Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 7: reduction to systems, matrix form, real distinct eigenvalues, complex eigenvalues, repeated eigenvalues, phase portraits and stability, matrix exponential and variation of parameters第 1 至 7 节：化为方程组、矩阵形式、实不同特征值、复特征值、重复特征值、相平面与稳定性、矩阵指数与常数变易法CALC IV
Reduce $y''' - 6y'' + 11y' - 6y = e^{t}$ to a first-order system; find the eigenvalues of the companion matrix; state why they equal the characteristic roots.将 $y''' - 6y'' + 11y' - 6y = e^{t}$ 化为一阶方程组，求伴随矩阵的特征值，并说明为何它们等于特征根。
Set $x_1 = y$, $x_2 = y'$, $x_3 = y''$. Differentiating: (M1)令 $x_1 = y$，$x_2 = y'$，$x_3 = y''$，微分得：(M1)
$$ x_1' = x_2, \qquad x_2' = x_3, \qquad x_3' = y''' = 6y'' - 11y' + 6y + e^{t} = 6x_3 - 11x_2 + 6x_1 + e^{t}. $$In matrix form $\mathbf{x}' = A\mathbf{x} + \mathbf{g}(t)$ with (M1)写成矩阵形式 $\mathbf{x}' = A\mathbf{x} + \mathbf{g}(t)$：(M1)
$$ A = \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 6 & -11 & 6\end{pmatrix}, \qquad \mathbf{g}(t) = \begin{pmatrix}0\\0\\e^{t}\end{pmatrix}. $$The matrix $A$ is the companion matrix of the characteristic polynomial $\lambda^3 - 6\lambda^2 + 11\lambda - 6$. (A1)矩阵 $A$ 是特征多项式 $\lambda^3 - 6\lambda^2 + 11\lambda - 6$ 的伴随矩阵。(A1)
For a companion matrix the characteristic polynomial is $\det(A - \lambda I) = -(\lambda^3 - 6\lambda^2 + 11\lambda - 6)$, which can be verified by cofactor expansion along the last row. (M1)对伴随矩阵，特征多项式为 $\det(A - \lambda I) = -(\lambda^3 - 6\lambda^2 + 11\lambda - 6)$，可沿最后一行展开余子式验证。(M1)
Test $\lambda = 1$: $1 - 6 + 11 - 6 = 0$. So $(\lambda - 1)$ is a factor. Dividing: $\lambda^3 - 6\lambda^2 + 11\lambda - 6 = (\lambda - 1)(\lambda^2 - 5\lambda + 6) = (\lambda - 1)(\lambda - 2)(\lambda - 3)$. (M1)代入 $\lambda = 1$：$1 - 6 + 11 - 6 = 0$，故 $(\lambda - 1)$ 为因子。除法得：$\lambda^3 - 6\lambda^2 + 11\lambda - 6 = (\lambda - 1)(\lambda^2 - 5\lambda + 6) = (\lambda - 1)(\lambda - 2)(\lambda - 3)$。(M1)
The eigenvalues of $A$ are $\lambda = 1, 2, 3$. (A1)$A$ 的特征值为 $\lambda = 1, 2, 3$。(A1)
The companion matrix $A$ is constructed precisely so that the characteristic polynomial $\det(A - \lambda I)$ equals (up to sign) the operator polynomial $\lambda^3 - 6\lambda^2 + 11\lambda - 6$ of the scalar ODE: the first $n-1$ rows encode the trivial equations $x_i' = x_{i+1}$, and the last row encodes the scalar ODE, so expanding the determinant recovers the same polynomial. (R1·A1)伴随矩阵 $A$ 的构造方式使得其特征多项式 $\det(A - \lambda I)$ （差一个符号）等于标量常微分方程的算子多项式 $\lambda^3 - 6\lambda^2 + 11\lambda - 6$：前 $n-1$ 行编码平凡方程 $x_i' = x_{i+1}$，最后一行编码标量常微分方程，展开行列式即还原同一多项式。(R1·A1)
For $A = \begin{pmatrix}1&3\\3&1\end{pmatrix}$, $\mathbf{x}(0)=(4,2)^T$: find eigenvalues and eigenvectors; solve the IVP; classify the portrait.对于 $A = \begin{pmatrix}1&3\\3&1\end{pmatrix}$，$\mathbf{x}(0)=(4,2)^T$：求特征值和特征向量，求解初值问题，并对相平面进行分类。
The characteristic equation is (M1)特征方程为 (M1)
$$ \det(A - \lambda I) = (1 - \lambda)^2 - 9 = \lambda^2 - 2\lambda - 8 = (\lambda - 4)(\lambda + 2) = 0. $$So $\lambda_1 = 4$ and $\lambda_2 = -2$. (A1)故 $\lambda_1 = 4$，$\lambda_2 = -2$。(A1)
For $\lambda_1 = 4$: $(A - 4I) = \begin{pmatrix}-3&3\\3&-3\end{pmatrix}$; the null space is spanned by $\mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix}$. Verify: $A\mathbf{v}_1 = \begin{pmatrix}4\\4\end{pmatrix} = 4\mathbf{v}_1$. (M1)对 $\lambda_1 = 4$：$(A - 4I) = \begin{pmatrix}-3&3\\3&-3\end{pmatrix}$，零空间由 $\mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix}$ 张成。验证：$A\mathbf{v}_1 = \begin{pmatrix}4\\4\end{pmatrix} = 4\mathbf{v}_1$。(M1)
