}$ 和 $te^{\alpha t}$ 均属于 $\ker L$。正确的试探解为 $y_p = At^{2}e^{\alpha t}$。(B1)
With $y_p = u_1 y_1 + u_2 y_2$: (a) impose $u_1' y_1 + u_2' y_2 = 0$ and explain. (b) Derive the second equation $u_1' y_1' + u_2' y_2' = g$. (c) Solve for $u_1'$, $u_2'$ by Cramer's rule and write $u_1$, $u_2$ as integrals. (d) Explain why $W \ne 0$ on $I$.设 $y_p = u_1 y_1 + u_2 y_2$：(a) 施加约束 $u_1' y_1 + u_2' y_2 = 0$ 并解释原因。(b) 推导第二个方程 $u_1' y_1' + u_2' y_2' = g$。(c) 用克拉默法则求 $u_1'$、$u_2'$，并将 $u_1$、$u_2$ 写成积分形式。(d) 解释为何 $W \ne 0$ 在 $I$ 上成立。
Differentiate $y_p = u_1 y_1 + u_2 y_2$: (M1)对 $y_p = u_1 y_1 + u_2 y_2$ 求导：(M1)
$$ y_p' = u_1' y_1 + u_1 y_1' + u_2' y_2 + u_2 y_2'. $$We have two unknown functions $u_1, u_2$ but only one equation $L[y_p] = g$, so we have one degree of freedom. We impose the constraint (A1)我们有两个未知函数 $u_1, u_2$，但只有一个方程 $L[y_p] = g$，因此有一个自由度。施加约束 (A1)：
$$ u_1' y_1 + u_2' y_2 = 0, $$which reduces $y_p'$ to $u_1 y_1' + u_2 y_2'$. This means $y_p''$ will involve only first derivatives of $u_1$ and $u_2$ (not second derivatives), keeping the resulting system linear and of first order. (R1)这将 $y_p'$ 化简为 $u_1 y_1' + u_2 y_2'$。这意味着 $y_p''$ 中只出现 $u_1$ 和 $u_2$ 的一阶导数（而非二阶导数），使所得方程组保持线性且为一阶。(R1)
Under the constraint, $y_p' = u_1 y_1' + u_2 y_2'$. Differentiate again: (M1)在约束下，$y_p' = u_1 y_1' + u_2 y_2'$。再次求导：(M1)
$$ y_p'' = u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''. $$Substitute $y_p$, $y_p'$, $y_p''$ into $y_p'' + p y_p' + q y_p = g$: (M1)将 $y_p$、$y_p'$、$y_p''$ 代入 $y_p'' + p y_p' + q y_p = g$：(M1)
$$ (u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2'') + p(u_1 y_1' + u_2 y_2') + q(u_1 y_1 + u_2 y_2) = g. $$Group the $u_i$ terms: $u_1(y_1'' + p y_1' + q y_1) + u_2(y_2'' + p y_2' + q y_2) + u_1' y_1' + u_2' y_2' = g$. (A1)整理 $u_i$ 各项：$u_1(y_1'' + p y_1' + q y_1) + u_2(y_2'' + p y_2' + q y_2) + u_1' y_1' + u_2' y_2' = g$。(A1)
Since $y_1$ and $y_2$ satisfy the homogeneous equation, $y_i'' + p y_i' + q y_i = 0$, so those groups vanish. The equation reduces to (A1)由于 $y_1$ 和 $y_2$ 满足齐次方程，$y_i'' + p y_i' + q y_i = 0$，故这些项消失。方程化简为 (A1)：
$$ u_1' y_1' + u_2' y_2' = g. $$The system is (M1):方程组为 (M1)：
$$ \begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix} \begin{pmatrix} u_1' \\ u_2' \end{pmatrix} = \begin{pmatrix} 0 \\ g \end{pmatrix}. $$The coefficient matrix has determinant $W = y_1 y_2' - y_2 y_1'$. By Cramer's rule: (A1)系数矩阵的行列式为 $W = y_1 y_2' - y_2 y_1'$。由克拉默法则：(A1)
