Unit D7: Systems of First-Order Linear ODEsUnit D7: 一阶线性微分方程组

Reduce $y'' + 3y' + 2y = 0$ to a first-order system.

Set $x_1 = y$ and $x_2 = y'$. Then $x_1' = x_2$, and solving the equation for $y''$ gives $x_2' = y'' = -3y' - 2y = -2x_1 - 3x_2$.

$$ \begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}. $$

The coefficient matrix is the companion matrix of the characteristic polynomial $\lambda^2 + 3\lambda + 2$. Its eigenvalues, $-1$ and $-2$, are exactly the roots of that polynomial.

Reduction runs only one way for a single scalar equation, but many genuine systems arise directly from modelling. Consider two tanks. Tank 1 holds salt amount $x_1(t)$, tank 2 holds $x_2(t)$, brine is pumped between them, and the mixing rates give

$$ x_1' = -2x_1 + x_2, \qquad x_2' = 2x_1 - 3x_2. $$

In matrix form this is $\mathbf{x}' = A\mathbf{x}$ with $A = \begin{pmatrix} -2 & 1 \\ 2 & -3 \end{pmatrix}$. There is no single second-order equation that we reduced; the system is the primary object. Note both column sums are not equal, so total salt is not conserved, which is physically correct here because each tank also drains to the outside.

We can still eliminate to a scalar equation if we wish. Differentiate the first equation and substitute the second: $x_1'' = -2x_1' + x_2' = -2x_1' + (2x_1 - 3x_2)$. From the first equation $x_2 = x_1' + 2x_1$, so $x_1'' = -2x_1' + 2x_1 - 3(x_1' + 2x_1) = -5x_1' - 4x_1$, giving $x_1'' + 5x_1' + 4x_1 = 0$. The roots $-1$ and $-4$ are exactly the eigenvalues of $A$, confirming the two viewpoints agree.

Reduce $y''' + 2y'' - y' + 4y = \sin t$ to a first-order system.

Set $x_1 = y$, $x_2 = y'$, $x_3 = y''$. Then $x_1' = x_2$, $x_2' = x_3$, and solving the equation for the top derivative gives $x_3' = y''' = -2y'' + y' - 4y + \sin t = -4x_1 + x_2 - 2x_3 + \sin t$.

$$ \mathbf{x}' = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -4 & 1 & -2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} 0 \\ 0 \\ \sin t \end{pmatrix}. $$

The forcing enters only the last component, the same row that carried the top derivative. This is the general pattern: a scalar forcing $g(t)$ becomes the vector $\mathbf{g} = (0,\dots,0,g(t))^{\mathsf T}$, and the matrix is always the companion matrix of the homogeneous part.

Claim: the companion matrix $C$ of $p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_0$ has characteristic polynomial $\det(\lambda I - C) = p(\lambda)$. We prove it by induction on $n$ using cofactor expansion along the first column.

Write $C$ with superdiagonal ones and bottom row $(-a_0, -a_1, \dots, -a_{n-1})$. Expanding $\det(\lambda I - C)$ down its first column, only two entries are nonzero: the top-left $\lambda$ and the bottom-left $a_0$. The $\lambda$ minor is the same determinant for the order $n-1$ companion matrix of $\lambda^{n-1} + a_{n-1}\lambda^{n-2} + \cdots + a_1$, which by induction equals $\lambda^{n-1} + a_{n-1}\lambda^{n-2} + \cdots + a_1$. The bottom-left cofactor, after tracking the sign $(-1)^{n+1}$ and the upper-triangular minor of all $-1$ entries, contributes exactly $a_0$.

$$ \det(\lambda I - C) = \lambda\big(\lambda^{n-1} + a_{n-1}\lambda^{n-2} + \cdots + a_1\big) + a_0 = \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1\lambda + a_0. $$

So the eigenvalues of the system matrix are precisely the characteristic roots of the scalar equation. This is why reducing to a system loses no information: the spectral data is identical. The base case $n=1$ is $\det(\lambda - (-a_0)) = \lambda + a_0$, which matches.