For $\lambda_2 = -2$: $(A + 2I) = \begin{pmatrix}3&3\\3&3\end{pmatrix}$; null space spanned by $\mathbf{v}_2 = \begin{pmatrix}1\\-1\end{pmatrix}$. Verify: $A\mathbf{v}_2 = \begin{pmatrix}-2\\2\end{pmatrix} = -2\mathbf{v}_2$. (A1)对 $\lambda_2 = -2$：$(A + 2I) = \begin{pmatrix}3&3\\3&3\end{pmatrix}$，零空间由 $\mathbf{v}_2 = \begin{pmatrix}1\\-1\end{pmatrix}$ 张成。验证：$A\mathbf{v}_2 = \begin{pmatrix}-2\\2\end{pmatrix} = -2\mathbf{v}_2$。(A1)
General solution: $\mathbf{x}(t) = c_1 e^{4t}\begin{pmatrix}1\\1\end{pmatrix} + c_2 e^{-2t}\begin{pmatrix}1\\-1\end{pmatrix}$. (M1)通解：$\mathbf{x}(t) = c_1 e^{4t}\begin{pmatrix}1\\1\end{pmatrix} + c_2 e^{-2t}\begin{pmatrix}1\\-1\end{pmatrix}$。(M1)
Apply $\mathbf{x}(0) = (4,2)^T$: $c_1 + c_2 = 4$ and $c_1 - c_2 = 2$. Adding: $2c_1 = 6$, so $c_1 = 3$, $c_2 = 1$. (A1)代入 $\mathbf{x}(0) = (4,2)^T$：$c_1 + c_2 = 4$，$c_1 - c_2 = 2$。相加得 $2c_1 = 6$，故 $c_1 = 3$，$c_2 = 1$。(A1)
$$ \mathbf{x}(t) = 3e^{4t}\begin{pmatrix}1\\1\end{pmatrix} + e^{-2t}\begin{pmatrix}1\\-1\end{pmatrix}. $$One eigenvalue is positive ($\lambda_1 = 4$) and one is negative ($\lambda_2 = -2$): the origin is a saddle, always unstable. (A1)一个特征值为正（$\lambda_1 = 4$），一个为负（$\lambda_2 = -2$）：原点为鞍点，始终不稳定。(A1)
As $t \to \infty$ the $e^{-2t}$ mode decays to zero and the $e^{4t}$ mode dominates, so the particular solution grows without bound along the direction $\begin{pmatrix}1\\1\end{pmatrix}$. (R1)当 $t \to \infty$ 时，$e^{-2t}$ 模态衰减至零，$e^{4t}$ 模态主导，特解沿方向 $\begin{pmatrix}1\\1\end{pmatrix}$ 无界增长。(R1)
Solve $\mathbf{x}' = \begin{pmatrix}-1&2\\-2&-1\end{pmatrix}\mathbf{x}$; write two real solutions; classify the portrait.求解 $\mathbf{x}' = \begin{pmatrix}-1&2\\-2&-1\end{pmatrix}\mathbf{x}$，写出两个实数解，并对相平面进行分类。
Characteristic equation: $\det(A - \lambda I) = (-1-\lambda)^2 + 4 = \lambda^2 + 2\lambda + 5 = 0$. By the quadratic formula, $\lambda = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i$. (M1) So $\alpha = -1$, $\beta = 2$. (A1)特征方程：$\det(A - \lambda I) = (-1-\lambda)^2 + 4 = \lambda^2 + 2\lambda + 5 = 0$。由求根公式，$\lambda = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i$。(M1) 故 $\alpha = -1$，$\beta = 2$。(A1)
For $\lambda = -1 + 2i$: $(A - \lambda I)\mathbf{v} = \begin{pmatrix}-2i&2\\-2&-2i\end{pmatrix}\mathbf{v} = \mathbf{0}$. Row 1 gives $-2iv_1 + 2v_2 = 0$, so $v_2 = iv_1$. Taking $v_1 = 1$: $\mathbf{v} = \begin{pmatrix}1\\i\end{pmatrix} = \underbrace{\begin{pmatrix}1\\0\end{pmatrix}}_{\mathbf{a}} + i\underbrace{\begin{pmatrix}0\\1\end{pmatrix}}_{\mathbf{b}}$. (M1)对 $\lambda = -1 + 2i$：$(A - \lambda I)\mathbf{v} = \begin{pmatrix}-2i&2\\-2&-2i\end{pmatrix}\mathbf{v} = \mathbf{0}$。第一行给出 $-2iv_1 + 2v_2 = 0$，故 $v_2 = iv_1$。取 $v_1 = 1$：$\mathbf{v} = \begin{pmatrix}1\\i\end{pmatrix} = \underbrace{\begin{pmatrix}1\\0\end{pmatrix}}_{\mathbf{a}} + i\underbrace{\begin{pmatrix}0\\1\end{pmatrix}}_{\mathbf{b}}$。(M1)
Verification: $A\mathbf{v} = \begin{pmatrix}-1+2i\\-2-i\end{pmatrix}$ and $\lambda\mathbf{v} = (-1+2i)\begin{pmatrix}1\\i\end{pmatrix} = \begin{pmatrix}-1+2i\\-2-i\end{pmatrix}$. They agree. (A1)验证：$A\mathbf{v} = \begin{pmatrix}-1+2i\\-2-i\end{pmatrix}$，$\lambda\mathbf{v} = (-1+2i)\begin{pmatrix}1\\i\end{pmatrix} = \begin{pmatrix}-1+2i\\-2-i\end{pmatrix}$，两者一致。(A1)
The two real linearly independent solutions are两个实线性无关解为
$$ \mathbf{x}_1(t) = e^{-t}\!\left(\mathbf{a}\cos 2t - \mathbf{b}\sin 2t\right) = e^{-t}\begin{pmatrix}\cos 2t\\-\sin 2t\end{pmatrix}, $$ $$ \mathbf{x}_2(t) = e^{-t}\!\left(\mathbf{a}\sin 2t + \mathbf{b}\cos 2t\right) = e^{-t}\begin{pmatrix}\sin 2t\\\cos 2t\end{pmatrix}. $$General solution: $\mathbf{x}(t) = c_1\mathbf{x}_1(t) + c_2\mathbf{x}_2(t)$.通解：$\mathbf{x}(t) = c_1\mathbf{x}_1(t) + c_2\mathbf{x}_2(t)$。
