Euler's Method for Coupled Systems

Rabbits and foxes affect each other: more rabbits means more food for foxes, and more foxes means fewer rabbits. A model of this needs two differential equations that depend on each other, a coupled system. These are rarely solvable by hand, but Euler’s method extends to them with almost no extra effort: you just step both variables forward at the same time. The same trick handles second-order equations like the motion of a spring. Working a few steps by hand is the best way to understand the method; in IB Mathematics AI HL exams the values themselves come from a spreadsheet or your GDC.

Key ideas

Euler’s method for one equation (recap)

For $\dfrac{dy}{dx} = f(x, y)$ with step size $h$:

$$x_{n+1} = x_n + h, \qquad y_{n+1} = y_n + h\,f(x_n, y_n)$$

New value = old value + step × slope at the old point.

Coupled systems

A coupled system has two quantities $x$ and $y$ that both change with time $t$, where each rate can depend on both quantities (and on $t$):

$$\frac{dx}{dt} = f_1(x, y, t), \qquad \frac{dy}{dt} = f_2(x, y, t)$$

Euler’s method updates both at once, using the old values in both formulas:

$$\begin{aligned} t_{n+1} &= t_n + h \\ x_{n+1} &= x_n + h\,f_1(x_n, y_n, t_n) \\ y_{n+1} &= y_n + h\,f_2(x_n, y_n, t_n) \end{aligned}$$

A table with columns $t$, $x$, $y$, $\dfrac{dx}{dt}$, $\dfrac{dy}{dt}$ keeps the work organized.

Using a spreadsheet or GDC

Beyond two or three steps, use technology. The guide expects spreadsheets for this, and in exams the values are generated with your GDC. In a spreadsheet, put $t$, $x$, $y$ in columns A, B, C, with the starting values in row 2. Then in row 3:

Cell	Formula (for step $h$ in cell F1)
A3	=A2+$F$1
B3	=B2+$F$1*(formula for dx/dt using A2, B2, C2)
C3	=C2+$F$1*(formula for dy/dt using A2, B2, C2)

Fill down as many rows as you need. Notice that both B3 and C3 refer to row 2 only. On a GDC, use the sequence or program mode in the same way, or a built-in Euler/differential equation tool if yours has one.

Second-order equations

A second-order equation $\dfrac{d^2x}{dt^2} = f\!\left(x, \dfrac{dx}{dt}, t\right)$ becomes a coupled first-order system if you give the velocity its own name. Let $y = \dfrac{dx}{dt}$. Then $\dfrac{d^2x}{dt^2} = \dfrac{dy}{dt}$, so

$$\frac{dx}{dt} = y, \qquad \frac{dy}{dt} = f(x, y, t)$$

You need two initial conditions: the starting value of $x$ and the starting value of $y = \dfrac{dx}{dt}$. Then use Euler’s method on the system as above. In context, $x$ is often a position and $y$ a velocity.

Accuracy

Each step follows a tangent line, so errors build up. A smaller step usually gives a better approximation but needs more steps. A good habit: repeat the calculation with half the step size. If the answers barely change, you can trust them more.

Worked examples

Example 1: Two steps by hand

Use Euler’s method with $h = 0.1$ to estimate $x(0.2)$ and $y(0.2)$ for

$$\frac{dx}{dt} = x - y, \qquad \frac{dy}{dt} = x + y, \qquad x(0) = 1,\ y(0) = 0$$

Solution.

$t$	$x$	$y$	$\dfrac{dx}{dt} = x - y$	$\dfrac{dy}{dt} = x + y$
$0$	$1$	$0$	$1$	$1$
$0.1$	$1 + 0.1(1) = 1.1$	$0 + 0.1(1) = 0.1$	$1$	$1.2$
$0.2$	$1.1 + 0.1(1) = 1.2$	$0.1 + 0.1(1.2) = 0.22$

So $x(0.2) \approx 1.2$ and $y(0.2) \approx 0.22$.

Example 2: A predator–prey model

On an island, the number of rabbits $x$ and foxes $y$ after $t$ years are modelled by

$$\frac{dx}{dt} = 0.5x - 0.01xy, \qquad \frac{dy}{dt} = -0.4y + 0.005xy$$

with $x(0) = 100$ and $y(0) = 20$.

(a) Explain the meaning of each term.

(b) Use Euler’s method with $h = 0.1$ to find the populations after $0.2$ years, by hand.

(c) Use technology with $h = 0.1$ to estimate both populations after $5$ years.

(d) Show that $(80, 50)$ is an equilibrium point and describe how the populations behave.

Solution.

(a) Without foxes, rabbits grow ($0.5x$). Meetings between rabbits and foxes (proportional to $xy$) reduce rabbits ($-0.01xy$) and help foxes grow ($+0.005xy$). Without rabbits, foxes die out ($-0.4y$).