For $\mathbf{x}' = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}\mathbf{x}$, verify that $\mathbf{x}^{(1)} = e^{3t}\binom{1}{2}$ and $\mathbf{x}^{(2)} = e^{-t}\binom{1}{-2}$ form a fundamental set.

Each is a solution (it is an eigenvector times $e^{\lambda t}$, checked in Section 3). The Wronskian is

$$ W = \det \begin{pmatrix} e^{3t} & e^{-t} \\ 2e^{3t} & -2e^{-t} \end{pmatrix} = -2e^{2t} - 2e^{2t} = -4e^{2t} \neq 0. $$

Since $W$ is never zero, the two solutions are linearly independent and the general solution is $\mathbf{x} = c_1 e^{3t}\binom{1}{2} + c_2 e^{-t}\binom{1}{-2}$.

Differentiate $W = \det \Psi$ column by column. The derivative of a determinant is the sum over columns of the determinant with that one column differentiated. Replacing the $k$th column $\mathbf{x}^{(k)}$ by $\mathbf{x}^{(k)\prime} = A\mathbf{x}^{(k)}$ and using multilinearity, each off-diagonal contribution of $A$ produces a repeated column and vanishes; only the diagonal entries $a_{kk}$ survive.

$$ W'(t) = \Big(\sum_k a_{kk}(t)\Big) W(t) = \operatorname{tr} A(t)\, W(t). $$

This scalar linear ODE has solution $W(t) = W(t_0)\exp\!\big(\int_{t_0}^t \operatorname{tr} A\big)$. The exponential factor is strictly positive, so the sign and the vanishing of $W$ are decided entirely by $W(t_0)$. Thus $W$ is everywhere zero or nowhere zero.

Continuing the previous system $\mathbf{x}' = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}\mathbf{x}$, solve the initial value problem with $\mathbf{x}(0) = \binom{2}{0}$.

A fundamental matrix is $\Psi(t) = \begin{pmatrix} e^{3t} & e^{-t} \\ 2e^{3t} & -2e^{-t} \end{pmatrix}$. At $t=0$, $\Psi(0) = \begin{pmatrix} 1 & 1 \\ 2 & -2 \end{pmatrix}$ with determinant $-4$, so

$$ \Psi(0)^{-1} = \frac{1}{-4}\begin{pmatrix} -2 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/4 \\ 1/2 & -1/4 \end{pmatrix}. $$

The constants are $\mathbf{c} = \Psi(0)^{-1}\mathbf{x}_0 = \begin{pmatrix} 1/2 & 1/4 \\ 1/2 & -1/4 \end{pmatrix}\binom{2}{0} = \binom{1}{1}$. Hence

$$ \mathbf{x}(t) = \Psi(t)\binom{1}{1} = e^{3t}\binom{1}{2} + e^{-t}\binom{1}{-2}. $$

Check at $t=0$: $\binom{1}{2} + \binom{1}{-2} = \binom{2}{0}$, the required initial data. The formula $\mathbf{x}(t) = \Psi(t)\Psi(t_0)^{-1}\mathbf{x}_0$ packages exactly this computation.

Suppose someone proposes $\mathbf{x}^{(1)} = e^{3t}\binom{1}{2}$ and $\mathbf{x}^{(2)} = 5e^{3t}\binom{1}{2}$ as a candidate fundamental set. Their Wronskian is

$$ W = \det\begin{pmatrix} e^{3t} & 5e^{3t} \\ 2e^{3t} & 10e^{3t} \end{pmatrix} = 10e^{6t} - 10e^{6t} = 0 \quad\text{for all } t. $$

Abel's theorem then guarantees $W \equiv 0$, which is consistent: the second solution is a scalar multiple of the first, so the pair spans only a one-dimensional subspace, not the full two-dimensional solution space. A fundamental set requires the columns to be linearly independent, which the nonvanishing of $W$ at a single point already certifies for solutions of the same linear system.

