Companion to the University-Style Practice Set大学风格练习题配套解析
Sections 1 to 6: superposition, characteristic equation (distinct real, complex, repeated roots), Wronskian and fundamental sets, reduction of orderCALC IV第 1 至 6 节：叠加原理、特征方程（不同实根、复数根、重复根）、朗斯基行列式与基本解组、降阶法CALC IV
(a) $y''-5y'+6y=0$, $y(0)=2$, $y'(0)=5$. (b) $y''+y'-12y=0$, $y(0)=1$, $y'(0)=-1$.（a）$y''-5y'+6y=0$，$y(0)=2$，$y'(0)=5$。（b）$y''+y'-12y=0$，$y(0)=1$，$y'(0)=-1$。
Characteristic equation: $r^{2} - 5r + 6 = 0$. Discriminant: $25 - 24 = 1 > 0$, so two distinct real roots. (M1)特征方程：$r^{2} - 5r + 6 = 0$。判别式：$25 - 24 = 1 > 0$，故有两个不同实根。(M1)
Factor: $(r-2)(r-3) = 0$, giving $r_1 = 2$, $r_2 = 3$. General solution: $y = c_1 e^{2t} + c_2 e^{3t}$.分解：$(r-2)(r-3) = 0$，得 $r_1 = 2$，$r_2 = 3$。通解：$y = c_1 e^{2t} + c_2 e^{3t}$。
Verification: $y'' - 5y' + 6y = (4c_1 e^{2t}+9c_2 e^{3t}) - 5(2c_1 e^{2t}+3c_2 e^{3t}) + 6(c_1 e^{2t}+c_2 e^{3t}) = 0$. (A1)验证：$y'' - 5y' + 6y = (4c_1 e^{2t}+9c_2 e^{3t}) - 5(2c_1 e^{2t}+3c_2 e^{3t}) + 6(c_1 e^{2t}+c_2 e^{3t}) = 0$。(A1)
Apply ICs. $y(0) = c_1 + c_2 = 2$. Then $y' = 2c_1 e^{2t} + 3c_2 e^{3t}$, so $y'(0) = 2c_1 + 3c_2 = 5$. (M1)代入初始条件：$y(0) = c_1 + c_2 = 2$。又 $y' = 2c_1 e^{2t} + 3c_2 e^{3t}$，故 $y'(0) = 2c_1 + 3c_2 = 5$。(M1)
Subtract twice the first equation from the second: $(2c_1+3c_2)-2(c_1+c_2)=5-4$, giving $c_2 = 1$, hence $c_1 = 1$. Particular solution: $y = e^{2t} + e^{3t}$. (A1)用第二式减去第一式的两倍：$(2c_1+3c_2)-2(c_1+c_2)=5-4$，得 $c_2 = 1$，故 $c_1 = 1$。特解：$y = e^{2t} + e^{3t}$。(A1)
Characteristic equation: $r^{2} + r - 12 = 0$. Discriminant: $1 + 48 = 49 > 0$. Factor: $(r-3)(r+4) = 0$, so $r_1 = 3$, $r_2 = -4$. (M1)特征方程：$r^{2} + r - 12 = 0$。判别式：$1 + 48 = 49 > 0$。分解：$(r-3)(r+4) = 0$，得 $r_1 = 3$，$r_2 = -4$。(M1)
General solution: $y = c_1 e^{3t} + c_2 e^{-4t}$. Substitution confirms it satisfies the ODE. (A1)通解：$y = c_1 e^{3t} + c_2 e^{-4t}$。代入验证满足方程。(A1)
$y(0) = c_1 + c_2 = 1$; $y' = 3c_1 e^{3t} - 4c_2 e^{-4t}$, so $y'(0) = 3c_1 - 4c_2 = -1$. (M1)$y(0) = c_1 + c_2 = 1$；$y' = 3c_1 e^{3t} - 4c_2 e^{-4t}$，故 $y'(0) = 3c_1 - 4c_2 = -1$。(M1)
System: $c_1 + c_2 = 1$ and $3c_1 - 4c_2 = -1$. Multiply the first by $4$ and add: $7c_1 = 3$, so $c_1 = \tfrac{3}{7}$ and $c_2 = 1 - \tfrac{3}{7} = \tfrac{4}{7}$. Check: $y'(0) = 3\cdot\tfrac{3}{7} - 4\cdot\tfrac{4}{7} = \tfrac{9-16}{7} = -1$. Particular solution: $y = \tfrac{3}{7}e^{3t} + \tfrac{4}{7}e^{-4t}$. (A1)方程组：$c_1 + c_2 = 1$ 且 $3c_1 - 4c_2 = -1$。第一式乘以 $4$ 后相加：$7c_1 = 3$，故 $c_1 = \tfrac{3}{7}$，$c_2 = 1 - \tfrac{3}{7} = \tfrac{4}{7}$。验证：$y'(0) = 3\cdot\tfrac{3}{7} - 4\cdot\tfrac{4}{7} = \tfrac{9-16}{7} = -1$。特解：$y = \tfrac{3}{7}e^{3t} + \tfrac{4}{7}e^{-4t}$。(A1)