(b)

$t$	$x$	$y$	$\dfrac{dx}{dt}$	$\dfrac{dy}{dt}$
$0$	$100$	$20$	$50 - 20 = 30$	$-8 + 10 = 2$
$0.1$	$103$	$20.2$	$30.694$	$2.323$
$0.2$	$106.0694$	$20.4323$

For example, at $t = 0.1$: $\dfrac{dx}{dt} = 0.5(103) - 0.01(103)(20.2) = 51.5 - 20.806 = 30.694$. After $0.2$ years there are about $106$ rabbits and $20$ foxes.

(c) With a spreadsheet (B3 =B2+0.1*(0.5*B2-0.01*B2*C2), C3 =C2+0.1*(-0.4*C2+0.005*B2*C2)), $50$ steps reach $t = 5$:

$$x(5) \approx 96.9, \qquad y(5) \approx 106 \quad \text{(3 s.f.)}$$

So after $5$ years there are about $97$ rabbits and $106$ foxes.

(d) At $(80, 50)$: $\dfrac{dx}{dt} = 40 - 0.01(80)(50) = 0$ and $\dfrac{dy}{dt} = -20 + 0.005(80)(50) = 0$. Neither population changes, so it is an equilibrium. Starting from $(100, 20)$, the populations oscillate around it: rabbits boom, then foxes boom (lagging behind), then rabbits crash, then foxes crash, and so on.

Euler’s method for the rabbit–fox model ($h = 0.1$). The populations oscillate, with foxes lagging behind rabbits; in the phase plane the path circles the equilibrium.

A comment on accuracy: for this model the true solutions repeat the same closed loop forever. The slow outward drift on the right is error from Euler’s method. With $h = 0.01$ the estimates at $t = 5$ become about $94.6$ rabbits and $102$ foxes, noticeably different from part (c), so the step $h = 0.1$ is only roughly accurate here.

Example 3: A second-order equation

A mass on a spring with some friction has displacement $x$ metres after $t$ seconds, where

$$\frac{d^2x}{dt^2} = -4x - 0.5\frac{dx}{dt}, \qquad x(0) = 1, \quad \frac{dx}{dt}(0) = 0$$

(a) Write this as a pair of coupled first-order equations.

(b) Use Euler’s method with $h = 0.1$ to estimate the displacement and velocity at $t = 0.3$.

Solution.

(a) Let $y = \dfrac{dx}{dt}$ (the velocity). Then

$$\frac{dx}{dt} = y, \qquad \frac{dy}{dt} = -4x - 0.5y, \qquad x(0) = 1,\ y(0) = 0$$

(b)

$t$	$x$	$y$	$\dfrac{dx}{dt} = y$	$\dfrac{dy}{dt} = -4x - 0.5y$
$0$	$1$	$0$	$0$	$-4$
$0.1$	$1$	$-0.4$	$-0.4$	$-4 + 0.2 = -3.8$
$0.2$	$0.96$	$-0.78$	$-0.78$	$-3.84 + 0.39 = -3.45$
$0.3$	$0.882$	$-1.125$

At $t = 0.3$ s the displacement is about $0.882$ m and the velocity about $-1.13$ m/s: the mass has been released and is speeding up back toward $x = 0$.

Common mistakes

Updating x first and then using the new x to update y. Both new values must be computed from the old pair $(x_n, y_n)$. In a spreadsheet, every formula in row $n + 1$ should refer only to row $n$.

Forgetting to multiply by the step size. The change in $x$ is $h \times \dfrac{dx}{dt}$, not $\dfrac{dx}{dt}$. With $h = 0.1$ a rate of $30$ rabbits per year adds only $3$ rabbits.

Missing the second initial condition. A second-order equation needs both $x(0)$ and $\dfrac{dx}{dt}(0)$. Without the starting velocity you can’t take the first step for $y$.

Writing dy/dt = x when converting a second-order equation. The first equation is always $\dfrac{dx}{dt} = y$ (the definition of $y$), and the second is the original equation with $\dfrac{d^2x}{dt^2}$ replaced by $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ replaced by $y$.

Forgetting to update t. If the rates depend on $t$, use the current $t_n$ at every step. It’s easy to leave $t = 0$ in every row by mistake.

Trusting a large step size. Euler estimates can drift far from the truth, especially for oscillating systems. Rerun with a smaller step and see whether the answer settles down before you interpret it.

Practice

1. (Warm-up) For $\dfrac{dx}{dt} = 2x - y$, $\dfrac{dy}{dt} = x + t$, with $x(0) = 1$ and $y(0) = 2$, use one step of Euler’s method with $h = 0.5$ to estimate $x(0.5)$ and $y(0.5)$.