$$ u_1' = \frac{\det\begin{pmatrix}0 & y_2 \\ g & y_2'\end{pmatrix}}{W} = \frac{-y_2 g}{W}, \qquad u_2' = \frac{\det\begin{pmatrix}y_1 & 0 \\ y_1' & g\end{pmatrix}}{W} = \frac{y_1 g}{W}. $$Integrating: (A1)积分得：(A1)
$$ u_1 = -\int \frac{y_2(t)\, g(t)}{W(t)}\, dt, \qquad u_2 = \int \frac{y_1(t)\, g(t)}{W(t)}\, dt. $$By Abel's theorem, if $y_1, y_2$ are solutions of $y'' + p(t)y' + q(t)y = 0$ with $p$ continuous on $I$, then (M1)由阿贝尔定理，若 $y_1, y_2$ 是 $y'' + p(t)y' + q(t)y = 0$ 的解且 $p$ 在 $I$ 上连续，则 (M1)：
$$ W(t) = W(t_0)\, e^{-\int_{t_0}^{t} p(s)\, ds}. $$If $y_1, y_2$ form a fundamental pair (i.e., $W(t_0) \ne 0$ at some point $t_0 \in I$), then $W(t) \ne 0$ for all $t \in I$ because the exponential is never zero. This guarantees the $2 \times 2$ system always has a unique solution for $u_1'$ and $u_2'$. (A1)若 $y_1, y_2$ 构成基本解对（即在某点 $t_0 \in I$ 处 $W(t_0) \ne 0$），则对所有 $t \in I$ 均有 $W(t) \ne 0$，因为指数函数永远不为零。这保证了 $2 \times 2$ 方程组对 $u_1'$ 和 $u_2'$ 始终有唯一解。(A1)
Find a particular solution of $y'' + y = \sec t$ on $-\tfrac{\pi}{2} < t < \tfrac{\pi}{2}$ using variation of parameters.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。
The characteristic equation is $r^{2} + 1 = 0$, so $r = \pm i$, giving (M1)特征方程为 $r^{2} + 1 = 0$，故 $r = \pm i$，给出 (M1)：
$$ y_c = C_1\cos t + C_2\sin t, \qquad y_1 = \cos t,\ y_2 = \sin t. $$The Wronskian: $W = y_1 y_2' - y_2 y_1' = \cos t \cdot \cos t - \sin t \cdot (-\sin t) = \cos^{2}t + \sin^{2}t = 1 \ne 0$ on the interval. (A1)朗斯基行列式：$W = y_1 y_2' - y_2 y_1' = \cos t \cdot \cos t - \sin t \cdot (-\sin t) = \cos^{2}t + \sin^{2}t = 1 \ne 0$，在该区间上成立。(A1)
With $g(t) = \sec t$ and $W = 1$: (M1)取 $g(t) = \sec t$，$W = 1$：(M1)
$$ u_1' = \frac{-y_2 g}{W} = -\sin t \cdot \sec t = -\frac{\sin t}{\cos t} = -\tan t. $$ $$ u_2' = \frac{y_1 g}{W} = \cos t \cdot \sec t = 1. $$(A1 for each, 2 marks)（各得A1，共2分）
Integrate $u_1'$ (M1):对 $u_1'$ 积分 (M1)：
$$ u_1 = \int (-\tan t)\, dt = \int \frac{-\sin t}{\cos t}\, dt = \ln|\cos t| + C. $$Take $C = 0$ (any particular choice works). (A1)取 $C = 0$（任意特定选择均可）。(A1)
$$ u_2 = \int 1\, dt = t. $$(A1)
$y_p = u_1 y_1 + u_2 y_2 = \cos t \ln|\cos t| + t\sin t$. (M1)$y_p = u_1 y_1 + u_2 y_2 = \cos t \ln|\cos t| + t\sin t$。(M1)
Verification: compute $y_p'$ and $y_p''$. We have $y_p = \cos t \ln|\cos t| + t\sin t$.验证：计算 $y_p'$ 和 $y_p''$。已知 $y_p = \cos t \ln|\cos t| + t\sin t$。
