Systems of First-Order Differential Equations

Why systems?

Many real-world models are coupled: predator–prey, chemical reactions, electrical circuits with multiple loops, mechanical systems with several degrees of freedom. These are naturally systems of first-order ODEs: $\mathbf{x}' = \mathbf{F}(t, \mathbf{x}), \qquad \mathbf{x} = (x_1, \ldots, x_n).$

For linear constant-coefficient systems, this simplifies to $\mathbf{x}' = A \mathbf{x}$ with a constant matrix $A$, and we get to bring the full machinery of linear algebra (eigenvalues, eigenvectors, matrix exponentials) to bear.

Any higher-order linear ODE can also be converted into a first-order system — that's what numerical solvers like Runge-Kutta assume as input — so mastering systems pays twice.

The given system

$\frac{dx}{dt} = x + y, \qquad \frac{dy}{dt} = -x + y.$

In matrix form: $\mathbf{x}' = A \mathbf{x}, \quad A = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}.$

Step-by-step solution

Step 1 — Eigenvalues of $A$.

$\det(A - \lambda I) = \det \begin{pmatrix} 1 - \lambda & 1 \\ -1 & 1 - \lambda \end{pmatrix} = (1 - \lambda)^{2} - (1)(-1) = (1 - \lambda)^{2} + 1.$

Set to zero: $(1 - \lambda)^{2} = -1 \implies 1 - \lambda = \pm i \implies \lambda = 1 \pm i.$

Complex conjugate eigenvalues with $\alpha = 1, \beta = 1$.

Step 2 — Eigenvector for $\lambda_1 = 1 + i$.

Solve $(A - (1 + i) I) \mathbf{v} = \mathbf{0}$: $\begin{pmatrix} -i & 1 \\ -1 & -i \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \mathbf{0}.$

Top row: $-i v_1 + v_2 = 0 \implies v_2 = i v_1$. Take $\mathbf{v} = (1, i)^{T}$.

Step 3 — Complex exponential solution and its real form.

The complex solution is $\mathbf{x}_c(t) = e^{(1 + i) t} \begin{pmatrix} 1 \\ i \end{pmatrix} = e^{t} (\cos t + i \sin t) \begin{pmatrix} 1 \\ i \end{pmatrix}.$

Expand: $= e^{t} \begin{pmatrix} \cos t + i \sin t \\ -\sin t + i \cos t \end{pmatrix}.$

Real and imaginary parts are each a real solution. The general real solution is any linear combination: $\boxed{\, \mathbf{x}(t) = e^{t} \begin{pmatrix} C_1 \cos t + C_2 \sin t \\ -C_1 \sin t + C_2 \cos t \end{pmatrix} \,}$

Phase-plane interpretation

Eigenvalues $\lambda = \alpha \pm i \beta$ with $\alpha = 1 > 0$, $\beta = 1$:

• $\beta \ne 0$ → spiral (rotational component).
• $\alpha > 0$ → spiral outward (unstable spiral).

Every non-trivial solution is a spiral emerging from the origin with angular speed $1$ (one rotation per $2 \pi$ time) and exponential growth rate $1$ (amplitude doubles roughly every $\ln 2 \approx 0.69$ time units).

Classification chart for 2D linear systems

By the eigenvalues of $A$:

• Real, same sign: node. Stable if both negative, unstable if both positive.
• Real, opposite signs: saddle. Unstable; one stable manifold + one unstable manifold.
• Repeated real root: degenerate or star node, depending on whether $A$ has a full basis of eigenvectors.
• Complex conjugate, $\alpha > 0$: unstable spiral. (This problem.)
• Complex conjugate, $\alpha < 0$: stable spiral.
• Complex conjugate, $\alpha = 0$: center — pure rotation, neither stable nor unstable.

The eigenvalues' positions in the complex plane (the root locus) tell you the whole story of the phase portrait. See #197 for pictures.

Initial value problem

$\mathbf{x}(0) = (1, 0)^{T}$: $\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} C_1 \\ C_2 \end{pmatrix} \implies C_1 = 1, \; C_2 = 0.$ $\mathbf{x}(t) = e^{t} \begin{pmatrix} \cos t \\ -\sin t \end{pmatrix}.$

Check:

• At $t = 0$: $(1, 0)^{T}$. ✓
• $\mathbf{x}'(0) = A \mathbf{x}(0) = A (1, 0)^{T} = (1, -1)^{T}$. Directly: $\mathbf{x}'(t) = e^{t} (\cos t - \sin t, -\sin t - \cos t)^{T}$, at $t = 0$: $(1, -1)^{T}$. ✓

As $t \to \infty$, amplitude grows like $e^{t}$ and the vector rotates clockwise (because the angular phase $-\sin t$ in the $y$-component is clockwise). Classic outward spiral.

Matrix exponential viewpoint

The general solution can be written as $\mathbf{x}(t) = e^{A t} \mathbf{x}_0$ where $e^{A t}$ is the matrix exponential: $e^{A t} = \sum_{n = 0}^{\infty} \frac{(A t)^{n}}{n!}.$

For our matrix $A$, the clean formula is $e^{A t} = e^{t} \begin{pmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{pmatrix}$ (a scaled rotation). Applied to $\mathbf{x}_0$, it gives the solution with initial value $\mathbf{x}_0$.

This form is the generalisation of the scalar formula $x(t) = e^{a t} x_0$ from $x' = a x$. Very compact and conceptually clean.

Decoupling via eigenvectors

If $A$ has a basis of eigenvectors $\mathbf{v}_1, \ldots, \mathbf{v}_n$ with eigenvalues $\lambda_1, \ldots, \lambda_n$, the system decomposes into $n$ uncoupled scalar ODEs via the transformation $\mathbf{x} = V \mathbf{y}$ where $V = [\mathbf{v}_1 | \ldots | \mathbf{v}_n]$: $\mathbf{y}' = V^{-1} A V \mathbf{y} = D \mathbf{y} \quad \text{(diagonal)}.$

Each $y_i$ satisfies $y_i' = \lambda_i y_i$, trivially solved.

From higher-order ODE to system

Any $n$-th order ODE $y^{(n)} = f(t, y, y', \ldots, y^{(n-1)})$ becomes a first-order system by setting $x_1 = y, x_2 = y', \ldots, x_n = y^{(n-1)}$: $\begin{aligned} x_1' &= x_2 \\ x_2' &= x_3 \\ &\vdots \\ x_n' &= f(t, x_1, \ldots, x_n). \end{aligned}$

This is what numerical ODE solvers always expect.

Common mistakes

• Forgetting to take real and imaginary parts. Complex eigenvalues give complex solutions; the physically relevant real solutions are the real and imaginary parts of the complex exponential.
• Wrong orientation of the spiral. The sign convention depends on which axis you plot; always verify with a specific time check.
• Mis-reading eigenvalue stability. Stability of $\mathbf{x}' = A \mathbf{x}$ is determined by real parts of eigenvalues of $A$, not the eigenvalues themselves. $\lambda = -1 + 10 i$ means stable (real part $< 0$) even though $|\lambda|$ is large.
• Confusing "system" with "ODE." A system is a vector-valued ODE. Some textbooks and online resources mix the terms.

Try it in the visualization

Plot the trajectory in the $(x, y)$ phase plane as a spiral curve, with an overlay of the eigenvalue pair in the complex plane. Drag the initial condition — watch the same spiral shape, just starting from a different place (all trajectories are scalings / rotations of each other). Slide the coefficients of $A$ to move the eigenvalues across the imaginary axis and see the spiral flip between outward, circular (centre), and inward.