Sections 1 to 7: power-series manipulations, ordinary vs singular points, series solutions and recurrence relations, Frobenius method and indicial equations, Euler, improved Euler (Heun), RK4, and global error order第1至7节：幂级数运算、常点与奇点、级数解与递推关系、Frobenius方法与指标方程、欧拉法、改进欧拉法（Heun法）、RK4及全局误差阶CALC IV
Name:姓名：Date:日期：
PART I · CORE TECHNIQUES第一部分 · 核心技巧Computational fluency · 28 marks计算熟练度 · 28分
Power Series, Ordinary Points, and Numerical Steps幂级数、常点与数值步骤
Show all working. For series problems, shift indices explicitly before equating coefficients. For numerical problems, carry at least four significant figures at each intermediate stage and compare with the exact value where requested.请写出完整解题过程。对于级数问题，在比较系数之前明确移位指标。对于数值问题，每一中间步骤保留至少四位有效数字，并与精确值比较（如题目要求）。
Q1MEDIUMCOREordinary vs singular points; radius of convergence常点与奇点；收敛半径[8 marks]
Consider the differential equation $(x^2 - 9)\,y'' + x\,y' + y = 0$. Write it in standard form $y'' + p(x)\,y' + q(x)\,y = 0$ and answer the following.考虑微分方程 $(x^2 - 9)\,y'' + x\,y' + y = 0$。将其化为标准形式 $y'' + p(x)\,y' + q(x)\,y = 0$，并回答以下问题。
(a)Identify all singular points and classify each as regular or irregular. Justify your answer.找出所有奇点，并将每个奇点分类为正则或非正则奇点。请说明理由。[3]
(b)State whether $x_0 = 0$ is an ordinary point. If it is, give a lower bound on the radius of convergence of a power-series solution centred there, and justify the bound.判断 $x_0 = 0$ 是否为常点。若是，给出以该点为中心的幂级数解的收敛半径下界，并说明理由。[3]
(c)State whether $x_0 = 1$ is an ordinary point, giving a lower bound on the radius of convergence for a series solution centred there.判断 $x_0 = 1$ 是否为常点，并给出以该点为中心的级数解的收敛半径下界。[2]
Q2MEDIUMCOREpower-series solution about an ordinary point; recurrence relation常点处的幂级数解；递推关系[10 marks]
Seek a power-series solution $y = \displaystyle\sum_{n=0}^{\infty} a_n x^n$ to the initial value problem
$$ y'' + y = 0, \quad y(0)=1,\quad y'(0)=0. $$对以下初值问题求幂级数解 $y = \displaystyle\sum_{n=0}^{\infty} a_n x^n$：
$$ y'' + y = 0, \quad y(0)=1,\quad y'(0)=0. $$
(a)Substitute $y = \sum a_n x^n$ and $y'' = \sum (n+2)(n+1)a_{n+2}\,x^n$ into the ODE. Collect all terms under a single summation and derive the recurrence relation for $a_{n+2}$ in terms of $a_n$.将 $y = \sum a_n x^n$ 和 $y'' = \sum (n+2)(n+1)a_{n+2}\,x^n$ 代入方程，将所有项合并到同一求和号下，推导 $a_{n+2}$ 关于 $a_n$ 的递推关系。[4]
(b)Apply the initial conditions to fix $a_0$ and $a_1$, then compute $a_2, a_3, a_4, a_5$ from the recurrence. Write the solution as a series up to and including the $x^5$ term.利用初始条件确定 $a_0$ 和 $a_1$，再由递推关系计算 $a_2, a_3, a_4, a_5$。将解写成包含至 $x^5$ 项的级数形式。[4]
(c)Identify the closed-form function your series represents and verify it satisfies the original ODE and both initial conditions.指出该级数所代表的闭合形式函数，并验证它满足原方程及两个初始条件。[2]
Q3MEDIUMCOREEuler steps and Heun step; comparing one-step errors欧拉步骤与Heun步骤；比较单步误差[10 marks]
Consider $y' = y - x$, $y(0) = 2$. The exact solution is $y = x + 1 + e^x$.考虑 $y' = y - x$，$y(0) = 2$。精确解为 $y = x + 1 + e^x$。
(a)Apply two steps of Euler's method with $h = 0.2$ to estimate $y(0.4)$. Show the slope and update at each step.用步长 $h = 0.2$ 执行两步欧拉法，估计 $y(0.4)$。请在每步中写出斜率和更新值。[4]
(b)Apply one step of the improved Euler (Heun) method with $h = 0.4$ to estimate $y(0.4)$. Show the predictor value and the corrector computation explicitly.用步长 $h = 0.4$ 执行一步改进欧拉法（Heun法），估计 $y(0.4)$。请明确写出预测值和校正计算过程。[4]
(c)Compute the exact value $y(0.4)$ and state the absolute error of each estimate from (a) and (b). Which is more accurate?计算精确值 $y(0.4)$，并给出 (a) 和 (b) 各估计值的绝对误差。哪种方法更精确？[2]