Here $p = \operatorname{tr} A = -2 < 0$, $q = \det A = 1 + 4 = 5 > 0$, and $\Delta = 4 - 20 = -16 < 0$: complex eigenvalues with negative real part. The origin is a stable spiral (spiral sink). (A1·R1)此处 $p = \operatorname{tr} A = -2 < 0$，$q = \det A = 1 + 4 = 5 > 0$，$\Delta = 4 - 20 = -16 < 0$：复特征值且实部为负。原点为稳定螺旋（螺旋汇）。(A1·R1)
Solve $\mathbf{x}' = \begin{pmatrix}-2&1\\0&-2\end{pmatrix}\mathbf{x}$; show the matrix is defective; find the generalized eigenvector and general solution.求解 $\mathbf{x}' = \begin{pmatrix}-2&1\\0&-2\end{pmatrix}\mathbf{x}$，证明矩阵有亏损，求广义特征向量和通解。
$\det(A - \lambda I) = (-2-\lambda)^2 = 0$ gives the double eigenvalue $\lambda = -2$. (M1)$\det(A - \lambda I) = (-2-\lambda)^2 = 0$ 给出二重特征值 $\lambda = -2$。(M1)
Compute $A - \lambda I = A + 2I = \begin{pmatrix}0&1\\0&0\end{pmatrix}$. Solving $(A+2I)\mathbf{v} = \mathbf{0}$ forces $v_2 = 0$ with $v_1$ free, so $\ker(A+2I) = \operatorname{span}\left\{\begin{pmatrix}1\\0\end{pmatrix}\right\}$, which is one-dimensional. Because the geometric multiplicity $(1)$ is less than the algebraic multiplicity $(2)$, the matrix is defective. The eigenvector is $\mathbf{v} = \begin{pmatrix}1\\0\end{pmatrix}$. (A1)计算 $A - \lambda I = A + 2I = \begin{pmatrix}0&1\\0&0\end{pmatrix}$。求解 $(A+2I)\mathbf{v} = \mathbf{0}$ 可得 $v_2 = 0$，$v_1$ 自由，故 $\ker(A+2I) = \operatorname{span}\left\{\begin{pmatrix}1\\0\end{pmatrix}\right\}$，为一维空间。几何重数（$1$）小于代数重数（$2$），故矩阵有亏损。特征向量为 $\mathbf{v} = \begin{pmatrix}1\\0\end{pmatrix}$。(A1)
Solve $(A + 2I)\mathbf{w} = \mathbf{v}$, that is $\begin{pmatrix}0&1\\0&0\end{pmatrix}\mathbf{w} = \begin{pmatrix}1\\0\end{pmatrix}$. This forces $w_2 = 1$ with $w_1$ free; take $\mathbf{w} = \begin{pmatrix}0\\1\end{pmatrix}$. (M1) Verify: $(A+2I)\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = \mathbf{v}$. (A1)求解 $(A + 2I)\mathbf{w} = \mathbf{v}$，即 $\begin{pmatrix}0&1\\0&0\end{pmatrix}\mathbf{w} = \begin{pmatrix}1\\0\end{pmatrix}$。这要求 $w_2 = 1$，$w_1$ 自由；取 $\mathbf{w} = \begin{pmatrix}0\\1\end{pmatrix}$。(M1) 验证：$(A+2I)\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = \mathbf{v}$。(A1)
The general solution is (M1)通解为 (M1)
$$ \mathbf{x}(t) = c_1 e^{-2t}\begin{pmatrix}1\\0\end{pmatrix} + c_2 e^{-2t}\!\left[t\begin{pmatrix}1\\0\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix}\right] = e^{-2t}\begin{pmatrix}c_1 + c_2 t\\ c_2\end{pmatrix}. $$Since $\lambda = -2 < 0$, the exponential decay dominates the polynomial growth in $t$: all trajectories tend to the origin. The origin is a stable improper (degenerate) node. (A1)由于 $\lambda = -2 < 0$，指数衰减主导 $t$ 的多项式增长：所有轨迹趋向原点。原点为稳定退化（非正常）结点。(A1)
(a) Substitute $\mathbf{x}=e^{\lambda t}\mathbf{v}$ into $\mathbf{x}'=A\mathbf{x}$ and derive $A\mathbf{v}=\lambda\mathbf{v}$. (b) Prove eigenvectors for distinct eigenvalues are linearly independent.(a) 将 $\mathbf{x}=e^{\lambda t}\mathbf{v}$ 代入 $\mathbf{x}'=A\mathbf{x}$，推导 $A\mathbf{v}=\lambda\mathbf{v}$。(b) 证明对应不同特征值的特征向量线性无关。
Compute the left-hand side using the scalar chain rule for $e^{\lambda t}$: (M1)利用 $e^{\lambda t}$ 的链式法则计算左端：(M1)
$$ \mathbf{x}'(t) = \frac{d}{dt}\!\left(e^{\lambda t}\mathbf{v}\right) = \lambda e^{\lambda t}\mathbf{v}. $$Compute the right-hand side (M1)计算右端 (M1)
$$ A\mathbf{x}(t) = A\!\left(e^{\lambda t}\mathbf{v}\right) = e^{\lambda t}A\mathbf{v}, $$using the fact that $A$ is constant so it passes through the scalar factor $e^{\lambda t}$. Setting $\mathbf{x}' = A\mathbf{x}$: (A1)利用 $A$ 为常数矩阵，可提出标量因子 $e^{\lambda t}$。令 $\mathbf{x}' = A\mathbf{x}$：(A1)
