Picture: two dials moving at once: Imagine two animal populations — rabbits R and foxes F — each changing at a rate that depends on both numbers. You have two rate equations running side by side: a coupled system.
Euler's method handles this by taking one small time-step h and nudging both quantities forward at the same time, using the rates calculated from the current row.
The golden rule: compute all the rates from the old values first, then update everything together — never feed a freshly-updated R into the fox rate for the same step.
IB-style question — rabbits and foxes
A reserve models its rabbit population R (hundreds) and fox population F (tens) by
dR/dt = 0.5R − 0.4F and dF/dt = 0.3R − 0.6F, where t is in years.
Initially R = 10 and F = 4.
Use Euler's method with step length h = 0.5 to estimate R and F after 1 year.
Step by step
- Reaching t = 1 from t = 0 in steps of 0.5 needs 1/0.5 = 2 steps. Start by finding both rates at the first row.
- Step 1: update BOTH from the old row, then advance t.
- Recompute the rates at the new row (t = 0.5).
- Step 2: update again to reach t = 1.
Final answer
After 1 year the model estimates R ≈ 13.8 (so ≈ 1380 rabbits) and F ≈ 4.77 (so ≈ 48 foxes). Both rates were always read off the SAME old row before stepping — that is what stops the two equations contaminating each other.
Lay it out as a table: Always draw a table with columns n, t, x (or R), y (or F), and the two derivatives. Fill the derivatives for a row first, then write the next row underneath.
On the GDC you can store this as a recurrence or a spreadsheet — but in the exam, neatly counting rows is what earns the marks and avoids the off-by-one trap.
Name the velocity, and the second-order DE splits in two: A second-order DE like d²x/dt² = g(t, x, dx/dt) describes things with acceleration — a spring, a falling object, an oscillating circuit.
The trick: give the first derivative its own name, v = dx/dt. Now d²x/dt² is just dv/dt, so the single second-order equation becomes a coupled pair:
• dx/dt = v (definition of v) • dv/dt = g(t, x, v) (the original DE, rewritten)
That is exactly the form Euler can step — apply the coupled rule to x and v together.
IB-style question — a vibrating spring
A mass on a spring obeys d²x/dt² = −4x, where x is the displacement from rest (cm) and t is in seconds.
It is released from rest at x = 3 cm, so x(0) = 3 and dx/dt = 0 at t = 0.
(a) Write the system as two first-order equations.
(b) Use Euler's method with h = 0.1 to estimate the displacement at t = 0.2 s.
Step by step
- (a) Let v = dx/dt. Then the second-order DE becomes a coupled pair.
- (b) Reaching t = 0.2 in steps of 0.1 is 2 steps. Row 0: x₀ = 3, v₀ = 0, so the acceleration is −4(3) = −12.
- Step 1: x uses the old v, v uses the old acceleration.
- New acceleration at row 1: −4(3) = −12. Step 2 to reach t = 0.2.
Final answer
(a) dx/dt = v, dv/dt = −4x. (b) The estimated displacement at t = 0.2 s is x ≈ 2.88 cm. Notice x barely moved on the first step because the spring started from rest (v₀ = 0) — Euler only lets the velocity build up gradually.
The derivative column you actually use: For a 2nd-order DE your table needs four working columns: n, t, x, v — plus the acceleration dv/dt = g(t, x, v) that drives the v-update.
Write the acceleration for a row before stepping. The x-update only ever needs v, but the v-update needs g, so it pays to compute g explicitly each row.