PART II · DEFINITIONS AND PROOF第二部分 · 定义与证明Rigorous arguments · 26 marks严格论证 · 26分
Recurrence Relations, Convergence, and Error Theory递推关系、收敛性与误差理论
These items are graded on the logic of the argument, not just the final answer. Derive recurrence relations by index-shifting explicitly before equating coefficients. In error-order arguments, state which Taylor-series term is responsible for the leading error.本部分按论证逻辑评分，而非仅看最终答案。推导递推关系时，请在比较系数前明确进行指标移位。在讨论误差阶时，请指出哪个泰勒级数项对应主误差项。
Q4HARDPROOFderiving the recurrence for $y'' - xy = 0$ (Airy's equation)推导 $y'' - xy = 0$（Airy方程）的递推关系[10 marks]
Consider Airy's equation $y'' - xy = 0$ with a series solution $y = \displaystyle\sum_{n=0}^{\infty} a_n x^n$ centred at $x_0 = 0$.考虑Airy方程 $y'' - xy = 0$，以 $x_0 = 0$ 为中心求级数解 $y = \displaystyle\sum_{n=0}^{\infty} a_n x^n$。
(a)Write $y''$ as a power series in $x^n$ and write $xy$ as a power series in $x^n$. Explain why the two series start at different powers and how to handle the constant term ($n=0$) of the combined equation separately.将 $y''$ 写成 $x^n$ 的幂级数，将 $xy$ 也写成 $x^n$ 的幂级数。解释为何两个级数的起始幂次不同，以及如何单独处理合并方程中的常数项（$n=0$）。[4]
(b)Derive the recurrence relation $a_{n+3}$ in terms of $a_n$ for $n \ge 0$ after accounting for the $n=0$ case. (Note: because $y''$ starts at index $n$ and $xy$ shifts the index, the recurrence links coefficients three apart.) Compute $a_3, a_4, a_5, a_6$ in terms of $a_0$ and $a_1$ (recall $a_2 = 0$).在处理 $n=0$ 情形后，推导 $n \ge 0$ 时 $a_{n+3}$ 关于 $a_n$ 的递推关系。（注：由于 $y''$ 从指标 $n$ 开始而 $xy$ 使指标移位，递推关系将相差三位的系数联系起来。）用 $a_0$ 和 $a_1$ 表示 $a_3, a_4, a_5, a_6$（已知 $a_2 = 0$）。[4]
(c)Write out the general solution as a sum of two independent series and explain why the radius of convergence is infinite.将通解写成两个线性无关级数之和，并解释收敛半径为无穷大的原因。[2]
Consider $2x\,y'' + y' + y = 0$.考虑 $2x\,y'' + y' + y = 0$。
(a)Show that $x = 0$ is a regular singular point. (Write the equation in standard form, identify $p$ and $q$, and apply the definition of a regular singular point.)证明 $x = 0$ 是正则奇点。（将方程化为标准形式，确定 $p$ 和 $q$，并应用正则奇点的定义。）[3]
(b)Substitute the Frobenius ansatz $y = x^r \displaystyle\sum_{n=0}^{\infty} a_n x^n$ (with $a_0 \ne 0$) and isolate the coefficient of the lowest power $x^{r-1}$ to derive and solve the indicial equation. State both roots.代入Frobenius试探解 $y = x^r \displaystyle\sum_{n=0}^{\infty} a_n x^n$（其中 $a_0 \ne 0$），提取最低次幂 $x^{r-1}$ 的系数，推导并求解指标方程。写出两个根。[3]
(c)Using the larger indicial root, write down the general recurrence relation for $a_n$ in terms of $a_{n-1}$ and compute $a_1$ and $a_2$ in terms of $a_0$.使用较大的指标根，写出 $a_n$ 关于 $a_{n-1}$ 的一般递推关系，并用 $a_0$ 计算 $a_1$ 和 $a_2$。[2]
Q6HARDPROOFderiving Euler's update from the tangent line; local truncation error从切线推导欧拉更新公式；局部截断误差[8 marks]
Let $y(x)$ be the exact solution to $y' = f(x,y)$, $y(x_0) = y_0$.设 $y(x)$ 是 $y' = f(x,y)$，$y(x_0) = y_0$ 的精确解。
(a)Expand $y(x_0 + h)$ in a Taylor series about $x_0$ through the $h^2$ term and use $y'(x_0) = f(x_0,y_0)$ to show that the Euler update $y_1 = y_0 + h\,f(x_0,y_0)$ agrees with the exact solution up to, but not including, the $h^2$ term. Hence state the local truncation error of Euler's method.将 $y(x_0 + h)$ 在 $x_0$ 处展开至 $h^2$ 项的泰勒级数，利用 $y'(x_0) = f(x_0,y_0)$ 证明欧拉更新公式 $y_1 = y_0 + h\,f(x_0,y_0)$ 与精确解在 $h^2$ 项之前一致。由此给出欧拉法的局部截断误差。[4]
(b)Explain, using the result of (a), why the global error of Euler's method over an interval of fixed length is $O(h)$ even though the local error is $O(h^2)$. Your explanation must state how many steps are taken and how the local errors accumulate.利用 (a) 的结果，解释为何欧拉法在固定长度区间上的全局误差为 $O(h)$，而局部误差为 $O(h^2)$。你的解释必须说明总步数以及局部误差如何累积。[2]