$$ \lambda e^{\lambda t}\mathbf{v} = e^{\lambda t}A\mathbf{v}. $$Since $e^{\lambda t} \neq 0$ for all $t \in \mathbb{R}$, we may divide both sides by $e^{\lambda t}$: (R1)由于对所有 $t \in \mathbb{R}$ 均有 $e^{\lambda t} \neq 0$，两边可除以 $e^{\lambda t}$：(R1)
$$ A\mathbf{v} = \lambda\mathbf{v}. $$This is exactly the eigenvalue equation, so $\mathbf{v}$ must be a nonzero eigenvector of $A$ with eigenvalue $\lambda$.这正是特征方程，故 $\mathbf{v}$ 必为 $A$ 关于特征值 $\lambda$ 的非零特征向量。
We prove the statement by strong induction on $k$. For $k = 1$: a single nonzero eigenvector is trivially independent. Assume the statement holds for any set of $k-1$ eigenvectors with distinct eigenvalues. (M1)对 $k$ 进行强归纳法证明。$k = 1$ 时：单个非零特征向量显然线性无关。设命题对任意 $k-1$ 个具有不同特征值的特征向量成立。(M1)
Suppose $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$. Apply the operator $(A - \lambda_k I)$ to both sides: (M1)设 $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$。对两边施加算子 $(A - \lambda_k I)$：(M1)
$$ \sum_{j=1}^{k} c_j (A - \lambda_k I)\mathbf{v}_j = \mathbf{0}. $$Since $A\mathbf{v}_j = \lambda_j\mathbf{v}_j$, we have $(A - \lambda_k I)\mathbf{v}_j = (\lambda_j - \lambda_k)\mathbf{v}_j$. For $j = k$ this is $(\lambda_k - \lambda_k)\mathbf{v}_k = \mathbf{0}$, so the $k$th term drops out: (A1)由 $A\mathbf{v}_j = \lambda_j\mathbf{v}_j$，得 $(A - \lambda_k I)\mathbf{v}_j = (\lambda_j - \lambda_k)\mathbf{v}_j$。当 $j = k$ 时为 $(\lambda_k - \lambda_k)\mathbf{v}_k = \mathbf{0}$，故第 $k$ 项消去：(A1)
$$ \sum_{j=1}^{k-1} c_j(\lambda_j - \lambda_k)\mathbf{v}_j = \mathbf{0}. $$By the induction hypothesis, $\mathbf{v}_1, \dots, \mathbf{v}_{k-1}$ are independent, so $c_j(\lambda_j - \lambda_k) = 0$ for each $j < k$. Since the eigenvalues are distinct, $\lambda_j \neq \lambda_k$, so $c_j = 0$ for all $j < k$. Substituting back: $c_k\mathbf{v}_k = \mathbf{0}$, and since $\mathbf{v}_k \neq \mathbf{0}$, also $c_k = 0$. Thus all coefficients are zero, proving independence. (R1)由归纳假设，$\mathbf{v}_1, \dots, \mathbf{v}_{k-1}$ 线性无关，故对每个 $j < k$ 有 $c_j(\lambda_j - \lambda_k) = 0$。由于特征值互不相同，$\lambda_j \neq \lambda_k$，故对所有 $j < k$ 有 $c_j = 0$。代回得 $c_k\mathbf{v}_k = \mathbf{0}$，又 $\mathbf{v}_k \neq \mathbf{0}$，故 $c_k = 0$。所有系数均为零，线性无关得证。(R1)
(a) Derive $W' = (\operatorname{tr} A)W$ and the explicit Wronskian formula. (b) Use it to prove $W \neq 0$ everywhere iff $W(t_0) \neq 0$ for some $t_0$.(a) 推导 $W' = (\operatorname{tr} A)W$ 及 Wronskian 行列式的显式公式。(b) 利用该公式证明：$W$ 处处非零当且仅当对某 $t_0$ 有 $W(t_0) \neq 0$。
The Wronskian $W = \det\Psi$ where $\Psi$ has columns $\mathbf{x}^{(j)}$. Use the fact that the derivative of a determinant is the sum over $j$ of the determinant with the $j$th column replaced by $(\mathbf{x}^{(j)})'$: (M1)Wronskian 行列式 $W = \det\Psi$，$\Psi$ 的列为 $\mathbf{x}^{(j)}$。利用行列式的求导法则：$W'$ 为各项之和，每项将第 $j$ 列替换为 $(\mathbf{x}^{(j)})'$：(M1)
$$ W'(t) = \sum_{j=1}^{n} \det\!\bigl[\mathbf{x}^{(1)} \;\cdots\; (\mathbf{x}^{(j)})' \;\cdots\; \mathbf{x}^{(n)}\bigr]. $$Since each $(\mathbf{x}^{(j)})' = A\mathbf{x}^{(j)}$, the $j$th column in the $j$th determinant is $A\mathbf{x}^{(j)} = \sum_k a_{kj}\mathbf{e}_k$ where $\mathbf{e}_k$ are standard basis vectors. Writing $A\mathbf{x}^{(j)}$ as a sum of columns of $\Psi$ weighted by the entries of $A$, every term where a column of $\Psi$ appears in two places in the same determinant produces a matrix with two equal columns and hence vanishes. (M1) The only surviving term for the $j$th summand comes from the $j$th row of $A$, contributing $a_{jj}\det\Psi = a_{jj}W$. Summing over $j$: (A1)由于每列满足 $(\mathbf{x}^{(j)})' = A\mathbf{x}^{(j)}$，第 $j$ 行列式中第 $j$ 列为 $A\mathbf{x}^{(j)} = \sum_k a_{kj}\mathbf{e}_k$（$\mathbf{e}_k$ 为标准基向量）。将 $A\mathbf{x}^{(j)}$ 写为 $\Psi$ 列的加权和时，同一行列式中某列出现两次的项因两列相同而消失。(M1) 第 $j$ 项仅有来自 $A$ 的第 $j$ 行的贡献，即 $a_{jj}\det\Psi = a_{jj}W$。对 $j$ 求和：(A1)