Solve $\mathbf{x}' = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}\mathbf{x}$.

Characteristic equation: $\det(A - \lambda I) = (1-\lambda)^2 - 4 = \lambda^2 - 2\lambda - 3 = 0$, so $\lambda = 3, -1$.

For $\lambda = 3$: $(A - 3I)\mathbf{v} = \begin{pmatrix} -2 & 1 \\ 4 & -2 \end{pmatrix}\mathbf{v} = 0$ gives $\mathbf{v}_1 = \binom{1}{2}$. For $\lambda = -1$: $\begin{pmatrix} 2 & 1 \\ 4 & 2 \end{pmatrix}\mathbf{v} = 0$ gives $\mathbf{v}_2 = \binom{1}{-2}$.

$$ \mathbf{x}(t) = c_1 e^{3t}\binom{1}{2} + c_2 e^{-t}\binom{1}{-2}. $$

One eigenvalue is positive and one negative, so the origin is a saddle point.

Suppose $\mathbf{v}_1, \dots, \mathbf{v}_k$ are eigenvectors for distinct eigenvalues $\lambda_1, \dots, \lambda_k$ and assume a dependence $c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = 0$ with the fewest nonzero terms. Apply $A - \lambda_k I$ to the relation. The $\mathbf{v}_k$ term dies, and each other term is scaled by $(\lambda_i - \lambda_k) \neq 0$, producing a shorter dependence, contradiction unless all $c_i = 0$.

Hence $n$ distinct eigenvalues yield $n$ independent eigenvectors. The corresponding solutions $e^{\lambda_i t}\mathbf{v}_i$ have Wronskian at $t=0$ equal to $\det[\mathbf{v}_1 \mid \cdots \mid \mathbf{v}_n] \neq 0$, so they form a fundamental set.

Solve $\mathbf{x}' = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}\mathbf{x}$ with $\mathbf{x}(0) = \binom{3}{1}$.

Characteristic equation: $(-2-\lambda)^2 - 1 = \lambda^2 + 4\lambda + 3 = 0$, so $\lambda = -1, -3$.

For $\lambda = -1$: $(A+I)\mathbf{v} = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\mathbf{v} = 0$ gives $\mathbf{v}_1 = \binom{1}{1}$. For $\lambda = -3$: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\mathbf{v} = 0$ gives $\mathbf{v}_2 = \binom{1}{-1}$.

General solution $\mathbf{x} = c_1 e^{-t}\binom{1}{1} + c_2 e^{-3t}\binom{1}{-1}$. Apply the initial data: $c_1\binom{1}{1} + c_2\binom{1}{-1} = \binom{3}{1}$ gives $c_1 + c_2 = 3$ and $c_1 - c_2 = 1$, so $c_1 = 2$, $c_2 = 1$.

$$ \mathbf{x}(t) = 2e^{-t}\binom{1}{1} + e^{-3t}\binom{1}{-1}. $$

Both eigenvalues are negative, so the origin is a stable node. As $t\to\infty$ the faster mode $e^{-3t}$ dies first, and trajectories approach the origin tangent to the slow eigendirection $\binom{1}{1}$.

Solve $\mathbf{x}' = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 0 & 3 \end{pmatrix}\mathbf{x}$.

The matrix is lower triangular, so the eigenvalues are the diagonal entries $\lambda = 1, 2, 3$, all real and distinct.

For $\lambda = 1$: $(A - I)\mathbf{v} = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 2 \end{pmatrix}\mathbf{v} = 0$. From rows two and three, $v_2 = -v_1$ and $v_3 = -v_1/2$. Taking $v_1 = 2$ gives $\mathbf{v}_1 = (2,-2,-1)^{\mathsf T}$.

For $\lambda = 2$: $\begin{pmatrix} -1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix}\mathbf{v} = 0$ forces $v_1 = 0$ and $v_3 = 0$ with $v_2$ free, so $\mathbf{v}_2 = (0,1,0)^{\mathsf T}$.

