Systems of first-order linear ODEs — matrix method: Exercises

Intuition How to use this page

These problems climb a ladder. L1 just asks you to recognize the machinery. L5 asks you to build something new. Try each one with pen and paper FIRST, then open the collapsible [!recall]- solution. Every worked step says WHAT we did and WHY. All the tools come from the parent note Systems of First-Order Linear ODEs — matrix method; when a new idea sneaks in, we rebuild it here.

Before we start, one reminder of the whole game so no symbol is unearned:

$x^{'} = A x$ means "the rate of change of the vector $x$ equals the matrix $A$ times $x$ ." A vector is just a list of numbers stacked up (a point, or an arrow from the origin).
$A v = λ v$ says "the arrow $v$ is only stretched (by the number $λ$ ) when $A$ acts on it — its direction is untouched." Such an arrow is an eigenvector; the stretch factor $λ$ is its eigenvalue. See Eigenvalues and Eigenvectors.
$det (A - λ I) = 0$ is the characteristic equation — the filter that finds those special $λ$ .

Level 1 — Recognition

Exercise 1.1

Which of these vector functions is a solution of $x^{'} = A x$ for some matrix $A$ , by form alone (i.e. is it built from the eigenvector template)? (a) $x (t) = e^{2 t} (3 1)$ (b) $x (t) = (e ^{2 t} + t e ^{2 t})$ (c) $x (t) = e^{- t} (- 5 2)$

Recall Solution

WHAT the template is. The parent note derived that $x = e^{λ t} v$ solves the system exactly when $(λ, v)$ is an eigenpair. So a "clean" single-mode solution must be one fixed vector multiplied by one scalar exponential.

(a) $e^{2 t} (3 1)$ : fixed direction $(3 1)$ , scaled by $e^{2 t}$ . ✅ Matches the template with $λ = 2$ , $v = (3 1)$ .
(c) $e^{- t} (- 5 2)$ : same shape, $λ = - 1$ , $v = (- 5 2)$ . ✅
(b) has a $+ t$ inside a component that is not multiplying a fixed eigenvector uniformly — the two components grow at genuinely different rates, so it is not a single exponential mode. It could be a piece of a defective solution $e^{λ t} (t v + w)$ , but as written it is not the plain eigenvector template. ❌ for "single-mode".

Answer: (a) and (c).

Exercise 1.2

For $A = (50 0 - 2)$ , write the eigenvalues and eigenvectors by inspection, and the general solution.

Recall Solution

WHY by inspection. $A$ is diagonal: it does not mix $x_{1}$ and $x_{2}$ at all. The two equations are already decoupled: $x_{1}^{'} = 5 x_{1}$ , $x_{2}^{'} = - 2 x_{2}$ . See Diagonalization.

The standard basis arrows $(0 1)$ and $(1 0)$ are unmoved in direction — only stretched: $A (0 1) = 5 (0 1), A (1 0) = - 2 (1 0) .$ So eigenpairs are $(5, (0 1))$ and $(- 2, (1 0))$ . $x (t) = c_{1} e^{5 t} (0 1) + c_{2} e^{- 2 t} (1 0) .$

Level 2 — Application

Exercise 2.1

Solve $x^{'} = A x$ , $A = (2112)$ , with $x (0) = (0 4)$ .

Recall Solution

Step 1 — characteristic equation. Why: a nonzero eigenvector exists only when $A - λ I$ is singular, i.e. $det = 0$ . $det (2 - λ 1 1 2 - λ) = (2 - λ)^{2} - 1 = 0.$ $(2 - λ)^{2} = 1 \Rightarrow 2 - λ = \pm 1 \Rightarrow λ = 1, 3.$

Step 2 — eigenvectors. Why: solve the null direction of the now-singular matrix.