(a) $y''+2y'+5y=0$, $y(0)=3$, $y'(0)=1$. (b) $y''-4y'+13y=0$, $y(0)=0$, $y'(0)=6$.（a）$y''+2y'+5y=0$，$y(0)=3$，$y'(0)=1$。（b）$y''-4y'+13y=0$，$y(0)=0$，$y'(0)=6$。
Characteristic equation: $r^{2}+2r+5=0$. Discriminant: $4 - 20 = -16 < 0$, so complex roots. (M1)特征方程：$r^{2}+2r+5=0$。判别式：$4 - 20 = -16 < 0$，故为复数根。(M1)
Roots: $r = \tfrac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i$, so $\lambda = -1$, $\mu = 2$. General solution: $y = e^{-t}(c_1\cos 2t + c_2\sin 2t)$. (A1)根：$r = \tfrac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i$，故 $\lambda = -1$，$\mu = 2$。通解：$y = e^{-t}(c_1\cos 2t + c_2\sin 2t)$。(A1)
The ODE is satisfied by structure (Euler's formula). Apply ICs: $y(0)=c_1=3$. (M1)方程由结构满足（欧拉公式）。代入初始条件：$y(0)=c_1=3$。(M1)
Differentiate: $y' = -e^{-t}(c_1\cos 2t + c_2\sin 2t) + e^{-t}(-2c_1\sin 2t + 2c_2\cos 2t)$. At $t=0$: $y'(0) = -c_1 + 2c_2 = -3 + 2c_2 = 1$, so $c_2 = 2$. Particular solution: $y = e^{-t}(3\cos 2t + 2\sin 2t)$. Verify: $y(0)=3$, $y'(0)=-3+4=1$. (A1)求导：$y' = -e^{-t}(c_1\cos 2t + c_2\sin 2t) + e^{-t}(-2c_1\sin 2t + 2c_2\cos 2t)$。在 $t=0$ 处：$y'(0) = -c_1 + 2c_2 = -3 + 2c_2 = 1$，故 $c_2 = 2$。特解：$y = e^{-t}(3\cos 2t + 2\sin 2t)$。验证：$y(0)=3$，$y'(0)=-3+4=1$。(A1)
Characteristic equation: $r^{2}-4r+13=0$. Discriminant: $16-52=-36<0$. Roots: $r = 2\pm 3i$, so $\lambda=2$, $\mu=3$. (M1)特征方程：$r^{2}-4r+13=0$。判别式：$16-52=-36<0$。根：$r = 2\pm 3i$，故 $\lambda=2$，$\mu=3$。(M1)
General solution: $y = e^{2t}(c_1\cos 3t + c_2\sin 3t)$. (A1)通解：$y = e^{2t}(c_1\cos 3t + c_2\sin 3t)$。(A1)
$y(0) = c_1 = 0$. $y' = 2e^{2t}(c_1\cos 3t + c_2\sin 3t) + e^{2t}(-3c_1\sin 3t + 3c_2\cos 3t)$. At $t=0$: $y'(0) = 2c_1 + 3c_2 = 0 + 3c_2 = 6$, so $c_2 = 2$. (M1)$y(0) = c_1 = 0$。$y' = 2e^{2t}(c_1\cos 3t + c_2\sin 3t) + e^{2t}(-3c_1\sin 3t + 3c_2\cos 3t)$。在 $t=0$ 处：$y'(0) = 2c_1 + 3c_2 = 0 + 3c_2 = 6$，故 $c_2 = 2$。(M1)
Particular solution: $y = 2e^{2t}\sin 3t$. Verify: $y(0)=0$, $y'(0)=2\cdot 3=6$. (A1)特解：$y = 2e^{2t}\sin 3t$。验证：$y(0)=0$，$y'(0)=2\cdot 3=6$。(A1)
(a) $y_1 = e^{-t}\cos 2t$, $y_2 = e^{-t}\sin 2t$ for $y''+2y'+5y=0$. (b) $y_1 = e^{3t}$, $y_2 = te^{3t}$ for $y''-6y'+9y=0$.（a）$y_1 = e^{-t}\cos 2t$，$y_2 = e^{-t}\sin 2t$，对应 $y''+2y'+5y=0$。（b）$y_1 = e^{3t}$，$y_2 = te^{3t}$，对应 $y''-6y'+9y=0$。
Compute $y_1' = -e^{-t}\cos 2t - 2e^{-t}\sin 2t = e^{-t}(-\cos 2t - 2\sin 2t)$ and $y_2' = -e^{-t}\sin 2t + 2e^{-t}\cos 2t = e^{-t}(-\sin 2t + 2\cos 2t)$. (M1)计算 $y_1' = -e^{-t}\cos 2t - 2e^{-t}\sin 2t = e^{-t}(-\cos 2t - 2\sin 2t)$ 和 $y_2' = -e^{-t}\sin 2t + 2e^{-t}\cos 2t = e^{-t}(-\sin 2t + 2\cos 2t)$。(M1)
$$ W = y_1 y_2' - y_2 y_1' = e^{-t}\cos 2t \cdot e^{-t}(-\sin 2t+2\cos 2t) - e^{-t}\sin 2t \cdot e^{-t}(-\cos 2t-2\sin 2t). $$Expand: $e^{-2t}[-\cos 2t\sin 2t + 2\cos^{2}2t + \sin 2t\cos 2t + 2\sin^{2}2t] = e^{-2t}\cdot 2(\cos^{2}2t+\sin^{2}2t) = 2e^{-2t}$. (A1)展开：$e^{-2t}[-\cos 2t\sin 2t + 2\cos^{2}2t + \sin 2t\cos 2t + 2\sin^{2}2t] = e^{-2t}\cdot 2(\cos^{2}2t+\sin^{2}2t) = 2e^{-2t}$。(A1)