For $\lambda = 3$: $\begin{pmatrix} -2 & 0 & 0 \\ 1 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\mathbf{v} = 0$ forces $v_1 = 0$ and $v_2 = 0$ with $v_3$ free, so $\mathbf{v}_3 = (0,0,1)^{\mathsf T}$.

$$ \mathbf{x}(t) = c_1 e^{t}\!\begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix} + c_2 e^{2t}\!\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + c_3 e^{3t}\!\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}. $$

All eigenvalues are positive, so the origin is a three-dimensional unstable node. Triangular matrices make the spectrum readable at a glance, a useful shortcut to remember on exams.

Solve $\mathbf{x}' = \begin{pmatrix} -1 & -2 \\ 2 & -1 \end{pmatrix}\mathbf{x}$.

$\det(A - \lambda I) = (-1-\lambda)^2 + 4 = 0$, so $\lambda = -1 \pm 2i$; here $\alpha = -1$, $\beta = 2$.

For $\lambda = -1 + 2i$: $(A - \lambda I)\mathbf{v} = \begin{pmatrix} -2i & -2 \\ 2 & -2i \end{pmatrix}\mathbf{v} = 0$ gives $\mathbf{v} = \binom{1}{-i} = \binom{1}{0} + i\binom{0}{-1}$, so $\mathbf{a} = \binom{1}{0}$, $\mathbf{b} = \binom{0}{-1}$.

$$ \mathbf{x}(t) = e^{-t}\!\left[ c_1\!\begin{pmatrix}\cos 2t \\ \sin 2t\end{pmatrix} + c_2\!\begin{pmatrix}\sin 2t \\ -\cos 2t\end{pmatrix} \right]. $$

Since $\alpha = -1 < 0$, trajectories spiral inward to the origin: a stable spiral (spiral sink).

Solve $\mathbf{x}' = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}\mathbf{x}$ with $\mathbf{x}(0) = \binom{1}{0}$.

$\det(A - \lambda I) = \lambda^2 + 4 = 0$, so $\lambda = \pm 2i$; here $\alpha = 0$, $\beta = 2$.

For $\lambda = 2i$: $(A - 2iI)\mathbf{v} = \begin{pmatrix} -2i & 2 \\ -2 & -2i \end{pmatrix}\mathbf{v} = 0$. The top row gives $-2i\,v_1 + 2v_2 = 0$, so $v_2 = i v_1$; take $\mathbf{v} = \binom{1}{i} = \binom{1}{0} + i\binom{0}{1}$, hence $\mathbf{a} = \binom{1}{0}$, $\mathbf{b} = \binom{0}{1}$.

The two real solutions are $\mathbf{x}_1 = \mathbf{a}\cos 2t - \mathbf{b}\sin 2t = \binom{\cos 2t}{-\sin 2t}$ and $\mathbf{x}_2 = \mathbf{a}\sin 2t + \mathbf{b}\cos 2t = \binom{\sin 2t}{\cos 2t}$, so

$$ \mathbf{x}(t) = c_1\binom{\cos 2t}{-\sin 2t} + c_2\binom{\sin 2t}{\cos 2t}. $$

Initial data $\binom{1}{0}$ forces $c_1 = 1$, $c_2 = 0$, giving $\mathbf{x}(t) = \binom{\cos 2t}{-\sin 2t}$. This traces the unit circle clockwise with period $\pi$: a closed orbit, confirming the center. Since $x_1^2 + x_2^2 = 1$ for all $t$, the trajectory is bounded but never approaches the origin, the signature of neutral stability.

Let $\mathbf{z}(t) = e^{\lambda t}\mathbf{v}$ be a complex solution of $\mathbf{x}' = A\mathbf{x}$ with $A$ real and $\lambda = \alpha + i\beta$, $\beta \neq 0$. Write $\mathbf{z} = \mathbf{u} + i\mathbf{w}$ with $\mathbf{u} = \operatorname{Re}\mathbf{z}$ and $\mathbf{w} = \operatorname{Im}\mathbf{z}$ both real.