$$ y_p' = -\sin t \ln|\cos t| + \cos t \cdot \frac{-\sin t}{\cos t} + \sin t + t\cos t = -\sin t \ln|\cos t| - \sin t + \sin t + t\cos t = t\cos t - \sin t\ln|\cos t|. $$ $$ y_p'' = \cos t - \sin t\cdot\frac{-\sin t}{\cos t} + (-\cos t\ln|\cos t| - \sin t\cdot\frac{-\sin t}{\cos t}) $$ $$ = \cos t + \frac{\sin^{2}t}{\cos t} - \cos t\ln|\cos t| + \frac{\sin^{2}t}{\cos t}. $$Then $y_p'' + y_p$:则 $y_p'' + y_p$：
$$ \left(\cos t + \frac{2\sin^{2}t}{\cos t} - \cos t\ln|\cos t|\right) + (\cos t\ln|\cos t| + t\sin t - t\sin t). $$Wait, the $t\sin t$ term from $y_p$ and the $t\sin t$ from $y_p''$ do not immediately cancel. Let us redo $y_p''$ carefully.注意，$y_p$ 中的 $t\sin t$ 项与 $y_p''$ 中的 $t\sin t$ 并不直接消去。重新仔细计算 $y_p''$。
$y_p' = t\cos t - \sin t\ln|\cos t|$. Differentiate:$y_p' = t\cos t - \sin t\ln|\cos t|$，求导：
$$ y_p'' = \cos t - t\sin t - \cos t\ln|\cos t| - \sin t\cdot\frac{-\sin t}{\cos t} = \cos t - t\sin t - \cos t\ln|\cos t| + \frac{\sin^{2}t}{\cos t}. $$Now $y_p'' + y_p$:则 $y_p'' + y_p$：
$$ \left(\cos t - t\sin t - \cos t\ln|\cos t| + \frac{\sin^{2}t}{\cos t}\right) + \left(\cos t\ln|\cos t| + t\sin t\right) $$ $$ = \cos t + \frac{\sin^{2}t}{\cos t} = \frac{\cos^{2}t + \sin^{2}t}{\cos t} = \frac{1}{\cos t} = \sec t. \checkmark $$(A1) The particular solution is confirmed.(A1) 特解得到验证。
Solve $y'' - y' - 6y = 12e^{-t}$, $y(0) = 2$, $y'(0) = -1$.求解 $y'' - y' - 6y = 12e^{-t}$，$y(0) = 2$，$y'(0) = -1$。
Characteristic equation: $r^{2} - r - 6 = (r - 3)(r + 2) = 0$, roots $r = 3$ and $r = -2$. (M1)特征方程：$r^{2} - r - 6 = (r - 3)(r + 2) = 0$，根为 $r = 3$ 和 $r = -2$。(M1)
$$ y_c = C_1 e^{3t} + C_2 e^{-2t}. \quad \text{(A1)} $$Since $r = -1$ is not a root of the characteristic equation ($r = 3$ and $r = -2$ are the roots), try $y_p = Ae^{-t}$. (M1) Then $y_p'' = Ae^{-t}$, $y_p' = -Ae^{-t}$, and由于 $r = -1$ 不是特征方程的根（根为 $r = 3$ 和 $r = -2$），取 $y_p = Ae^{-t}$。(M1) 则 $y_p'' = Ae^{-t}$，$y_p' = -Ae^{-t}$，且
$$ y_p'' - y_p' - 6y_p = Ae^{-t} - (-Ae^{-t}) - 6Ae^{-t} = (1 + 1 - 6)Ae^{-t} = -4Ae^{-t} = 12e^{-t}. $$So $A = -3$ and $y_p = -3e^{-t}$. (A1)故 $A = -3$，$y_p = -3e^{-t}$。(A1)
Verification: $(-3e^{-t})'' - (-3e^{-t})' - 6(-3e^{-t}) = -3e^{-t} - 3e^{-t} + 18e^{-t} = 12e^{-t}$. (B1)验证：$(-3e^{-t})'' - (-3e^{-t})' - 6(-3e^{-t}) = -3e^{-t} - 3e^{-t} + 18e^{-t} = 12e^{-t}$。(B1)
The complete solution is $y = C_1 e^{3t} + C_2 e^{-2t} - 3e^{-t}$. Apply ICs to this full expression, not to $y_c$ alone.完整解为 $y = C_1 e^{3t} + C_2 e^{-2t} - 3e^{-t}$。将初始条件代入此完整表达式，而非仅代入 $y_c$。