Since $W = 2e^{-2t} \ne 0$ for all $t$, the pair is linearly independent and forms a fundamental set for $y''+2y'+5y=0$. (R1)由于对所有 $t$ 都有 $W = 2e^{-2t} \ne 0$，故该解对线性无关，构成 $y''+2y'+5y=0$ 的基本解组。(R1)
$y_1 = e^{3t}$, $y_1' = 3e^{3t}$; $y_2 = te^{3t}$, $y_2' = e^{3t} + 3te^{3t}$. (M1)$y_1 = e^{3t}$，$y_1' = 3e^{3t}$；$y_2 = te^{3t}$，$y_2' = e^{3t} + 3te^{3t}$。(M1)
$$ W = e^{3t}(e^{3t}+3te^{3t}) - te^{3t}\cdot 3e^{3t} = e^{6t} + 3te^{6t} - 3te^{6t} = e^{6t}. $$(A1) Since $W = e^{6t} > 0$ for all $t$, the pair is linearly independent and forms a fundamental set. (R1)(A1) 由于对所有 $t$ 都有 $W = e^{6t} > 0$，故该解对线性无关，构成基本解组。(R1)
$y''+6y'+9y=0$, $y(0)=4$, $y'(0)=-6$. (a) General solution and Wronskian. (b) Particular solution verified by substitution.$y''+6y'+9y=0$，$y(0)=4$，$y'(0)=-6$。（a）通解与朗斯基行列式。（b）代入验证特解。
Characteristic equation: $r^{2}+6r+9=0$. Discriminant: $36-36=0$, a repeated root. Factor: $(r+3)^{2}=0$, so $r=-3$ (double). (M1)特征方程：$r^{2}+6r+9=0$。判别式：$36-36=0$，为重根。分解：$(r+3)^{2}=0$，故 $r=-3$（二重根）。(M1)
The two independent solutions are $y_1 = e^{-3t}$ and $y_2 = te^{-3t}$. General solution: $y = (c_1+c_2 t)e^{-3t}$. (A1)两个线性无关解为 $y_1 = e^{-3t}$ 和 $y_2 = te^{-3t}$。通解：$y = (c_1+c_2 t)e^{-3t}$。(A1)
Wronskian: $y_1' = -3e^{-3t}$, $y_2' = e^{-3t} - 3te^{-3t}$. $W = e^{-3t}(e^{-3t}-3te^{-3t}) - te^{-3t}(-3e^{-3t}) = e^{-6t} - 3te^{-6t} + 3te^{-6t} = e^{-6t} \ne 0$. (A1)朗斯基行列式：$y_1' = -3e^{-3t}$，$y_2' = e^{-3t} - 3te^{-3t}$。$W = e^{-3t}(e^{-3t}-3te^{-3t}) - te^{-3t}(-3e^{-3t}) = e^{-6t} - 3te^{-6t} + 3te^{-6t} = e^{-6t} \ne 0$。(A1)
$y(0) = c_1 = 4$. Differentiate: $y' = c_2 e^{-3t} + (c_1+c_2 t)(-3)e^{-3t}$. At $t=0$: $y'(0) = c_2 - 3c_1 = c_2 - 12 = -6$, so $c_2 = 6$. (M1)$y(0) = c_1 = 4$。求导：$y' = c_2 e^{-3t} + (c_1+c_2 t)(-3)e^{-3t}$。在 $t=0$ 处：$y'(0) = c_2 - 3c_1 = c_2 - 12 = -6$，故 $c_2 = 6$。(M1)
Particular solution: $y = (4+6t)e^{-3t}$. (A1)特解：$y = (4+6t)e^{-3t}$。(A1)
Verification: $y' = 6e^{-3t} - 3(4+6t)e^{-3t} = (6-12-18t)e^{-3t} = (-6-18t)e^{-3t}$. $y'' = (-18)e^{-3t} - (-6-18t)(3)e^{-3t} = (-18+18+54t)e^{-3t} = 54te^{-3t}$. Then $y''+6y'+9y = 54te^{-3t} + 6(-6-18t)e^{-3t} + 9(4+6t)e^{-3t} = (54t - 36 - 108t + 36 + 54t)e^{-3t} = 0$. (B1) Also $y(0)=4$ and $y'(0)=-6$ confirmed.验证：$y' = 6e^{-3t} - 3(4+6t)e^{-3t} = (6-12-18t)e^{-3t} = (-6-18t)e^{-3t}$。$y'' = (-18)e^{-3t} - (-6-18t)(3)e^{-3t} = (-18+18+54t)e^{-3t} = 54te^{-3t}$。则 $y''+6y'+9y = 54te^{-3t} + 6(-6-18t)e^{-3t} + 9(4+6t)e^{-3t} = (54t - 36 - 108t + 36 + 54t)e^{-3t} = 0$。(B1) 同时确认 $y(0)=4$，$y'(0)=-6$。