Since $A$ is real, $\mathbf{z}' = A\mathbf{z}$ separates into real and imaginary parts: $\mathbf{u}' + i\mathbf{w}' = A\mathbf{u} + iA\mathbf{w}$. Matching real and imaginary parts gives $\mathbf{u}' = A\mathbf{u}$ and $\mathbf{w}' = A\mathbf{w}$ separately, so each part is itself a real solution.

Independence: suppose $c_1\mathbf{u} + c_2\mathbf{w} = \mathbf{0}$ for all $t$ with real constants. Then $c_1\mathbf{u} + c_2\mathbf{w} = \operatorname{Re}\!\big((c_1 - i c_2)\mathbf{z}\big) = \mathbf{0}$. Evaluating at $t=0$ gives $\operatorname{Re}\!\big((c_1 - ic_2)\mathbf{v}\big) = \mathbf{0}$ and differentiating once and evaluating at $0$ gives $\operatorname{Re}\!\big((c_1 - ic_2)\lambda\mathbf{v}\big) = \mathbf{0}$. Because $\beta \neq 0$, the scalar $\lambda$ is non-real, and these two conditions force $c_1 - ic_2 = 0$, hence $c_1 = c_2 = 0$. The real and imaginary parts are therefore linearly independent and span the real solution space of the conjugate pair.

Solve $\mathbf{x}' = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\mathbf{x}$.

$\det(A - \lambda I) = (3-\lambda)(-1-\lambda) + 4 = \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2$, so $\lambda = 1$ is a double root.

Eigenvectors: $(A - I)\mathbf{v} = \begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix}\mathbf{v} = 0$ gives the single direction $\mathbf{v} = \binom{2}{1}$, so $A$ is defective.

Generalized eigenvector: solve $(A - I)\mathbf{w} = \mathbf{v}$, that is $\begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix}\mathbf{w} = \binom{2}{1}$. Taking $\mathbf{w} = \binom{1}{0}$ works.

$$ \mathbf{x}(t) = c_1 e^{t}\binom{2}{1} + c_2 e^{t}\!\left[ t\binom{2}{1} + \binom{1}{0} \right]. $$

Because $\lambda = 1 > 0$, this is an unstable improper (degenerate) node.

Solve $\mathbf{x}' = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}\mathbf{x}$.

$\det(A - \lambda I) = (\lambda + 3)^2$, so $\lambda = -3$ is a double root. But here $A - \lambda I = A + 3I = \mathbf{0}$, so every nonzero vector is an eigenvector and the eigenspace is the whole plane. The matrix is a scalar multiple of the identity, the model complete case.

We may choose the standard basis $\mathbf{v}_1 = \binom{1}{0}$, $\mathbf{v}_2 = \binom{0}{1}$:

$$ \mathbf{x}(t) = c_1 e^{-3t}\binom{1}{0} + c_2 e^{-3t}\binom{0}{1} = e^{-3t}\binom{c_1}{c_2}. $$

No $t$ factor appears because the eigenspace is two-dimensional. Every trajectory is a straight ray into the origin along whatever direction $\binom{c_1}{c_2}$ the initial data picks, so this is a stable star (proper) node. Contrast Worked Example 5.1, where a one-dimensional eigenspace forced the $t$-multiplied second solution.

Solve $\mathbf{x}' = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}\mathbf{x}$ with $\mathbf{x}(0) = \binom{1}{2}$.

$\det(A - \lambda I) = (\lambda + 1)^2$, so $\lambda = -1$ is a double root. Eigenvectors solve $(A + I)\mathbf{v} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\mathbf{v} = 0$, forcing $v_2 = 0$, so the eigenspace is one-dimensional, spanned by $\mathbf{v} = \binom{1}{0}$. The matrix is defective.