$$ W'(t) = \left(\sum_{j=1}^n a_{jj}\right)W(t) = \operatorname{tr}(A)\,W(t). $$This is a separable scalar ODE. Solving with initial value $W(t_0)$: (A1)这是一个可分离的标量常微分方程。以初值 $W(t_0)$ 求解：(A1)
$$ W(t) = W(t_0)\exp\!\bigl(\operatorname{tr}(A)(t - t_0)\bigr). $$The exponential function $\exp(\operatorname{tr}(A)(t-t_0))$ is never zero for any real $t$. (M1) Therefore from the formula: (A1)指数函数 $\exp(\operatorname{tr}(A)(t-t_0))$ 对任意实数 $t$ 恒非零。(M1) 由此公式：(A1)
$$ W(t) = 0 \;\iff\; W(t_0) = 0. $$If $W(t_0) \neq 0$ for even one $t_0$, then $W(t_0) \cdot (\text{nonzero}) = W(t) \neq 0$ for all $t$. (M1)若存在某 $t_0$ 使 $W(t_0) \neq 0$，则 $W(t_0) \cdot (\text{非零}) = W(t) \neq 0$ 对所有 $t$ 成立。(M1)
Conversely, if $W(t^*) = 0$ at any single $t^*$, then the formula forces $W(t_0) = 0$ for that same $t_0$, so $W \equiv 0$ everywhere. This means checking independence at a single instant is both necessary and sufficient: one evaluation tells the whole story. (R1)反之，若存在某 $t^*$ 使 $W(t^*) = 0$，则公式迫使对应 $t_0$ 有 $W(t_0) = 0$，故 $W \equiv 0$ 处处成立。这意味着在单一时刻检验线性无关性既是必要条件也是充分条件：一次计算即可说明全局情况。(R1)
(a) Classify four matrices using $(p,q,\Delta)$ only. (b) Prove asymptotic stability iff $p<0$ and $q>0$.(a) 仅用 $(p,q,\Delta)$ 对四个矩阵进行分类。(b) 证明渐近稳定当且仅当 $p<0$ 且 $q>0$。
(i) $A = \begin{pmatrix}2&-1\\5&-4\end{pmatrix}$: $p = 2 + (-4) = -2$, $q = (2)(-4) - (-1)(5) = -8 + 5 = -3 < 0$.
Since $q < 0$ the eigenvalues are real with opposite signs: the origin is a saddle (always unstable). (A1+A1)由于 $q < 0$，特征值为实数且异号：原点为鞍点（始终不稳定）。(A1+A1)
(ii) $A = \begin{pmatrix}-3&2\\-2&-1\end{pmatrix}$: $p = -4$, $q = (-3)(-1) - (2)(-2) = 3 + 4 = 7 > 0$, $\Delta = 16 - 28 = -12 < 0$.
Complex eigenvalues ($\Delta < 0$) with $p < 0$: stable spiral. (A1+A1)复特征值（$\Delta < 0$）且 $p < 0$：稳定螺旋。(A1+A1)
(iii) $A = \begin{pmatrix}1&-2\\2&1\end{pmatrix}$: $p = 2$, $q = 1 + 4 = 5 > 0$, $\Delta = 4 - 20 = -16 < 0$.
Complex eigenvalues with $p > 0$: unstable spiral. (A1+A1)复特征值且 $p > 0$：不稳定螺旋。(A1+A1)
(iv) $A = \begin{pmatrix}0&-2\\2&0\end{pmatrix}$: $p = 0$, $q = 0 + 4 = 4 > 0$, $\Delta = 0 - 16 = -16 < 0$.
Complex eigenvalues with $p = 0$: purely imaginary, so the origin is a center (neutrally stable). (A1+A1)复特征值且 $p = 0$：纯虚特征值，原点为中心（中性稳定）。(A1+A1)
The eigenvalues of a $2\times 2$ matrix satisfy $\lambda_1 + \lambda_2 = p$ and $\lambda_1\lambda_2 = q$. Asymptotic stability requires both eigenvalues to have strictly negative real part. (M1)$2\times 2$ 矩阵的特征值满足 $\lambda_1 + \lambda_2 = p$，$\lambda_1\lambda_2 = q$。渐近稳定要求两个特征值的实部均严格为负。(M1)
Case 1: real eigenvalues情形 1：实特征值 ($\Delta \geq 0$). Both eigenvalues real. Both negative requires their sum $p = \lambda_1 + \lambda_2 < 0$ and their product $q = \lambda_1\lambda_2 > 0$ (positive product of two real numbers of the same sign). (M1)两个特征值均为实数。两者均为负数要求其和 $p = \lambda_1 + \lambda_2 < 0$，其积 $q = \lambda_1\lambda_2 > 0$（同号两实数之积为正）。(M1)
Case 2: complex eigenvalues情形 2：复特征值 ($\Delta < 0$). The eigenvalues are $\lambda = \alpha \pm i\beta$ with $\alpha = p/2$ and $q = \alpha^2 + \beta^2 > 0$ automatically. Both have real part $\alpha$, which is negative exactly when $p = 2\alpha < 0$. (M1)特征值为 $\lambda = \alpha \pm i\beta$，其中 $\alpha = p/2$，$q = \alpha^2 + \beta^2 > 0$ 自动成立。两者实部均为 $\alpha$，恰好在 $p = 2\alpha < 0$ 时为负。(M1)
In both cases: both eigenvalues have negative real part $\iff$ $p < 0$ and $q > 0$. (A1·R1)两种情形下：两个特征值实部均为负 $\iff$ $p < 0$ 且 $q > 0$。(A1·R1)
