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 = Mx, its exact solution is built from the eigenvalues and eigenvectors of M — each eigenpair contributes an exponential term.

📘 What you need to know

Writing the system in matrix form

Matrix form of a coupled system
=
ab
cd
x
y
 i.e.  = Mx ✓ matrix form and the eigen-solution are in the booklet
Reading off M: the system dxdt = 2x − 3y, dydt = x + 4y has coefficient matrix
2−3
14
— 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 = 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

  1. Write = Mx and identify M.
  2. Find the eigenvalues λ₁, λ₂ and eigenvectors p1, p2 (often given).
  3. Write x = Aeλ₁tp1 + Beλ₂tp2.
  4. Substitute the initial condition at t = 0 (where e0 = 1).
  5. Solve the simultaneous equations for A and B.

Worked examples

Examples 1–4 use dxdt = 4xy, dydt = 2x + y, with x = 2, y = 1 initially. The matrix

4−1
21
has eigenvalues 3 and 2 with eigenvectors
1
1
and
1
2
.

WE 1

Write the system in matrix form.

Coefficients of x and y form the rows.

=
4−1
21
x
y
WE 2

Write down the general solution.

Drop each eigenvalue and eigenvector into the formula.

x = A e3t
1
1
+ B e2t
1
2
WE 3

Use the initial condition to find A and B.

At t = 0 both exponentials are 1.

at t = 0: A
1
1
+ B
1
2
=
2
1
A + B = 2,   A + 2B = 1 A = 3, B = −1
WE 4

State the exact solution, and hence x(t) and y(t).

Put A = 3, B = −1 back in and read off each component.

x = 3e3t
1
1
− e2t
1
2
x = 3e3t − e2t,   y = 3e3t − 2e2t
WE 5

Write dxdt = 2x − 3y, dydt = x + 4y in matrix form.

Each equation gives a row of M.

=
2−3
14
x
y

💡 Top tips

⚠ Common mistakes

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 xy 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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