$L[y]=y''+p(t)y'+q(t)y$. (a) Prove $L[c_1 y_1+c_2 y_2]=c_1 L[y_1]+c_2 L[y_2]$. (b) Deduce the superposition principle and explain why it fails for $y''+(y')^{2}=0$.$L[y]=y''+p(t)y'+q(t)y$。（a）证明 $L[c_1 y_1+c_2 y_2]=c_1 L[y_1]+c_2 L[y_2]$。（b）由此推导叠加原理，并说明为何对 $y''+(y')^{2}=0$ 该原理不成立。
Let $u = c_1 y_1 + c_2 y_2$. Since differentiation is linear, $u' = c_1 y_1' + c_2 y_2'$ and $u'' = c_1 y_1'' + c_2 y_2''$. (M1)令 $u = c_1 y_1 + c_2 y_2$。由于微分运算是线性的，$u' = c_1 y_1' + c_2 y_2'$，$u'' = c_1 y_1'' + c_2 y_2''$。(M1)
$$ L[u] = u'' + p(t)u' + q(t)u = (c_1 y_1'' + c_2 y_2'') + p(t)(c_1 y_1' + c_2 y_2') + q(t)(c_1 y_1 + c_2 y_2). $$Collecting by subscript: $= c_1(y_1''+p(t)y_1'+q(t)y_1) + c_2(y_2''+p(t)y_2'+q(t)y_2) = c_1 L[y_1] + c_2 L[y_2]$. (A1) The key step is that multiplication by $p(t)$ and $q(t)$ distributes across the linear combination, which is why we require these coefficients to be functions of $t$ only, not of $y$. (R1)按下标合并：$= c_1(y_1''+p(t)y_1'+q(t)y_1) + c_2(y_2''+p(t)y_2'+q(t)y_2) = c_1 L[y_1] + c_2 L[y_2]$。(A1) 关键步骤在于 $p(t)$ 和 $q(t)$ 的乘法可分配到线性组合中，这正是要求这些系数仅为 $t$ 的函数而非 $y$ 的函数的原因。(R1)
If $L[y_1]=0$ and $L[y_2]=0$, then by part (a): $L[c_1 y_1+c_2 y_2] = c_1 \cdot 0 + c_2 \cdot 0 = 0$. (M1) So $c_1 y_1 + c_2 y_2$ is also a solution. (A1)若 $L[y_1]=0$ 且 $L[y_2]=0$，由（a）得：$L[c_1 y_1+c_2 y_2] = c_1 \cdot 0 + c_2 \cdot 0 = 0$。(M1) 故 $c_1 y_1 + c_2 y_2$ 也是解。(A1)
For $y'' + (y')^{2} = 0$, the analogous operator $N[y]=y''+(y')^{2}$ satisfies $N[c_1 y_1 + c_2 y_2] = c_1 y_1'' + c_2 y_2'' + (c_1 y_1' + c_2 y_2')^{2}$. The squared term expands to $c_1^{2}(y_1')^{2} + 2c_1 c_2 y_1' y_2' + c_2^{2}(y_2')^{2}$, which is not $c_1 N[y_1] + c_2 N[y_2] = c_1\bigl(y_1''+(y_1')^{2}\bigr) + c_2\bigl(y_2''+(y_2')^{2}\bigr)$: the extraneous cross term $2c_1 c_2 y_1' y_2'$ cannot be absorbed. Hence $N$ is not linear, and superposition fails. (R1)对于 $y'' + (y')^{2} = 0$，类比算子 $N[y]=y''+(y')^{2}$ 满足 $N[c_1 y_1 + c_2 y_2] = c_1 y_1'' + c_2 y_2'' + (c_1 y_1' + c_2 y_2')^{2}$。平方项展开为 $c_1^{2}(y_1')^{2} + 2c_1 c_2 y_1' y_2' + c_2^{2}(y_2')^{2}$，这不等于 $c_1 N[y_1] + c_2 N[y_2] = c_1\bigl(y_1''+(y_1')^{2}\bigr) + c_2\bigl(y_2''+(y_2')^{2}\bigr)$：多余的交叉项 $2c_1 c_2 y_1' y_2'$ 无法被吸收。故 $N$ 非线性，叠加原理失效。(R1)
(a) Dependent pair implies $W=0$. (b) Derive Abel's identity $W'=-p(t)W$. (c) Apply to $y''+4y'+4y=0$ and verify.（a）线性相关解对意味着 $W=0$。（b）推导阿贝尔恒等式 $W'=-p(t)W$。（c）应用于 $y''+4y'+4y=0$ 并验证。
Suppose $y_2 = k y_1$ for a constant $k$. Then $y_2' = k y_1'$. (M1)设 $y_2 = k y_1$，$k$ 为常数。则 $y_2' = k y_1'$。(M1)
$$ W(y_1,y_2) = y_1 y_2' - y_2 y_1' = y_1(ky_1') - (ky_1)y_1' = ky_1 y_1' - ky_1 y_1' = 0. $$(A1) Since $W=0$ at every $t$, in particular $W$ is identically zero.(A1) 由于对每个 $t$ 都有 $W=0$，特别地，$W$ 恒为零。