Solve $\mathbf{x}' = \begin{pmatrix}-1&2\\-2&-1\end{pmatrix}\mathbf{x}$, $\mathbf{x}(0)=(1,0)^T$; verify; classify the portrait and describe the trajectory.求解 $\mathbf{x}' = \begin{pmatrix}-1&2\\-2&-1\end{pmatrix}\mathbf{x}$，$\mathbf{x}(0)=(1,0)^T$；验证答案；对相平面进行分类并描述轨迹。
The characteristic equation is $\lambda^2 + 2\lambda + 5 = 0$, giving $\lambda = -1 \pm 2i$ (as in Q3). So $\alpha = -1$, $\beta = 2$. (M1·A1)特征方程为 $\lambda^2 + 2\lambda + 5 = 0$，给出 $\lambda = -1 \pm 2i$（与第 3 题相同）。故 $\alpha = -1$，$\beta = 2$。(M1·A1)
For $\lambda = -1 + 2i$ the eigenvector is $\mathbf{v} = \begin{pmatrix}1\\i\end{pmatrix} = \mathbf{a} + i\mathbf{b}$ with $\mathbf{a} = \begin{pmatrix}1\\0\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}0\\1\end{pmatrix}$ (verified in Q3). (M1) The two real solutions are (A1)对 $\lambda = -1 + 2i$，特征向量为 $\mathbf{v} = \begin{pmatrix}1\\i\end{pmatrix} = \mathbf{a} + i\mathbf{b}$，其中 $\mathbf{a} = \begin{pmatrix}1\\0\end{pmatrix}$，$\mathbf{b} = \begin{pmatrix}0\\1\end{pmatrix}$（已在第 3 题验证）。(M1) 两个实数解为：(A1)
$$ \mathbf{x}_1(t) = e^{-t}\begin{pmatrix}\cos 2t\\-\sin 2t\end{pmatrix}, \qquad \mathbf{x}_2(t) = e^{-t}\begin{pmatrix}\sin 2t\\\cos 2t\end{pmatrix}. $$At $t = 0$: $c_1\mathbf{x}_1(0) + c_2\mathbf{x}_2(0) = c_1\begin{pmatrix}1\\0\end{pmatrix} + c_2\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}$. (M1) So $c_1 = 1$, $c_2 = 0$. (A1)在 $t = 0$ 时：$c_1\mathbf{x}_1(0) + c_2\mathbf{x}_2(0) = c_1\begin{pmatrix}1\\0\end{pmatrix} + c_2\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}$。(M1) 故 $c_1 = 1$，$c_2 = 0$。(A1)
$$ \mathbf{x}(t) = e^{-t}\begin{pmatrix}\cos 2t\\-\sin 2t\end{pmatrix}. $$Differentiate component-wise: (M1)逐分量求导：(M1)
$$ x_1' = e^{-t}(-\cos 2t - 2\sin 2t), \qquad x_2' = e^{-t}(\sin 2t - 2\cos 2t). $$Check row 1 of $A\mathbf{x}$: $-x_1 + 2x_2 = e^{-t}(-\cos 2t + 2(-\sin 2t)) = e^{-t}(-\cos 2t - 2\sin 2t) = x_1'$. (B1)验证 $A\mathbf{x}$ 第一行：$-x_1 + 2x_2 = e^{-t}(-\cos 2t + 2(-\sin 2t)) = e^{-t}(-\cos 2t - 2\sin 2t) = x_1'$。(B1)
Check row 2: $-2x_1 - x_2 = e^{-t}(-2\cos 2t - (-\sin 2t)) = e^{-t}(-2\cos 2t + \sin 2t) = x_2'$. Both rows check out.验证第二行：$-2x_1 - x_2 = e^{-t}(-2\cos 2t - (-\sin 2t)) = e^{-t}(-2\cos 2t + \sin 2t) = x_2'$。两行均验证正确。
With $\alpha = -1 < 0$ and $\beta = 2 \neq 0$ the portrait is a stable spiral (spiral sink). (A1) The trajectory starting at $(1,0)^T$ spirals inward toward the origin with angular frequency $\beta = 2$ radians per unit time, and $\|\mathbf{x}(t)\| = e^{-t} \to 0$ as $t \to \infty$, so every trajectory eventually reaches the origin. (R1)由 $\alpha = -1 < 0$，$\beta = 2 \neq 0$，相平面为稳定螺旋（螺旋汇）。(A1) 从 $(1,0)^T$ 出发的轨迹以角频率 $\beta = 2$（弧度每单位时间）向内螺旋趋向原点，$\|\mathbf{x}(t)\| = e^{-t} \to 0$（$t \to \infty$），故所有轨迹最终均趋向原点。(R1)
For $x_1'=-3x_1+x_2$, $x_2'=2x_1-2x_2$, $x_1(0)=3$, $x_2(0)=0$: matrix form, portrait classification, eigenvalue solution, long-time behaviour.对于 $x_1'=-3x_1+x_2$，$x_2'=2x_1-2x_2$，$x_1(0)=3$，$x_2(0)=0$：写成矩阵形式，对相平面分类，用特征值法求解，并描述长时行为。
The system is $\mathbf{x}' = A\mathbf{x}$ with $A = \begin{pmatrix}-3&1\\2&-2\end{pmatrix}$. (M1)方程组为 $\mathbf{x}' = A\mathbf{x}$，其中 $A = \begin{pmatrix}-3&1\\2&-2\end{pmatrix}$。(M1)
Compute $p = \operatorname{tr} A = -5$, $q = \det A = (-3)(-2) - (1)(2) = 6 - 2 = 4 > 0$, $\Delta = p^2 - 4q = 25 - 16 = 9 > 0$. (A1)计算 $p = \operatorname{tr} A = -5$，$q = \det A = (-3)(-2) - (1)(2) = 6 - 2 = 4 > 0$，$\Delta = p^2 - 4q = 25 - 16 = 9 > 0$。(A1)
$q > 0$ and $\Delta > 0$ means real eigenvalues of the same sign; since $p < 0$ both are negative: the origin is a stable node (sink). (R1)$q > 0$ 且 $\Delta > 0$ 表明实特征值同号；由于 $p < 0$，两者均为负：原点为稳定结点（汇）。(R1)