Generalized eigenvector: solve $(A + I)\mathbf{w} = \mathbf{v}$, that is $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\mathbf{w} = \binom{1}{0}$, which gives $w_2 = 1$ with $w_1$ free; take $\mathbf{w} = \binom{0}{1}$.

$$ \mathbf{x}(t) = c_1 e^{-t}\binom{1}{0} + c_2 e^{-t}\!\left[ t\binom{1}{0} + \binom{0}{1} \right]. $$

At $t=0$: $c_1\binom{1}{0} + c_2\binom{0}{1} = \binom{1}{2}$, so $c_1 = 1$, $c_2 = 2$. Therefore $\mathbf{x}(t) = e^{-t}\binom{1 + 2t}{2}$. Even though the polynomial factor $1 + 2t$ grows, the dominant $e^{-t}$ decay wins and $\mathbf{x}(t) \to \mathbf{0}$: a stable improper node.

Let $\lambda$ be a double eigenvalue with a one-dimensional eigenspace spanned by $\mathbf{v}$. We look for a second solution of the form $\mathbf{x}_2(t) = e^{\lambda t}\big(t\,\mathbf{v} + \mathbf{w}\big)$ and determine what $\mathbf{w}$ must satisfy. Differentiate using the product rule:

$$ \mathbf{x}_2' = \lambda e^{\lambda t}(t\mathbf{v} + \mathbf{w}) + e^{\lambda t}\mathbf{v} = e^{\lambda t}\big(\lambda t\,\mathbf{v} + \lambda\mathbf{w} + \mathbf{v}\big). $$

We require $\mathbf{x}_2' = A\mathbf{x}_2 = e^{\lambda t}\big(t\,A\mathbf{v} + A\mathbf{w}\big)$. Cancel $e^{\lambda t}$ and match the coefficient of $t$ and the constant term separately.

Coefficient of $t$: $\lambda\mathbf{v} = A\mathbf{v}$, which holds because $\mathbf{v}$ is an eigenvector. Constant term: $\lambda\mathbf{w} + \mathbf{v} = A\mathbf{w}$, that is $(A - \lambda I)\mathbf{w} = \mathbf{v}$. This is exactly the generalized eigenvector equation. It is solvable because $\mathbf{v}$ lies in the range of $(A - \lambda I)$ whenever the eigenvalue is defective with algebraic multiplicity two, a fact guaranteed by the structure of the Jordan block for $\lambda$.

Finally $\mathbf{x}_1 = e^{\lambda t}\mathbf{v}$ and $\mathbf{x}_2$ are independent: at $t=0$ their values are $\mathbf{v}$ and $\mathbf{w}$, and $\mathbf{w} \notin \operatorname{span}\{\mathbf{v}\}$ since otherwise $(A-\lambda I)\mathbf{w}$ would be a multiple of $(A - \lambda I)\mathbf{v} = \mathbf{0}$, contradicting $(A - \lambda I)\mathbf{w} = \mathbf{v} \neq \mathbf{0}$.

Classify the origin for $A = \begin{pmatrix} -3 & 2 \\ -2 & -3 \end{pmatrix}$.

$p = \operatorname{tr} A = -6$, $q = \det A = 9 + 4 = 13$. Discriminant $p^2 - 4q = 36 - 52 = -16 < 0$, so the eigenvalues are complex, $\lambda = -3 \pm 2i$.

Complex with $p < 0$ and $q > 0$: a stable spiral. Trajectories spiral inward to the origin.

For a $2 \times 2$ matrix the eigenvalues satisfy $\lambda_1 + \lambda_2 = p$ and $\lambda_1 \lambda_2 = q$. Asymptotic stability requires both eigenvalues to have negative real part.

If the eigenvalues are real, both negative means their sum is negative ($p < 0$) and their product is positive ($q > 0$). If they are complex conjugates $\alpha \pm i\beta$, then $p = 2\alpha$ and $q = \alpha^2 + \beta^2 > 0$ automatically, so $\alpha < 0$ is equivalent to $p < 0$.