Differentiate $W = y_1 y_2' - y_2 y_1'$: (M1)对 $W = y_1 y_2' - y_2 y_1'$ 求导：(M1)
$$ W' = y_1' y_2' + y_1 y_2'' - y_2' y_1' - y_2 y_1'' = y_1 y_2'' - y_2 y_1''. $$Since $y_1$ and $y_2$ both solve $y'' + p(t)y' + q(t)y = 0$, we have $y_1'' = -p y_1' - q y_1$ and $y_2'' = -p y_2' - q y_2$. (M1) Substituting: (A1)由于 $y_1$ 和 $y_2$ 都满足 $y'' + p(t)y' + q(t)y = 0$，有 $y_1'' = -p y_1' - q y_1$ 和 $y_2'' = -p y_2' - q y_2$。(M1) 代入：(A1)
$$ W' = y_1(-py_2'-qy_2) - y_2(-py_1'-qy_1) = -p(y_1 y_2' - y_2 y_1') - q(y_1 y_2 - y_2 y_1) = -pW. $$The ODE $W' = -p(t)W$ is separable: $\dfrac{dW}{W} = -p(t)\,dt$. Integrating from $t_0$ to $t$: (M1)常微分方程 $W' = -p(t)W$ 是可分离的：$\dfrac{dW}{W} = -p(t)\,dt$。从 $t_0$ 到 $t$ 积分：(M1)
$$ W(t) = W(t_0)\,e^{-\int_{t_0}^{t}p(s)\,ds}. $$Since the exponential factor is never zero, $W(t)$ has the same sign as $W(t_0)$ everywhere on $I$. Hence $W$ is either identically zero (if $W(t_0)=0$) or never zero. (A1)由于指数因子从不为零，$W(t)$ 在 $I$ 上处处与 $W(t_0)$ 同号。故 $W$ 在 $I$ 上恒为零（当 $W(t_0)=0$ 时）或处处不为零。(A1)
Here $p(t)=4$, so Abel's identity gives $W(t) = Ce^{-4t}$ for some constant $C$. (M1)此处 $p(t)=4$，由阿贝尔恒等式得 $W(t) = Ce^{-4t}$，$C$ 为某常数。(M1)
The characteristic equation is $r^{2}+4r+4=(r+2)^{2}=0$, so $r=-2$ (double). The fundamental pair is $y_1=e^{-2t}$, $y_2=te^{-2t}$. (A1)特征方程为 $r^{2}+4r+4=(r+2)^{2}=0$，故 $r=-2$（二重根）。基本解对为 $y_1=e^{-2t}$，$y_2=te^{-2t}$。(A1)
Direct computation: $y_1'=-2e^{-2t}$, $y_2'=e^{-2t}-2te^{-2t}$. $W = e^{-2t}(e^{-2t}-2te^{-2t}) - te^{-2t}(-2e^{-2t}) = e^{-4t} - 2te^{-4t}+2te^{-4t} = e^{-4t}$. So $C=1$ in this case. (B1)直接计算：$y_1'=-2e^{-2t}$，$y_2'=e^{-2t}-2te^{-2t}$。$W = e^{-2t}(e^{-2t}-2te^{-2t}) - te^{-2t}(-2e^{-2t}) = e^{-4t} - 2te^{-4t}+2te^{-4t} = e^{-4t}$。故此情形下 $C=1$。(B1)
$t^{2}y''-3ty'+4y=0$ ($t>0$), $y_1=t^{2}$. (a) Verify $y_1$. (b) Substitute $y_2=v(t)t^{2}$ and reduce. (c) Solve for $v$ and confirm $\{y_1,y_2\}$ is a fundamental set.$t^{2}y''-3ty'+4y=0$（$t>0$），$y_1=t^{2}$。（a）验证 $y_1$。（b）代入 $y_2=v(t)t^{2}$ 并化简。（c）求解 $v$，确认 $\{y_1,y_2\}$ 为基本解组。
$y_1 = t^{2}$, $y_1' = 2t$, $y_1'' = 2$. Substituting: (M1)$y_1 = t^{2}$，$y_1' = 2t$，$y_1'' = 2$。代入：(M1)
$$ t^{2}(2) - 3t(2t) + 4t^{2} = 2t^{2} - 6t^{2} + 4t^{2} = 0. $$(A1) Confirmed.(A1) 验证成立。
Set $y_2 = v(t)t^{2}$. Then $y_2' = v't^{2} + 2tv$ and $y_2'' = v''t^{2} + 4tv' + 2v$. (M1)令 $y_2 = v(t)t^{2}$。则 $y_2' = v't^{2} + 2tv$，$y_2'' = v''t^{2} + 4tv' + 2v$。(M1)
Substitute into $t^{2}y''-3ty'+4y=0$: (M1)代入 $t^{2}y''-3ty'+4y=0$：(M1)
$$ t^{2}(v''t^{2}+4tv'+2v) - 3t(v't^{2}+2tv) + 4vt^{2} = 0. $$ $$ t^{4}v'' + 4t^{3}v' + 2t^{2}v - 3t^{3}v' - 6t^{2}v + 4t^{2}v = 0. $$ $$ t^{4}v'' + t^{3}v' + (2-6+4)t^{2}v = 0 \implies t^{4}v'' + t^{3}v' = 0. $$Divide by $t^{3}$ (valid for $t>0$): $tv'' + v' = 0$. (A1) Setting $w = v'$ gives the first-order ODE $tw' + w = 0$. (A1)除以 $t^{3}$（$t>0$ 时有效）：$tv'' + v' = 0$。(A1) 令 $w = v'$ 得一阶常微分方程 $tw' + w = 0$。(A1)
