Differential Equations · step 13 of 14

Systems of linear ODEs via eigenvalues

For the system x′ = Ax, each eigenvalue–eigenvector pair (λ, v) gives a solution ve^(λt), and the general solution is their linear combination. The eigenvalues also classify the phase portrait completely: both negative gives a stable node, opposite signs a saddle, and complex eigenvalues a spiral whose stability follows the sign of the real part.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Find the eigenvalues of ASolve det(A − λI) = 0.
Find an eigenvector for eachSolve (A − λI)v = 0 — the system is singular by construction, so a free variable appears.
Assemble the general solutionx = c₁v₁e^(λ₁t) + c₂v₂e^(λ₂t) for distinct real eigenvalues.
Classify the equilibriumBoth λ < 0: stable node. Both > 0: unstable node. Opposite signs: saddle. Complex with negative real part: stable spiral.

Worked example

Solve x′ = Ax for A = [[1, 2], [0, 3]] and classify the origin.

A is triangular, so the eigenvalues are 1 and 3.
For λ = 1: (A − I) = [[0,2],[0,2]], giving y = 0 and v₁ = (1,0).
For λ = 3: (A − 3I) = [[−2,2],[0,0]], giving x = y and v₂ = (1,1).
Both eigenvalues are positive.

Answer. x = c₁(1,0)e^t + c₂(1,1)e^(3t); the origin is an unstable node, with all trajectories fleeing it.

Where marks get dropped

These are the specific errors that cost credit on systems of linear odes via eigenvalues questions — QED's rubric penalises each of them separately.

Pairing an eigenvector with the wrong eigenvalue in the solution. Each v must be multiplied by its own e^(λt).
Stopping at complex eigenvalues. Take real and imaginary parts to get two real solutions, giving the spiral form.
Forgetting the repeated-eigenvalue case, which needs a generalised eigenvector and a term of the form (vt + w)e^(λt).

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Systems of linear ODEs via eigenvalues — frequently asked questions

What is a saddle point?

An equilibrium with eigenvalues of opposite sign: trajectories approach along one eigenvector direction and escape along the other. It is always unstable.

How do complex eigenvalues appear in the phase portrait?

As rotation. λ = α ± βi gives spirals with rotation rate β, spiralling inwards when α < 0 and outwards when α > 0. Pure imaginary gives closed orbits — a centre.

How does this relate to second-order equations?

Any second-order equation converts to a 2×2 system by setting x₁ = y and x₂ = y′. The eigenvalues are exactly the characteristic roots.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

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