$\det(A - \lambda I) = (-3-\lambda)(-2-\lambda) - 2 = \lambda^2 + 5\lambda + 4 = (\lambda+1)(\lambda+4) = 0$. So $\lambda_1 = -1$ and $\lambda_2 = -4$. (M1·A1)$\det(A - \lambda I) = (-3-\lambda)(-2-\lambda) - 2 = \lambda^2 + 5\lambda + 4 = (\lambda+1)(\lambda+4) = 0$。故 $\lambda_1 = -1$，$\lambda_2 = -4$。(M1·A1)
For $\lambda_1 = -1$: $(A+I) = \begin{pmatrix}-2&1\\2&-1\end{pmatrix}$, null space spanned by $\mathbf{v}_1 = \begin{pmatrix}1\\2\end{pmatrix}$. Verify: $A\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}-3+2\\2-4\end{pmatrix} = \begin{pmatrix}-1\\-2\end{pmatrix} = -1\cdot\mathbf{v}_1$. (M1)对 $\lambda_1 = -1$：$(A+I) = \begin{pmatrix}-2&1\\2&-1\end{pmatrix}$，零空间由 $\mathbf{v}_1 = \begin{pmatrix}1\\2\end{pmatrix}$ 张成。验证：$A\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}-3+2\\2-4\end{pmatrix} = \begin{pmatrix}-1\\-2\end{pmatrix} = -1\cdot\mathbf{v}_1$。(M1)
For $\lambda_2 = -4$: $(A+4I) = \begin{pmatrix}1&1\\2&2\end{pmatrix}$, null space spanned by $\mathbf{v}_2 = \begin{pmatrix}1\\-1\end{pmatrix}$. Verify: $A\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}-3-1\\2+2\end{pmatrix} = \begin{pmatrix}-4\\4\end{pmatrix} = -4\cdot\mathbf{v}_2$. (A1)对 $\lambda_2 = -4$：$(A+4I) = \begin{pmatrix}1&1\\2&2\end{pmatrix}$，零空间由 $\mathbf{v}_2 = \begin{pmatrix}1\\-1\end{pmatrix}$ 张成。验证：$A\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}-3-1\\2+2\end{pmatrix} = \begin{pmatrix}-4\\4\end{pmatrix} = -4\cdot\mathbf{v}_2$。(A1)
General solution: $\mathbf{x}(t) = c_1 e^{-t}\begin{pmatrix}1\\2\end{pmatrix} + c_2 e^{-4t}\begin{pmatrix}1\\-1\end{pmatrix}$. Apply $\mathbf{x}(0) = (3,0)^T$: (M1)通解：$\mathbf{x}(t) = c_1 e^{-t}\begin{pmatrix}1\\2\end{pmatrix} + c_2 e^{-4t}\begin{pmatrix}1\\-1\end{pmatrix}$。代入 $\mathbf{x}(0) = (3,0)^T$：(M1)
$$ c_1 + c_2 = 3, \qquad 2c_1 - c_2 = 0 \;\Rightarrow\; c_2 = 2c_1. $$Substituting: $3c_1 = 3$, so $c_1 = 1$ and $c_2 = 2$. (A1)代入得：$3c_1 = 3$，故 $c_1 = 1$，$c_2 = 2$。(A1)
$$ \mathbf{x}(t) = e^{-t}\begin{pmatrix}1\\2\end{pmatrix} + 2e^{-4t}\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}e^{-t} + 2e^{-4t}\\ 2e^{-t} - 2e^{-4t}\end{pmatrix}. $$Check: at $t=0$, $\begin{pmatrix}1+2\\2-2\end{pmatrix} = \begin{pmatrix}3\\0\end{pmatrix}$. Correct. (B1)验证：在 $t=0$ 时，$\begin{pmatrix}1+2\\2-2\end{pmatrix} = \begin{pmatrix}3\\0\end{pmatrix}$，正确。(B1)
As $t \to \infty$, the fast mode $e^{-4t}$ decays to zero much faster than the slow mode $e^{-t}$. The dominant behaviour is $\mathbf{x}(t) \approx e^{-t}\begin{pmatrix}1\\2\end{pmatrix}$ for large $t$. (A1)当 $t \to \infty$ 时，快模态 $e^{-4t}$ 的衰减远快于慢模态 $e^{-t}$。对足够大的 $t$，主导行为为 $\mathbf{x}(t) \approx e^{-t}\begin{pmatrix}1\\2\end{pmatrix}$。(A1)
Both $x_1(t)$ and $x_2(t)$ tend to zero, meaning all salt is eventually lost from the system. The ratio $x_2/x_1 \to 2$ as $t \to \infty$, so the two compartments approach the proportion $1:2$ dictated by the slow eigenvector before the system goes to zero. (R1)$x_1(t)$ 和 $x_2(t)$ 均趋向零，意味着系统中的盐分最终全部流失。$t \to \infty$ 时 $x_2/x_1 \to 2$，故两容器在系统趋零之前按慢特征向量所规定的 $1:2$ 比例趋近。(R1)
For $A = \begin{pmatrix}0&1\\-2&-3\end{pmatrix}$: (a) diagonalize; (b) compute $e^{At}$; (c) solve the forced IVP by variation of parameters.设 $A = \begin{pmatrix}0&1\\-2&-3\end{pmatrix}$：(a) 对角化；(b) 计算 $e^{At}$；(c) 用常数变易法求解强迫初值问题。
Characteristic equation: $\det(A - \lambda I) = -\lambda(-3-\lambda) + 2 = \lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2) = 0$. So $\lambda_1 = -1$ and $\lambda_2 = -2$. (M1)特征方程：$\det(A - \lambda I) = -\lambda(-3-\lambda) + 2 = \lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2) = 0$。故 $\lambda_1 = -1$，$\lambda_2 = -2$。(M1)