(c)An RK4 run over the interval $[0,1]$ with step size $h = 0.1$ produces a global error of $3.2 \times 10^{-6}$. Estimate the global error when $h$ is halved to $0.05$. Justify your answer from the order of RK4.在区间 $[0,1]$ 上用步长 $h = 0.1$ 运行RK4，得到全局误差为 $3.2 \times 10^{-6}$。估计步长减半至 $h = 0.05$ 时的全局误差。请根据RK4的误差阶说明理由。[2]
PART III · APPLICATIONS AND SYNTHESIS第三部分 · 应用与综合Extended problems · 28 marks综合应用题 · 28分
Euler Equations, Full RK4, and Numerical Comparison Tables欧拉方程、完整RK4与数值比较表
Set up each problem cleanly before computing. In numerical tables, carry four decimal places throughout. In series problems, verify the recurrence term-by-term against the ODE before declaring the answer.计算前请先清晰建立模型。数值表中全程保留四位小数。级数问题中，在给出最终答案前，请逐项代入方程验证递推关系。
Q7HARDAPPLIEDEuler equation; indicial equation; general solution by root type欧拉方程；指标方程；按根的类型求通解[8 marks]
(a)Write the equation in standard Euler form and read off the parameters $\alpha$ and $\beta$. Substitute the trial solution $y = x^r$ and derive the indicial equation. Find both roots $r_1$ and $r_2$.将方程化为标准欧拉形式，读出参数 $\alpha$ 和 $\beta$。代入试探解 $y = x^r$，推导指标方程，求出两根 $r_1$ 和 $r_2$。[3]
(b)Write the general solution for $x > 0$ and state the Wronskian condition that confirms linear independence of the two basis solutions.写出 $x > 0$ 时的通解，并给出确认两个基础解线性无关的朗斯基行列式条件。[3]
(a)Carry out one step of Euler's method with $h = 0.4$ to estimate $y(0.4)$.用步长 $h = 0.4$ 执行一步欧拉法，估计 $y(0.4)$。[2]
(b)Carry out one complete step of the classical RK4 method with $h = 0.4$. Compute $k_1, k_2, k_3, k_4$ in turn, showing the $x$-argument and $y$-argument of $f$ at each stage, then compute $y_1$.用步长 $h = 0.4$ 执行一步完整的经典RK4方法。依次计算 $k_1, k_2, k_3, k_4$，在每个阶段写出 $f$ 的 $x$ 自变量和 $y$ 自变量，然后计算 $y_1$。[6]
(c)Compute the exact value $y(0.4) = 3e^{0.4} - (0.4)^2 - 2(0.4) - 2$ (use $e^{0.4} \approx 1.49182$) and find the absolute error of the Euler estimate and the RK4 estimate. Comment on the difference in accuracy.计算精确值 $y(0.4) = 3e^{0.4} - (0.4)^2 - 2(0.4) - 2$（取 $e^{0.4} \approx 1.49182$），求欧拉估计值和RK4估计值的绝对误差。评述两种方法的精度差异。[2]
(a)Derive the recurrence relation linking $a_{n+2}$ to $a_n$ for $n \ge 0$. State $a_0$ and $a_1$ from the initial conditions.推导 $n \ge 0$ 时 $a_{n+2}$ 关于 $a_n$ 的递推关系。由初始条件给出 $a_0$ 和 $a_1$。[3]
(b)Compute $a_2, a_4, a_6$ from the recurrence and write the even solution up to the $x^6$ term. Verify that $a_3 = a_5 = 0$ (since $a_1 = 0$).由递推关系计算 $a_2, a_4, a_6$，写出包含至 $x^6$ 项的偶次解。验证 $a_3 = a_5 = 0$（由于 $a_1 = 0$）。[2]
(c)Show that the even solution terminates (becomes a polynomial) and identify it explicitly.证明偶次解终止（变为多项式），并明确写出该多项式。[1]
Q10HARDAPPLIEDEuler vs RK4 table; step-size and global error scaling欧拉法与RK4比较表；步长与全局误差缩放[4 marks]
(a)Complete the table below by carrying out three steps of Euler's method and three steps of RK4, filling in the $y_n$ column for each method. Then compute the absolute error at each step.填写下表：执行三步欧拉法和三步RK4，为每种方法填写 $y_n$ 列，然后计算每步的绝对误差。
$n$
$x_n$
Euler $y_n$
Euler error
RK4 $y_n$
RK4 error
Exact $e^{-2x_n}$
0
0.0
1.0000
0
1.0000
0
1.0000
1
0.2
0.6703
2
0.4
0.4493
3
0.6
0.3012
[3]
(b)If the step size $h = 0.2$ is halved to $h = 0.1$, by approximately what factor does the Euler global error decrease, and by what factor does the RK4 global error decrease?若步长从 $h = 0.2$ 减半至 $h = 0.1$，欧拉法的全局误差大约降低为原来的几分之几？RK4的全局误差又降低为原来的几分之几？[1]