In every case, both eigenvalues have negative real part if and only if $p < 0$ and $q > 0$. This is the Routh-Hurwitz criterion specialized to two dimensions.

Classify the origin for each matrix using only $(p, q)$ and the discriminant $\Delta = p^2 - 4q$.

(a) $A = \begin{pmatrix} -1 & 4 \\ -1 & -1 \end{pmatrix}$: $p = -2$, $q = 1 + 4 = 5$, $\Delta = 4 - 20 = -16 < 0$. Complex eigenvalues with $p<0$: a stable spiral.

(b) $A = \begin{pmatrix} 4 & 1 \\ 1 & 4 \end{pmatrix}$: $p = 8$, $q = 16 - 1 = 15$, $\Delta = 64 - 60 = 4 > 0$. Real same-sign eigenvalues with $p > 0$: an unstable node.

(c) $A = \begin{pmatrix} 1 & 5 \\ 1 & -1 \end{pmatrix}$: $p = 0$, $q = -1 - 5 = -6 < 0$. Negative determinant: real opposite-sign eigenvalues, a saddle. The trace being zero is irrelevant here because a saddle is decided by $q < 0$ alone.

(d) $A = \begin{pmatrix} 0 & -3 \\ 3 & 0 \end{pmatrix}$: $p = 0$, $q = 9 > 0$, $\Delta = -36 < 0$. Purely imaginary eigenvalues: a center, neutrally stable.

On the curve $\Delta = p^2 - 4q = 0$ the eigenvalues are real and equal, the repeated-root case of Section 5. Consider the one-parameter family $A_q = \begin{pmatrix} -2 & 1 \\ -q & 0 \end{pmatrix}$, which has $p = -2$ and $\det A_q = q$.

For $q < 1$ the discriminant $\Delta = 4 - 4q > 0$, so the eigenvalues are real and distinct: a stable node (since $p<0$, $q>0$ when $0 1$, $\Delta < 0$ and the origin becomes a stable spiral. Exactly at $q = 1$, $\Delta = 0$ and $\lambda = -1$ is a double root, the borderline degenerate node. The qualitative picture changes from straight-line approach to rotational approach as $q$ crosses the parabola $q = p^2/4 = 1$.

This is why the trace-determinant plane is the standard summary chart: the line $q = 0$ separates saddles from nodes, the line $p = 0$ (with $q>0$) holds the centers, and the parabola $q = p^2/4$ separates nodes from spirals.

Solve $\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\mathbf{x} + \binom{0}{1}$ with $\mathbf{x}(0) = \mathbf{0}$.

Here $A^2 = 0$, so $e^{At} = I + At = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$. With $\mathbf{x}_0 = \mathbf{0}$,

$$ \mathbf{x}(t) = \int_0^t e^{A(t-s)}\binom{0}{1}\,ds = \int_0^t \begin{pmatrix} 1 & t-s \\ 0 & 1 \end{pmatrix}\binom{0}{1}\,ds = \int_0^t \binom{t-s}{1}\,ds = \binom{t^2/2}{t}. $$

This matches the elementary solution: the second component integrates the forcing to $t$, and the first integrates that to $t^2/2$.

Compute the matrix exponential for $A = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}$, the saddle from Section 3.

We found eigenvalues $\lambda = 3, -1$ with eigenvectors $\binom{1}{2}$ and $\binom{1}{-2}$. Set

$$ P = \begin{pmatrix} 1 & 1 \\ 2 & -2 \end{pmatrix}, \quad D = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}, \quad P^{-1} = \frac{1}{-4}\begin{pmatrix} -2 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/4 \\ 1/2 & -1/4 \end{pmatrix}. $$