At $t = 0$: $y(0) = C_1 + C_2 - 3 = 2 \Rightarrow C_1 + C_2 = 5$. (M1)在 $t = 0$ 处：$y(0) = C_1 + C_2 - 3 = 2 \Rightarrow C_1 + C_2 = 5$。(M1)
Differentiate: $y' = 3C_1 e^{3t} - 2C_2 e^{-2t} + 3e^{-t}$. At $t = 0$: $3C_1 - 2C_2 + 3 = -1 \Rightarrow 3C_1 - 2C_2 = -4$. (M1)求导：$y' = 3C_1 e^{3t} - 2C_2 e^{-2t} + 3e^{-t}$。在 $t = 0$ 处：$3C_1 - 2C_2 + 3 = -1 \Rightarrow 3C_1 - 2C_2 = -4$。(M1)
Solve the system $C_1 + C_2 = 5$ and $3C_1 - 2C_2 = -4$: multiply the first by 2 to get $2C_1 + 2C_2 = 10$, then add the second equation: $5C_1 = 6$, so $C_1 = \tfrac{6}{5}$ and $C_2 = 5 - \tfrac{6}{5} = \tfrac{19}{5}$. (A1 each)解方程组 $C_1 + C_2 = 5$ 和 $3C_1 - 2C_2 = -4$：将第一式乘以2得 $2C_1 + 2C_2 = 10$，加上第二式：$5C_1 = 6$，故 $C_1 = \tfrac{6}{5}$，$C_2 = 5 - \tfrac{6}{5} = \tfrac{19}{5}$。(各得A1)
$$ y = \tfrac{6}{5}e^{3t} + \tfrac{19}{5}e^{-2t} - 3e^{-t}. $$$y(0) = \tfrac{6}{5} + \tfrac{19}{5} - 3 = \tfrac{25}{5} - 3 = 5 - 3 = 2$. $y'(0) = 3 \cdot \tfrac{6}{5} - 2 \cdot \tfrac{19}{5} + 3 = \tfrac{18}{5} - \tfrac{38}{5} + \tfrac{15}{5} = \tfrac{-5}{5} = -1$. Both initial conditions confirmed. (B1)$y(0) = \tfrac{6}{5} + \tfrac{19}{5} - 3 = \tfrac{25}{5} - 3 = 5 - 3 = 2$。$y'(0) = 3 \cdot \tfrac{6}{5} - 2 \cdot \tfrac{19}{5} + 3 = \tfrac{18}{5} - \tfrac{38}{5} + \tfrac{15}{5} = \tfrac{-5}{5} = -1$。两个初始条件均得到验证。(B1)
A damped oscillator satisfies $x'' + 4x' + 13x = 10\cos 3t$. (a) Find the transient $x_c(t)$, show it decays, and state $\omega_0$. (b) Find the steady-state $x_p(t)$. (c) Explain why resonance is absent and state the pure-resonance condition for the undamped system.一个阻尼振子满足 $x'' + 4x' + 13x = 10\cos 3t$。(a) 求瞬态响应 $x_c(t)$，证明其衰减，并写出 $\omega_0$。(b) 求稳态响应 $x_p(t)$。(c) 解释为何不发生共振，并写出无阻尼系统的纯共振条件。
Characteristic equation: $r^{2} + 4r + 13 = 0$, so $r = \dfrac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i$. (M1)特征方程：$r^{2} + 4r + 13 = 0$，故 $r = \dfrac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i$。(M1)
$$ x_c = e^{-2t}(C_1\cos 3t + C_2\sin 3t). $$(A1) Since the real part of the roots is $-2 < 0$, the factor $e^{-2t} \to 0$ as $t \to \infty$, so $x_c(t) \to 0$ regardless of the constants. The system is damped and the transient decays. The natural (free) oscillation frequency of the damped system is $\omega_0 = 3$ rad/s. (R1)(A1) 由于根的实部为 $-2 < 0$，因子 $e^{-2t} \to 0$（当 $t \to \infty$ 时），故无论常数取何值，$x_c(t) \to 0$。系统有阻尼，瞬态响应衰减。阻尼系统自由振荡的固有频率为 $\omega_0 = 3$ rad/s。(R1)
