IB Maths AI HL
Coupled & Second Order Differential Equations
Paper 1 & 2
~7 min read
Coupled Differential Equations
When two quantities change with respect to time and each rate depends on both quantities, you have a coupled system. Written in matrix form ẋ = Mx, its exact solution is built from the eigenvalues and eigenvectors of M — each eigenpair contributes an exponential term.
📘 What you need to know
- The system: dxdt = ax + by, dydt = cx + dy.
- Matrix form: ẋ = Mx, with M = the coefficient matrix and the dot meaning ddt.
- Exact solution: x = Aeλ₁tp1 + Beλ₂tp2, using eigenvalues λ and eigenvectors p.
- Real, distinct, non-zero: the exam only asks for exact solutions in this case.
- Find A, B: substitute the initial condition and solve simultaneous equations.
- A, B are constants of integration; the eigen-formula is in the booklet.
Writing the system in matrix form
Reading off M: the system
dxdt = 2
x − 3
y,
dydt =
x + 4
y has coefficient matrix
— top row from the first equation, bottom row from the second.
The exact solution
Exact solution (real, distinct, non-zero eigenvalues)
x = Aeλ₁tp1 + Beλ₂tp2
✓ given in the formula booklet
🤔 Why do eigenvalues and eigenvectors give the solution?
Along an eigenvector direction p, multiplying by M just scales by λ. So a trial solution eλtp has derivative λeλtp = M(eλtp), satisfying ẋ = Mx exactly. Two distinct eigenpairs give two independent solutions, and a general linear combination of them — with constants A, B — covers every solution.
🧠 “Each eigenpair → one eλt term”
Every eigenvalue–eigenvector pair contributes a piece eλtp. Add them with constants A, B, then use the starting values to pin those constants down.
🧭 Recipe — solving a coupled system
- Write ẋ = Mx and identify M.
- Find the eigenvalues λ₁, λ₂ and eigenvectors p1, p2 (often given).
- Write x = Aeλ₁tp1 + Beλ₂tp2.
- Substitute the initial condition at t = 0 (where e0 = 1).
- Solve the simultaneous equations for A and B.
Worked examples
Examples 1–4 use dxdt = 4x − y, dydt = 2x + y, with x = 2, y = 1 initially. The matrix has eigenvalues 3 and 2 with eigenvectors and .
WE 1Write the system in matrix form.
Coefficients of x and y form the rows.
WE 2Write down the general solution.
Drop each eigenvalue and eigenvector into the formula.
WE 3Use the initial condition to find A and B.
At t = 0 both exponentials are 1.
at t = 0: A + B =
A + B = 2, A + 2B = 1
A = 3, B = −1 WE 4State the exact solution, and hence x(t) and y(t).
Put A = 3, B = −1 back in and read off each component.
x = 3e3t − e2t
x = 3e3t − e2t, y = 3e3t − 2e2t WE 5Write dxdt = 2x − 3y, dydt = x + 4y in matrix form.
Each equation gives a row of M.
💡 Top tips
- Rows of M come straight from the two equations’ coefficients.
- One eλt term per eigenpair — match each eigenvalue with its eigenvector.
- Substitute at t = 0 so every exponential becomes 1.
- Solve simultaneously for A and B.
- Read components: the top row of the vector is x, the bottom is y.
- Check by differentiating — your solution should satisfy both equations.
⚠ Common mistakes
- Pairing the wrong eigenvector with an eigenvalue.
- Forgetting e0 = 1 when substituting the initial condition.
- Mislabelling rows/columns of M.
- Solving the simultaneous equations carelessly — sign slips in A, B.
- Mixing up x and y when reading the vector components.
Next up — Phase Portraits. You can now solve a coupled system exactly when the eigenvalues are real and distinct. The next topic visualises these systems: a phase portrait sketches typical solution trajectories in the x–y plane, and the same eigenvalues and eigenvectors that built the exact solution control its whole shape — the eigenvector lines and whether trajectories flow towards or away from the origin.
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