$λ = 3$ : $(- 1 1 1 - 1) v = 0 \Rightarrow - v_{1} + v_{2} = 0 \Rightarrow v = (1 1) .$
$λ = 1$ : $(1111) v = 0 \Rightarrow v_{1} + v_{2} = 0 \Rightarrow v = (- 1 1) .$

Step 3 — general solution. $x = c_{1} e^{3 t} (1 1) + c_{2} e^{t} (- 1 1) .$

Step 4 — apply IC $x (0) = (0 4)$ : $c_{1} (1 1) + c_{2} (- 1 1) = (0 4)$ . $c_{1} + c_{2} = 4$ , $c_{1} - c_{2} = 0 \Rightarrow c_{1} = c_{2} = 2.$ $x (t) = 2 e^{3 t} (1 1) + 2 e^{t} (- 1 1) .$

Exercise 2.2

Solve $x^{'} = A x$ , $A = (- 3 0 4 - 1)$ (general solution only).

Recall Solution

Step 1. $A$ is upper-triangular, so its eigenvalues are the diagonal entries: read $det (A - λ I) = (- 3 - λ) (- 1 - λ) = 0 \Rightarrow λ = - 3, - 1.$

Step 2.

$λ = - 3$ : $(0042) v = 0 \Rightarrow 4 v_{2} = 0 \Rightarrow v_{2} = 0 \Rightarrow v = (0 1) .$
$λ = - 1$ : $(- 2 0 40) v = 0 \Rightarrow - 2 v_{1} + 4 v_{2} = 0 \Rightarrow v_{1} = 2 v_{2} \Rightarrow v = (1 2) .$

Step 3. $x = c_{1} e^{- 3 t} (0 1) + c_{2} e^{- t} (1 2) .$ Both exponents are negative → every trajectory decays to the origin (a stable node), see Phase Portraits and Stability.

Level 3 — Analysis

Exercise 3.1

$A = (12 - 2 1)$ . Find real solutions and describe the trajectory geometry.

Recall Solution

Step 1. $det (A - λ I) = (1 - λ)^{2} + 4 = 0 \Rightarrow (1 - λ)^{2} = - 4 \Rightarrow 1 - λ = \pm 2 i \Rightarrow λ = 1 \pm 2 i .$ So $α = 1$ (real part), $β = 2$ (imaginary part). Why care about $α, β$ ? By Euler's Formula, $e^{(α + i β) t} = e^{α t} (cos β t + i sin β t)$ : $α$ sets growth/decay, $β$ sets rotation speed.

Step 2 — eigenvector for $λ = 1 + 2 i$ . $(A - λ I) = (- 2 i 2 - 2 - 2 i)$ . Top row: $- 2 i v_{1} - 2 v_{2} = 0 \Rightarrow v_{2} = - i v_{1} .$ Take $v_{1} = 1$ : $v = (- i 1)$ . Split into real and imaginary parts: $v = (0 1) + i (- 1 0)$ , so $p = (0 1)$ , $q = (- 1 0)$ .

Step 3 — assemble two real solutions using the parent formula with $α = 1, β = 2$ : $x_{1} = e^{t} (p cos 2 t - q sin 2 t) = e^{t} (s i n 2 t c o s 2 t),$ $x_{2} = e^{t} (p sin 2 t + q cos 2 t) = e^{t} (- c o s 2 t s i n 2 t) .$ $x = e^{t} [c_{1} (sin 2 t cos 2 t) + c_{2} (- cos 2 t sin 2 t)] .$

Geometry. $α = 1 > 0$ means the radius grows: an outward spiral (unstable spiral / focus). Rotation period is $2 π / β = π$ . See the figure below.

Figure — Systems of first-order linear ODEs — matrix method

Exercise 3.2

For $A = (0 - 4 10)$ , decide the type of the equilibrium at the origin without writing the full solution, then confirm by finding $λ$ .

Recall Solution

Reasoning first. The trace of $A$ (sum of diagonal) is $0 + 0 = 0$ ; the determinant is $(0) (0) - (1) (- 4) = 4 > 0$ . For a $2 \times 2$ system, eigenvalues satisfy $λ^{2} - (tr) λ + det = 0$ , i.e. $λ^{2} + 4 = 0$ . Zero trace + positive determinant ⇒ purely imaginary eigenvalues ⇒ a center (closed orbits, neither growing nor decaying).