For $\lambda_1 = -1$: $(A + I) = \begin{pmatrix}1&1\\-2&-2\end{pmatrix}$, eigenvector $\mathbf{v}_1 = \begin{pmatrix}1\\-1\end{pmatrix}$. Check: $A\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}-1\\1\end{pmatrix} = -1\cdot\begin{pmatrix}1\\-1\end{pmatrix}$. (A1)对 $\lambda_1 = -1$：$(A + I) = \begin{pmatrix}1&1\\-2&-2\end{pmatrix}$，特征向量 $\mathbf{v}_1 = \begin{pmatrix}1\\-1\end{pmatrix}$。验证：$A\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}-1\\1\end{pmatrix} = -1\cdot\begin{pmatrix}1\\-1\end{pmatrix}$。(A1)
For $\lambda_2 = -2$: $(A + 2I) = \begin{pmatrix}2&1\\-2&-1\end{pmatrix}$, eigenvector $\mathbf{v}_2 = \begin{pmatrix}1\\-2\end{pmatrix}$. Check: $A\begin{pmatrix}1\\-2\end{pmatrix} = \begin{pmatrix}-2\\4\end{pmatrix} = -2\cdot\begin{pmatrix}1\\-2\end{pmatrix}$. (A1)对 $\lambda_2 = -2$：$(A + 2I) = \begin{pmatrix}2&1\\-2&-1\end{pmatrix}$，特征向量 $\mathbf{v}_2 = \begin{pmatrix}1\\-2\end{pmatrix}$。验证：$A\begin{pmatrix}1\\-2\end{pmatrix} = \begin{pmatrix}-2\\4\end{pmatrix} = -2\cdot\begin{pmatrix}1\\-2\end{pmatrix}$。(A1)
Set $P = \begin{pmatrix}1&1\\-1&-2\end{pmatrix}$, $D = \begin{pmatrix}-1&0\\0&-2\end{pmatrix}$. Then $\det P = -2 - (-1) = -1$, so (M1)令 $P = \begin{pmatrix}1&1\\-1&-2\end{pmatrix}$，$D = \begin{pmatrix}-1&0\\0&-2\end{pmatrix}$。则 $\det P = -2 - (-1) = -1$，故 (M1)
$$ P^{-1} = \frac{1}{-1}\begin{pmatrix}-2&-1\\1&1\end{pmatrix} = \begin{pmatrix}2&1\\-1&-1\end{pmatrix}. $$Verify: $PP^{-1} = \begin{pmatrix}1&1\\-1&-2\end{pmatrix}\begin{pmatrix}2&1\\-1&-1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}$. (B1)验证：$PP^{-1} = \begin{pmatrix}1&1\\-1&-2\end{pmatrix}\begin{pmatrix}2&1\\-1&-1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}$。(B1)
$e^{At} = P\,e^{Dt}\,P^{-1}$ with $e^{Dt} = \begin{pmatrix}e^{-t}&0\\0&e^{-2t}\end{pmatrix}$: (M1)$e^{At} = P\,e^{Dt}\,P^{-1}$，其中 $e^{Dt} = \begin{pmatrix}e^{-t}&0\\0&e^{-2t}\end{pmatrix}$：(M1)
$$ e^{At} = \begin{pmatrix}1&1\\-1&-2\end{pmatrix}\begin{pmatrix}e^{-t}&0\\0&e^{-2t}\end{pmatrix}\begin{pmatrix}2&1\\-1&-1\end{pmatrix} = \begin{pmatrix}e^{-t}&e^{-2t}\\-e^{-t}&-2e^{-2t}\end{pmatrix}\begin{pmatrix}2&1\\-1&-1\end{pmatrix}. $$Multiplying:相乘得：
$$ e^{At} = \begin{pmatrix}2e^{-t}-e^{-2t} & e^{-t}-e^{-2t}\\ -2e^{-t}+2e^{-2t} & -e^{-t}+2e^{-2t}\end{pmatrix}. $$Check $e^{A\cdot 0} = \begin{pmatrix}2-1&1-1\\-2+2&-1+2\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$. (A1)验证 $e^{A\cdot 0} = \begin{pmatrix}2-1&1-1\\-2+2&-1+2\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$。(A1)
With $\mathbf{g} = (2, 0)^T$ and $\mathbf{x}(0) = \mathbf{0}$, the formula gives $\mathbf{x}(t) = \int_0^t e^{A(t-s)}\mathbf{g}\,ds$. The first column of $e^{A(t-s)}$ is $\begin{pmatrix}2e^{-(t-s)}-e^{-2(t-s)}\\-2e^{-(t-s)}+2e^{-2(t-s)}\end{pmatrix}$, and $e^{A(t-s)}\mathbf{g} = 2\cdot\text{(first column)}$. (M1)取 $\mathbf{g} = (2, 0)^T$，$\mathbf{x}(0) = \mathbf{0}$，公式给出 $\mathbf{x}(t) = \int_0^t e^{A(t-s)}\mathbf{g}\,ds$。$e^{A(t-s)}$ 的第一列为 $\begin{pmatrix}2e^{-(t-s)}-e^{-2(t-s)}\\-2e^{-(t-s)}+2e^{-2(t-s)}\end{pmatrix}$，故 $e^{A(t-s)}\mathbf{g} = 2\cdot\text{（第一列）}$。(M1)
Integrating component by component:逐分量积分：
$$ x_1(t) = 2\int_0^t\!\bigl(2e^{-(t-s)}-e^{-2(t-s)}\bigr)ds = 2\bigl[2(1-e^{-t}) - \tfrac{1}{2}(1-e^{-2t})\bigr] = 3 - 4e^{-t} + e^{-2t}. $$ $$ x_2(t) = 2\int_0^t\!\bigl(-2e^{-(t-s)}+2e^{-2(t-s)}\bigr)ds = 2\bigl[-2(1-e^{-t}) + (1-e^{-2t})\bigr] = -2 + 4e^{-t} - 2e^{-2t}. $$So $\mathbf{x}(t) = \begin{pmatrix}3-4e^{-t}+e^{-2t}\\-2+4e^{-t}-2e^{-2t}\end{pmatrix}$. Check at $t=0$: $(3-4+1, -2+4-2)^T = (0,0)^T$. (A1)故 $\mathbf{x}(t) = \begin{pmatrix}3-4e^{-t}+e^{-2t}\\-2+4e^{-t}-2e^{-2t}\end{pmatrix}$。在 $t=0$ 处验证：$(3-4+1, -2+4-2)^T = (0,0)^T$。(A1)