The equation $tw' + w = 0$ is separable: $\dfrac{dw}{w} = -\dfrac{dt}{t}$, so $\ln|w| = -\ln t + C_0$, giving $w = \dfrac{A}{t}$. (M1)方程 $tw' + w = 0$ 是可分离的：$\dfrac{dw}{w} = -\dfrac{dt}{t}$，故 $\ln|w| = -\ln t + C_0$，得 $w = \dfrac{A}{t}$。(M1)
Integrate: $v = A\ln t + B$. Taking $A=1$, $B=0$ (we need only one new $v$), $v = \ln t$, so $y_2 = t^{2}\ln t$. (A1)积分：$v = A\ln t + B$。取 $A=1$，$B=0$（只需一个新的 $v$），$v = \ln t$，故 $y_2 = t^{2}\ln t$。(A1)
Wronskian: $y_1=t^{2}$, $y_2=t^{2}\ln t$, $y_1'=2t$, $y_2'=2t\ln t+t$. (M1)朗斯基行列式：$y_1=t^{2}$，$y_2=t^{2}\ln t$，$y_1'=2t$，$y_2'=2t\ln t+t$。(M1)
$$ W = t^{2}(2t\ln t+t) - t^{2}\ln t\cdot 2t = 2t^{3}\ln t + t^{3} - 2t^{3}\ln t = t^{3}. $$Since $W = t^{3} > 0$ for $t>0$, the pair $\{t^{2},\, t^{2}\ln t\}$ is linearly independent and forms a fundamental set. (A1)由于 $t>0$ 时 $W = t^{3} > 0$，解对 $\{t^{2},\, t^{2}\ln t\}$ 线性无关，构成基本解组。(A1)
$y''+2y'+10y=0$, $y(0)=4$, $y'(0)=-2$. (a) Roots and general solution. (b) Apply ICs. (c) Amplitude-phase form. (d) Damping classification.$y''+2y'+10y=0$，$y(0)=4$，$y'(0)=-2$。（a）特征根与通解。（b）代入初始条件。（c）振幅相位形式。（d）阻尼分类。
Characteristic equation: $r^{2}+2r+10=0$. Discriminant: $4-40=-36<0$. Roots: $r = \dfrac{-2\pm\sqrt{-36}}{2} = -1\pm 3i$, so $\lambda=-1$, $\mu=3$. (M1)特征方程：$r^{2}+2r+10=0$。判别式：$4-40=-36<0$。根：$r = \dfrac{-2\pm\sqrt{-36}}{2} = -1\pm 3i$，故 $\lambda=-1$，$\mu=3$。(M1)
General solution: $y = e^{-t}(c_1\cos 3t + c_2\sin 3t)$. (A1)通解：$y = e^{-t}(c_1\cos 3t + c_2\sin 3t)$。(A1)
$y(0) = c_1 = 4$. Differentiate using the product rule: $y' = -e^{-t}(c_1\cos 3t+c_2\sin 3t) + e^{-t}(-3c_1\sin 3t+3c_2\cos 3t)$. (M1)$y(0) = c_1 = 4$。用乘积法则求导：$y' = -e^{-t}(c_1\cos 3t+c_2\sin 3t) + e^{-t}(-3c_1\sin 3t+3c_2\cos 3t)$。(M1)
At $t=0$: $y'(0) = -c_1 + 3c_2 = -4 + 3c_2 = -2$, so $3c_2 = 2$, giving $c_2 = \tfrac{2}{3}$. (A1) Particular solution: $y = e^{-t}\!\left(4\cos 3t + \tfrac{2}{3}\sin 3t\right)$. Verify: $y(0)=4$, $y'(0)=-4+3\cdot\tfrac{2}{3}=-4+2=-2$. (A1)在 $t=0$ 处：$y'(0) = -c_1 + 3c_2 = -4 + 3c_2 = -2$，故 $3c_2 = 2$，$c_2 = \tfrac{2}{3}$。(A1) 特解：$y = e^{-t}\!\left(4\cos 3t + \tfrac{2}{3}\sin 3t\right)$。验证：$y(0)=4$，$y'(0)=-4+3\cdot\tfrac{2}{3}=-4+2=-2$。(A1)
Write $4\cos 3t + \tfrac{2}{3}\sin 3t = R\cos(3t-\phi)$ where $R = \sqrt{4^{2}+(\tfrac{2}{3})^{2}} = \sqrt{16+\tfrac{4}{9}} = \sqrt{\tfrac{148}{9}} = \tfrac{2\sqrt{37}}{3}$ and $\tan\phi = \dfrac{2/3}{4} = \dfrac{1}{6}$, so $\phi = \arctan\tfrac{1}{6}$. (M1)将 $4\cos 3t + \tfrac{2}{3}\sin 3t = R\cos(3t-\phi)$，其中 $R = \sqrt{4^{2}+(\tfrac{2}{3})^{2}} = \sqrt{16+\tfrac{4}{9}} = \sqrt{\tfrac{148}{9}} = \tfrac{2\sqrt{37}}{3}$，$\tan\phi = \dfrac{2/3}{4} = \dfrac{1}{6}$，故 $\phi = \arctan\tfrac{1}{6}$。(M1)