Confirm. $λ^{2} + 4 = 0 \Rightarrow λ = \pm 2 i$ . Here $α = 0$ : no growth, so trajectories are closed ellipses around the origin. ✅

Level 4 — Synthesis

Exercise 4.1

Solve the defective system $x^{'} = A x$ , $A = (4 - 1 12)$ , with $x (0) = (0 1)$ .

Recall Solution

Step 1. $det (A - λ I) = (4 - λ) (2 - λ) - (1) (- 1) = (4 - λ) (2 - λ) + 1.$ Expand: $λ^{2} - 6 λ + 8 + 1 = λ^{2} - 6 λ + 9 = (λ - 3)^{2} = 0 \Rightarrow λ = 3$ (double).

Step 2 — eigenvector. $(A - 3 I) = (1 - 1 1 - 1)$ . Row: $v_{1} + v_{2} = 0 \Rightarrow v = (- 1 1)$ . Only one independent eigenvector → defective (see parent note).

Step 3 — generalized eigenvector $w$ with $(A - 3 I) w = v$ : $(1 - 1 1 - 1) (w _{2} w _{1}) = (- 1 1) \Rightarrow w_{1} + w_{2} = 1.$ Pick $w_{1} = 1, w_{2} = 0$ : $w = (0 1)$ . Why the freedom? Any $w$ differing by a multiple of $v$ works; it only re-labels $c_{1}$ .

Step 4 — two independent solutions. $x_{1} = e^{3 t} (- 1 1), x_{2} = e^{3 t} (t (- 1 1) + (0 1)) .$ $x = c_{1} e^{3 t} (- 1 1) + c_{2} e^{3 t} (t (- 1 1) + (0 1)) .$

Step 5 — apply IC $x (0) = (0 1)$ . At $t = 0$ the $t$ -term vanishes: $c_{1} (- 1 1) + c_{2} (0 1) = (0 1) .$ Bottom component: $- c_{1} = 0 \Rightarrow c_{1} = 0$ . Top: $c_{1} + c_{2} = 1 \Rightarrow c_{2} = 1.$ $x (t) = e^{3 t} (t (- 1 1) + (0 1)) = e^{3 t} (- t t + 1) .$

Exercise 4.2

Build a $2 \times 2$ real matrix $A$ whose system $x^{'} = A x$ has the general solution $x = c_{1} e^{- t} (1 1) + c_{2} e^{2 t} (- 1 1) .$

Recall Solution

WHAT we reverse-engineer. We are told the eigenpairs: $(- 1, (1 1))$ and $(2, (- 1 1))$ . Any $A$ acting as $A v_{i} = λ_{i} v_{i}$ works, so we use Diagonalization: $A = P D P^{- 1}$ where $P$ has the eigenvectors as columns and $D$ is diagonal of eigenvalues. $P = (11 1 - 1), D = (- 1 0 02) .$ $det P = - 2$ , so $P^{- 1} = \frac{1}{- 2} (- 1 - 1 - 1 1) = (\frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2}) .$

Compute $A = P D P^{- 1}$ : $D P^{- 1} = (- 1 0 02) (\frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2}) = (- \frac{1}{2} 1 - \frac{1}{2} - 1) .$ $A = P (D P^{- 1}) = (11 1 - 1) (- \frac{1}{2} 1 - \frac{1}{2} - 1) = (\frac{1}{2} - \frac{3}{2} - \frac{3}{2} \frac{1}{2}) .$ $A = (\frac{1}{2} - \frac{3}{2} - \frac{3}{2} \frac{1}{2}) .$ Check: $tr A = 1 = λ_{1} + λ_{2} = (- 1) + 2$ ✅ and $det A = \frac{1}{4} - \frac{9}{4} = - 2 = λ_{1} λ_{2}$ ✅.

Level 5 — Mastery

Exercise 5.1

A system has $A = (0 - 1 1 - 2)$ . Classify the equilibrium fully (type + stability), solve it, and connect the answer to the second-order ODE $y^{''} + 2 y^{'} + y = 0$ .

