MAST20029 · Engineering Mathematics
Systems of First-Order ODEs: Eigenvalue Solution
Section 2 opens by solving linear systems ẋ = Ax with eigenvalues and eigenvectors of A. The three cases — distinct real, repeated/deficient (needing a generalised eigenvector), and complex conjugate (taking real and imaginary parts) — each use a general-solution form provided on the formula sheet. This is examined as a full Q4-type question, often paired with a condition on the matrix (for instance, the parameter values that make the origin a centre).
What this chapter covers
- 011. The system ẋ = Ax and solving it through the eigenvalues of A
- 022. Characteristic equation det(A − λI) = 0 and finding eigenvectors
- 033. Distinct real case: x(t) = α₁ w₁ e^(λ₁t) + α₂ w₂ e^(λ₂t)
- 044. Repeated (deficient) case: a generalised eigenvector u and the t e^(λt) term
- 055. Complex case λ = a ± bi: take Re and Im of w e^(λt) for real solutions
- 066. Reading parameter conditions from the eigenvalues (e.g. a centre needs Re λ = 0)
- 077. Applying initial conditions to fix the constants α₁, α₂
- 088. Eigenvectors as the straight-line orbit directions (link to phase portraits)
Centre condition and the complex-eigenvalue solution
- 1 mark(a) Write A = [[a, −b], [1, a]] and the characteristic equation det(A − λI) = (a − λ)² + b = 0, so λ = a ± √(−b).
- 2 marksA centre needs pure imaginary eigenvalues, i.e. real part exactly zero: a = 0, and −b < 0 so that the root is imaginary, i.e. b > 0. Therefore a = 0, b > 0.
- 2 marks(b) With a = 0, b = 4: λ = ±2i. For λ = 2i solve (A − 2iI)w = 0: −2i w₁ − 4 w₂ = 0, so taking w₁ = 2 gives w = (2, −i).
- 2 marksForm w e^(2it) = (2, −i)(cos 2t + i sin 2t) and take real and imaginary parts (the provided complex-case formula): x(t) = α₁ (2 cos 2t, sin 2t) + α₂ (2 sin 2t, −cos 2t).
- 1 markState the general solution as above (closed orbits, consistent with a centre).
Key terms
- Linear system ẋ = Ax
- A coupled first-order system written in matrix form; its solutions are built from the eigenvalues and eigenvectors of the constant matrix A.
- Characteristic equation
- det(A − λI) = 0, whose roots are the eigenvalues λ of A. Each eigenvalue yields an eigenvector w solving (A − λI)w = 0.
- Distinct-real case
- When A has two different real eigenvalues, the general solution is α₁ w₁ e^(λ₁t) + α₂ w₂ e^(λ₂t) — two independent exponential modes along the eigenvectors.
- Generalised eigenvector
- For a repeated (deficient) eigenvalue with only one eigenvector, the extra solution uses a vector u solving (A − λI)u = w, giving a (w t + u) e^(λt) term.
- Complex-eigenvalue case
- When λ = a ± bi, real solutions come from taking the real and imaginary parts of w e^(λt): x(t) = α₁ Re(w e^(λt)) + α₂ Im(w e^(λt)).
- Centre condition
- The origin is a centre (closed orbits) precisely when the eigenvalues are pure imaginary, λ = ±bi — that is, the real part is exactly zero.
Systems of First-Order ODEs: Eigenvalue Solution FAQ
How do I solve a linear system ẋ = Ax?
Find the eigenvalues from det(A − λI) = 0, then the eigenvectors from (A − λI)w = 0, and assemble the general solution using the case that matches the eigenvalues: distinct real, repeated/deficient, or complex. The formula sheet gives the general-solution form for each case, so the work is in computing eigenvalues and eigenvectors correctly.
What do I do with complex eigenvalues?
Use one of the conjugate pair, say λ = a + bi with eigenvector w, form w e^(λt), expand with Euler's formula, and take the real and imaginary parts to get two real independent solutions: x(t) = α₁ Re(w e^(λt)) + α₂ Im(w e^(λt)). The final real answer must contain no i.
What happens when an eigenvalue is repeated?
If a repeated eigenvalue has two independent eigenvectors (geometric multiplicity 2) you proceed as in the distinct case. If it is deficient (only one eigenvector w), you need a generalised eigenvector u solving (A − λI)u = w, giving a second solution of the form (w t + u) e^(λt).
How do I find the condition for a centre?
A centre occurs exactly when the eigenvalues are pure imaginary, λ = ±bi, which means the real part is zero. Set up det(A − λI) = 0 with any parameters, require the real part of the roots to be zero and the discriminant to make them imaginary, and solve for the parameter values.
Exam move
Make the eigenvalue routine second nature: characteristic equation, eigenvalues, eigenvectors, then pick the matching general-solution form from the sheet. Practise all three cases — distinct real, deficient (generalised eigenvector), and complex (real and imaginary parts) — because the exam can ask any of them, and the complex case is where most i-related slips happen. For parameter questions, translate the desired behaviour into an eigenvalue condition (a centre means real part exactly zero; stability means negative real parts). Always apply initial conditions cleanly at the end, and connect eigenvectors to the straight-line orbits you will draw in the next chapter.