The particular solution in amplitude-phase form is $y = \dfrac{2\sqrt{37}}{3}\,e^{-t}\cos\!\left(3t-\arctan\tfrac{1}{6}\right)$. (A1)振幅相位形式的特解为 $y = \dfrac{2\sqrt{37}}{3}\,e^{-t}\cos\!\left(3t-\arctan\tfrac{1}{6}\right)$。(A1)
The quasi-period (period of the oscillatory factor) is $T = \dfrac{2\pi}{\mu} = \dfrac{2\pi}{3}$. (A1)准周期（振荡因子的周期）为 $T = \dfrac{2\pi}{\mu} = \dfrac{2\pi}{3}$。(A1)
Since the discriminant $b^{2}-4mk = 4-40 < 0$ (or equivalently, the characteristic roots are complex with nonzero imaginary part), the system is underdamped. (M1) The factor $e^{-t}$ is a decaying envelope, so $y(t)\to 0$ as $t\to\infty$, while the $\cos$ factor produces oscillation with decreasing amplitude. (A1)由于判别式 $b^{2}-4mk = 4-40 < 0$（等价地，特征根为虚部非零的复数），系统为欠阻尼。(M1) 因子 $e^{-t}$ 是衰减包络，故 $y(t)\to 0$（当 $t\to\infty$），而 $\cos$ 因子产生振幅不断减小的振荡。(A1)
(a) $y''+5y'+4y=0$, $y(0)=2$, $y'(0)=0$ (overdamped). (b) $y''+4y'+4y=0$, $y(0)=1$, $y'(0)=-5$ (critically damped).（a）$y''+5y'+4y=0$，$y(0)=2$，$y'(0)=0$（过阻尼）。（b）$y''+4y'+4y=0$，$y(0)=1$，$y'(0)=-5$（临界阻尼）。
Characteristic equation: $r^{2}+5r+4=0$. Discriminant: $25-16=9>0$, two distinct real roots. Factor: $(r+1)(r+4)=0$, so $r_1=-1$, $r_2=-4$. (M1)特征方程：$r^{2}+5r+4=0$。判别式：$25-16=9>0$，两个不同实根。分解：$(r+1)(r+4)=0$，故 $r_1=-1$，$r_2=-4$。(M1)
General solution: $y = c_1 e^{-t} + c_2 e^{-4t}$. (A1)通解：$y = c_1 e^{-t} + c_2 e^{-4t}$。(A1)
Apply ICs: $y(0)=c_1+c_2=2$. $y'=-c_1 e^{-t}-4c_2 e^{-4t}$, so $y'(0)=-c_1-4c_2=0$, giving $c_1=-4c_2$. (M1) Substituting: $-4c_2+c_2=2 \Rightarrow -3c_2=2 \Rightarrow c_2=-\tfrac{2}{3}$, $c_1=\tfrac{8}{3}$. (A1)代入初始条件：$y(0)=c_1+c_2=2$。$y'=-c_1 e^{-t}-4c_2 e^{-4t}$，故 $y'(0)=-c_1-4c_2=0$，得 $c_1=-4c_2$。(M1) 代入：$-4c_2+c_2=2 \Rightarrow -3c_2=2 \Rightarrow c_2=-\tfrac{2}{3}$，$c_1=\tfrac{8}{3}$。(A1)
Particular solution: $y=\tfrac{8}{3}e^{-t}-\tfrac{2}{3}e^{-4t}$. Both exponentials decay to zero, and since $c_1>0$ and $c_2<0$, $y$ starts at $2$, and $y(t) > 0$ for all $t>0$ (check $y=\tfrac{2}{3}(4e^{-t}-e^{-4t})$; the factor is $4e^{-t}-e^{-4t}=e^{-t}(4-e^{-3t})$, which is positive for all $t>0$ since $e^{-3t}<4$). Hence $y\to 0$ monotonically. (R1)特解：$y=\tfrac{8}{3}e^{-t}-\tfrac{2}{3}e^{-4t}$。两个指数项均衰减至零，由于 $c_1>0$ 且 $c_2<0$，$y$ 从 $2$ 出发，对所有 $t>0$ 均有 $y(t) > 0$（验证：$y=\tfrac{2}{3}(4e^{-t}-e^{-4t})$，因子为 $4e^{-t}-e^{-4t}=e^{-t}(4-e^{-3t})$，由于 $e^{-3t}<4$ 对所有 $t>0$ 成立，故为正）。因此 $y$ 单调趋向 $0$。(R1)
Characteristic equation: $r^{2}+4r+4=(r+2)^{2}=0$, repeated root $r=-2$. General solution: $y=(c_1+c_2 t)e^{-2t}$. (M1)特征方程：$r^{2}+4r+4=(r+2)^{2}=0$，重根 $r=-2$。通解：$y=(c_1+c_2 t)e^{-2t}$。(M1)