Recall Solution

Step 1 — the connection. Let $x_{1} = y$ , $x_{2} = y^{'}$ . Then $x_{1}^{'} = x_{2}$ and $x_{2}^{'} = y^{''} = - 2 y^{'} - y = - x_{1} - 2 x_{2}$ , giving exactly $x^{'} = A x$ with this $A$ . This is the standard trick that turns any second-order ODE into a first-order system.

Step 2 — eigenvalues. $tr A = - 2$ , $det A = (0) (- 2) - (1) (- 1) = 1$ . Characteristic: $λ^{2} + 2 λ + 1 = (λ + 1)^{2} = 0 \Rightarrow λ = - 1$ (double). Note this is the same characteristic polynomial as $y^{''} + 2 y^{'} + y = 0$ .

Step 3 — defective? $(A + I) = (1 - 1 1 - 1)$ , rank 1, so only one eigenvector $v$ : $v_{1} + v_{2} = 0 \Rightarrow v = (- 1 1)$ . Defective.

Step 4 — generalized eigenvector. $(A + I) w = v$ : $w_{1} + w_{2} = 1$ . Take $w = (0 1)$ . $x = c_{1} e^{- t} (- 1 1) + c_{2} e^{- t} (t (- 1 1) + (0 1)) .$

Step 5 — classification. Both modes carry $e^{- t}$ (with a polynomial $t$ factor): $λ < 0$ , repeated, one eigenvector ⇒ improper (degenerate) stable node. Everything decays to the origin. This matches $y = (c_{1} + c_{2} t) e^{- t}$ , the critically-damped solution of the second-order ODE — same physics, two languages.

Exercise 5.2

Using the matrix exponential, show that $x (t) = e^{A t} x_{0}$ solves $x^{'} = A x$ , $x (0) = x_{0}$ , and evaluate $e^{A t}$ for $A = (20 0 - 1)$ .

Recall Solution

Step 1 — what $e^{A t}$ means. By analogy with the scalar series $e^{a t} = \sum_{k \geq 0} \frac{( a t ) ^{k}}{k !}$ , we define the Matrix Exponential $e^{A t} = \sum_{k = 0}^{\infty} \frac{( A t ) ^{k}}{k !} = I + A t + \frac{A ^{2} t ^{2}}{2 !} + \dots .$ This converges for every square $A$ .

Step 2 — why it solves the system. Differentiate term by term (allowed since the series converges nicely): $\frac{d}{d t} e^{A t} = A + A^{2} t + \frac{A ^{3} t ^{2}}{2 !} + \dots = A (I + A t + \frac{A ^{2} t ^{2}}{2 !} + \dots) = A e^{A t} .$ So $x = e^{A t} x_{0}$ gives $x^{'} = A e^{A t} x_{0} = A x$ . ✅ And at $t = 0$ , $e^{A \cdot 0} = I$ , so $x (0) = x_{0}$ . ✅

Step 3 — evaluate for diagonal $A$ . For a diagonal matrix, powers act entrywise: $A^{k} = (2^{k} 0 0 (- 1)^{k})$ . Summing each entry as its own scalar exponential series: $e^{A t} = (\sum_{k} \frac{( 2 t ) ^{k}}{k !} 0 0 \sum_{k} \frac{( - t ) ^{k}}{k !}) = (e^{2 t} 0 0 e^{- t}) .$ Then $x (t) = e^{A t} x_{0} = (x _{0, 2} e ^{- t} x _{0, 1} e ^{2 t})$ — exactly what the decoupled equations $x_{1}^{'} = 2 x_{1}, x_{2}^{'} = - x_{2}$ predict.

Recall One-line summary of the ladder

Recognize the template ::: $x = e^{λ t} v$ needs $A v = λ v$ Apply it ::: real distinct $λ$ → superpose exponentials Analyze ::: complex $λ = α \pm i β$ → spiral, stability from sign of $α$ Synthesize ::: defective → generalized eigenvector $e^{λ t} (t v + w)$ ; reverse-engineer with $A = P D P^{- 1}$ Master ::: 2nd-order ODEs are systems; $e^{A t}$ packages all solutions at once