Then $e^{At} = P\,e^{Dt}\,P^{-1}$ with $e^{Dt} = \operatorname{diag}(e^{3t}, e^{-t})$. Multiplying,

$$ e^{At} = \begin{pmatrix} 1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} e^{3t} & 0 \\ 0 & e^{-t} \end{pmatrix} \begin{pmatrix} 1/2 & 1/4 \\ 1/2 & -1/4 \end{pmatrix} = \begin{pmatrix} \tfrac12(e^{3t}+e^{-t}) & \tfrac14(e^{3t}-e^{-t}) \\[4pt] e^{3t}-e^{-t} & \tfrac12(e^{3t}+e^{-t}) \end{pmatrix}. $$

Check: at $t=0$ this is $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$, as required of any matrix exponential. Applying it to $\mathbf{x}_0 = \binom{2}{0}$ reproduces $e^{3t}\binom{1}{2} + e^{-t}\binom{1}{-2}$ from Worked Example 2.2.

Solve $\mathbf{x}' = \begin{pmatrix} -2 & 0 \\ 0 & -3 \end{pmatrix}\mathbf{x} + \binom{4}{6}$ with $\mathbf{x}(0) = \mathbf{0}$.

Because $A$ is diagonal, the system decouples: $x_1' = -2x_1 + 4$ and $x_2' = -3x_2 + 6$. Each is a scalar linear ODE. The equilibrium (particular) solution is found by setting derivatives to zero: $x_1^* = 2$, $x_2^* = 2$.

Using variation of parameters with $e^{At} = \operatorname{diag}(e^{-2t}, e^{-3t})$,

$$ \mathbf{x}(t) = \int_0^t e^{A(t-s)}\binom{4}{6}\,ds = \binom{4\int_0^t e^{-2(t-s)}\,ds}{6\int_0^t e^{-3(t-s)}\,ds} = \binom{2(1 - e^{-2t})}{2(1 - e^{-3t})}. $$

As $t\to\infty$ the transient $e^{-2t}, e^{-3t}$ decay and $\mathbf{x}(t) \to \binom{2}{2}$, the equilibrium. This decomposition into a decaying homogeneous transient plus a steady particular state is the structural meaning of the variation of parameters formula.

We verify directly from the series that $\Phi(t) = e^{At}$ satisfies $\Phi' = A\Phi$ and $\Phi(0) = I$, so that $\mathbf{x}(t) = e^{At}\mathbf{x}_0$ solves the IVP.

First, $\Phi(0) = \sum_{k\ge 0}\frac{(A\cdot 0)^k}{k!} = I$ since only the $k=0$ term survives. The series $\sum_k \frac{(At)^k}{k!}$ converges absolutely for every $t$ because, in any submultiplicative matrix norm, $\big\|\frac{(At)^k}{k!}\big\| \le \frac{(\|A\|\,|t|)^k}{k!}$, whose sum is $e^{\|A\|\,|t|}$. Absolute convergence licenses term-by-term differentiation.

$$ \frac{d}{dt}e^{At} = \frac{d}{dt}\sum_{k=0}^\infty \frac{A^k t^k}{k!} = \sum_{k=1}^\infty \frac{A^k t^{k-1}}{(k-1)!} = A\sum_{j=0}^\infty \frac{A^j t^j}{j!} = A\,e^{At}, $$

where we shifted the index $j = k-1$ and factored one $A$ out the front. Therefore with $\mathbf{x}(t) = e^{At}\mathbf{x}_0$ we get $\mathbf{x}' = A e^{At}\mathbf{x}_0 = A\mathbf{x}$ and $\mathbf{x}(0) = \mathbf{x}_0$. By the uniqueness theorem of Section 1 this is the only solution. Finally, for the forced equation, differentiating $\mathbf{x}(t) = e^{At}\mathbf{x}_0 + \int_0^t e^{A(t-s)}\mathbf{g}(s)\,ds$ with the Leibniz rule gives $\mathbf{x}' = A e^{At}\mathbf{x}_0 + e^{A\cdot 0}\mathbf{g}(t) + \int_0^t A e^{A(t-s)}\mathbf{g}(s)\,ds = A\mathbf{x} + \mathbf{g}(t)$, confirming the variation of parameters formula.