The forcing frequency $\omega = 3$. Check whether $\cos 3t$ or $\sin 3t$ appears in $x_c$: they do, but they appear multiplied by $e^{-2t}$, not as bare sinusoids. The bare trial $A\cos 3t + B\sin 3t$ is not a solution of the homogeneous equation, so there is no resonance. (M1 for the check)驱动频率 $\omega = 3$。检验 $\cos 3t$ 或 $\sin 3t$ 是否出现在 $x_c$ 中：确实出现，但它们乘以了 $e^{-2t}$，并非单独的正弦余弦函数。裸的试探解 $A\cos 3t + B\sin 3t$ 不是齐次方程的解，故不存在共振。(M1，检验得分)
Try $x_p = A\cos 3t + B\sin 3t$. Compute: (M1)取 $x_p = A\cos 3t + B\sin 3t$，计算：(M1)
$$ x_p'' = -9A\cos 3t - 9B\sin 3t, \qquad x_p' = -3A\sin 3t + 3B\cos 3t. $$ $$ x_p'' + 4x_p' + 13x_p = (-9A + 12B + 13A)\cos 3t + (-9B - 12A + 13B)\sin 3t = (4A + 12B)\cos 3t + (4B - 12A)\sin 3t. $$Set equal to $10\cos 3t + 0\sin 3t$: (A1)令其等于 $10\cos 3t + 0\sin 3t$：(A1)
$$ 4A + 12B = 10, \qquad 4B - 12A = 0. $$From the second equation: $B = 3A$. Substituting: $4A + 36A = 40A = 10$, so $A = \tfrac{1}{4}$ and $B = \tfrac{3}{4}$. (A1)由第二式：$B = 3A$。代入：$4A + 36A = 40A = 10$，故 $A = \tfrac{1}{4}$，$B = \tfrac{3}{4}$。(A1)
$$ x_p = \tfrac{1}{4}\cos 3t + \tfrac{3}{4}\sin 3t. $$We can also write this as $x_p = \tfrac{1}{4}(\cos 3t + 3\sin 3t)$ or in amplitude-phase form as $x_p = \tfrac{\sqrt{10}}{4}\cos(3t - \phi)$ with $\phi = \arctan 3$.也可写为 $x_p = \tfrac{1}{4}(\cos 3t + 3\sin 3t)$，或写成幅相形式 $x_p = \tfrac{\sqrt{10}}{4}\cos(3t - \phi)$，其中 $\phi = \arctan 3$。
The damped oscillation frequency is 3 rad/s (the imaginary part of $-2 \pm 3i$), but the undamped natural frequency of the spring-mass system is $\omega_n = \sqrt{13}$ rad/s (from $x'' + 13x = 0$). The driving frequency $\omega = 3 \ne \sqrt{13}$, so even without damping there would be no resonance here. With damping present, the characteristic roots have negative real part ($-2 \pm 3i$), so $\cos 3t$ and $\sin 3t$ alone are never in $\ker L$ and no $t$-multiplier is needed. True resonance (requiring the modification rule) occurs only in the undamped version $x'' + 13x = 10\cos\omega t$ when $\omega = \sqrt{13}$ rad/s. In the damped case the steady-state amplitude is finite for all $\omega$. (A1)阻尼振荡频率为3 rad/s（$-2 \pm 3i$ 的虚部），但弹簧质量系统的无阻尼固有频率为 $\omega_n = \sqrt{13}$ rad/s（来自 $x'' + 13x = 0$）。驱动频率 $\omega = 3 \ne \sqrt{13}$，故即使无阻尼此处也不发生共振。有阻尼时，特征根的实部为负（$-2 \pm 3i$），$\cos 3t$ 和 $\sin 3t$ 单独不属于 $\ker L$，无需乘以 $t$。真正的共振（需要修正规则）仅在无阻尼版本 $x'' + 13x = 10\cos\omega t$ 中，当 $\omega = \sqrt{13}$ rad/s 时发生。有阻尼时，稳态幅值对所有 $\omega$ 均有界。(A1)