$y(0)=c_1=1$. Differentiate: $y'=c_2 e^{-2t}+(c_1+c_2 t)(-2)e^{-2t}$. At $t=0$: $y'(0)=c_2-2c_1=c_2-2=-5$, so $c_2=-3$. (M1) Particular solution: $y=(1-3t)e^{-2t}$. (A1)$y(0)=c_1=1$。求导：$y'=c_2 e^{-2t}+(c_1+c_2 t)(-2)e^{-2t}$。在 $t=0$ 处：$y'(0)=c_2-2c_1=c_2-2=-5$，故 $c_2=-3$。(M1) 特解：$y=(1-3t)e^{-2t}$。(A1)
Verify: $y(0)=1$, $y'(0)=-3-2(1)=-5$. To find the zero crossing: set $1-3t=0$, giving $t=\tfrac{1}{3}>0$. (R1) For $0
$y''+4y=0$ on $[0,L]$. (a) General solution. (b) BCs $y(0)=y(L)=0$: nontrivial solutions iff $L=n\pi/2$. (c) Contrast IVP uniqueness with BVP.$[0,L]$ 上的 $y''+4y=0$。（a）通解。（b）边界条件 $y(0)=y(L)=0$：当且仅当 $L=n\pi/2$ 时有非平凡解。（c）对比初值问题的唯一性与边值问题的情况。
Characteristic equation: $r^{2}+4=0$, so $r=\pm 2i$, giving $\lambda=0$, $\mu=2$. (M1) General solution: $y = c_1\cos 2x + c_2\sin 2x$. (A1)特征方程：$r^{2}+4=0$，故 $r=\pm 2i$，得 $\lambda=0$，$\mu=2$。(M1) 通解：$y = c_1\cos 2x + c_2\sin 2x$。(A1)
Apply $y(0)=0$: $c_1\cos 0 + c_2\sin 0 = c_1 = 0$. (M1) So $y = c_2\sin 2x$. Apply $y(L)=0$: $c_2\sin 2L = 0$. (M1)代入 $y(0)=0$：$c_1\cos 0 + c_2\sin 0 = c_1 = 0$。(M1) 故 $y = c_2\sin 2x$。代入 $y(L)=0$：$c_2\sin 2L = 0$。(M1)
For a nontrivial solution we need $c_2\ne 0$, so $\sin 2L = 0$, i.e. $2L = n\pi$ for positive integers $n$, i.e. $L = \tfrac{n\pi}{2}$. (A1)要得到非平凡解，需 $c_2\ne 0$，故 $\sin 2L = 0$，即对正整数 $n$ 有 $2L = n\pi$，即 $L = \tfrac{n\pi}{2}$。(A1)
The corresponding nontrivial solutions are $y_n = \sin 2x$ (with $c_2$ arbitrary and nonzero) for each $n$; equivalently $y = A\sin\!\left(\tfrac{n\pi x}{L}\right)$. These are the eigenfunctions (modes) of the BVP. (A1)对应的非平凡解为 $y_n = \sin 2x$（$c_2$ 为任意非零值），对每个 $n$；等价地 $y = A\sin\!\left(\tfrac{n\pi x}{L}\right)$。这些是边值问题的特征函数（振型）。(A1)
For an IVP, both conditions $y(t_0)=\alpha$ and $y'(t_0)=\beta$ are imposed at the same point $t_0$. The system of equations for $c_1, c_2$ has coefficient determinant exactly $W(t_0)$, which is nonzero when $\{y_1,y_2\}$ is a fundamental set, guaranteeing a unique solution. (M1)对于初值问题，条件 $y(t_0)=\alpha$ 和 $y'(t_0)=\beta$ 均在同一点 $t_0$ 处施加。关于 $c_1, c_2$ 的方程组的系数行列式恰为 $W(t_0)$，当 $\{y_1,y_2\}$ 为基本解组时不为零，保证唯一解。(M1)
For a BVP, the two conditions are at different points: $y(0)=0$ and $y(L)=0$. After applying $y(0)=0$ to get $c_1=0$, the remaining equation $c_2\sin 2L=0$ may have the trivial solution $c_2=0$ only (when $\sin 2L\ne 0$, giving only the zero solution) or infinitely many solutions (when $\sin 2L=0$). The Wronskian at $t_0$ controls the initial-data system at one point; it says nothing about whether conditions at a remote second point are compatible. (A1)对于边值问题，两个条件在不同点处：$y(0)=0$ 和 $y(L)=0$。由 $y(0)=0$ 得 $c_1=0$ 后，剩余方程 $c_2\sin 2L=0$ 可能只有平凡解 $c_2=0$（当 $\sin 2L\ne 0$ 时，仅有零解），也可能有无穷多解（当 $\sin 2L=0$ 时）。$t_0$ 处的朗斯基行列式控制的是单点处的初始数据方程组，对于远端第二点处的条件是否相容，它什么也说明不了。(A1)