Solution

At $t = 0$: $\dfrac{dx}{dt} = 2(1) - 2 = 0$ and $\dfrac{dy}{dt} = 1 + 0 = 1$.

$$x(0.5) \approx 1 + 0.5(0) = 1, \qquad y(0.5) \approx 2 + 0.5(1) = 2.5$$

2. (Warm-up) Write $\dfrac{d^2x}{dt^2} = 3x - 2\dfrac{dx}{dt} + t$, with $x(0) = 4$ and $\dfrac{dx}{dt}(0) = -1$, as a system of coupled first-order equations with initial conditions.

Solution

Let $y = \dfrac{dx}{dt}$:

$$\frac{dx}{dt} = y, \qquad \frac{dy}{dt} = 3x - 2y + t, \qquad x(0) = 4,\ y(0) = -1$$

3. (Warm-up) For $\dfrac{dx}{dt} = y$, $\dfrac{dy}{dt} = -x$, with $x(0) = 0$ and $y(0) = 1$, use Euler’s method with $h = 0.1$ to estimate $x(0.2)$ and $y(0.2)$.

Solution

$t$	$x$	$y$	$\dfrac{dx}{dt} = y$	$\dfrac{dy}{dt} = -x$
$0$	$0$	$1$	$1$	$0$
$0.1$	$0.1$	$1$	$1$	$-0.1$
$0.2$	$0.2$	$0.99$

$x(0.2) \approx 0.2$ and $y(0.2) \approx 0.99$. (The exact solution is $x = \sin t$, $y = \cos t$, giving $0.199$ and $0.980$ to 3 s.f.)

4. (Core) Two species compete for food. Their populations (in thousands) satisfy

$$\frac{dx}{dt} = x(3 - x - y), \qquad \frac{dy}{dt} = y(2 - x - y)$$

with $x(0) = 1$ and $y(0) = 0.5$. Use Euler’s method with $h = 0.1$ to estimate both populations at $t = 0.2$, to 3 s.f.

Solution

$t$	$x$	$y$	$\dfrac{dx}{dt}$	$\dfrac{dy}{dt}$
$0$	$1$	$0.5$	$1(1.5) = 1.5$	$0.5(0.5) = 0.25$
$0.1$	$1.15$	$0.525$	$1.15(1.325) = 1.52375$	$0.525(0.325) = 0.170625$
$0.2$	$1.302375$	$0.5420625$

$x(0.2) \approx 1.30$ thousand and $y(0.2) \approx 0.542$ thousand (3 s.f.).

5. (Core) The displacement $x$ of an undamped spring satisfies $\dfrac{d^2x}{dt^2} = -9x$, with $x(0) = 1$ and $\dfrac{dx}{dt}(0) = 0$.

• (a) Use Euler’s method with $h = 0.1$ to estimate $x(0.3)$.
• (b) The exact solution is $x = \cos 3t$. Find the percentage error in your estimate, and suggest how to improve it.

Solution

(a) Let $y = \dfrac{dx}{dt}$, so $\dfrac{dx}{dt} = y$ and $\dfrac{dy}{dt} = -9x$.

$t$	$x$	$y$	$\dfrac{dx}{dt} = y$	$\dfrac{dy}{dt} = -9x$
$0$	$1$	$0$	$0$	$-9$
$0.1$	$1$	$-0.9$	$-0.9$	$-9$
$0.2$	$0.91$	$-1.8$	$-1.8$	$-8.19$
$0.3$	$0.73$	$-2.619$

$x(0.3) \approx 0.73$.

(b) Exactly, $x(0.3) = \cos 0.9 \approx 0.6216$ (radians). The percentage error is $\dfrac{0.73 - 0.6216}{0.6216} \times 100\% \approx 17.4\%$. Use a smaller step size (with technology).

6. (Core) A predator–prey system is modelled by

$$\frac{dx}{dt} = 0.6x - 0.02xy, \qquad \frac{dy}{dt} = -0.3y + 0.004xy$$

where $x$ is the number of prey and $y$ the number of predators after $t$ months, with $x(0) = 50$ and $y(0) = 10$.

• (a) Find the equilibrium point with $x, y \gt 0$.
• (b) Write down the first step of Euler’s method with $h = 0.05$.
• (c) Use technology with $h = 0.05$ to estimate $x(2)$ and $y(2)$.
• (d) Describe what is likely to happen to the predators shortly after $t = 2$.

Solution

(a) $\dfrac{dx}{dt} = x(0.6 - 0.02y) = 0$ gives $y = 30$; $\dfrac{dy}{dt} = y(-0.3 + 0.004x) = 0$ gives $x = 75$. The equilibrium is $(75, 30)$.

(b) At $t = 0$: $\dfrac{dx}{dt} = 30 - 10 = 20$ and $\dfrac{dy}{dt} = -3 + 2 = -1$. So $x(0.05) \approx 50 + 0.05(20) = 51$ and $y(0.05) \approx 10 + 0.05(-1) = 9.95$.

(c) $40$ steps of a spreadsheet give $x(2) \approx 112$ and $y(2) \approx 10.1$ (3 s.f.).

(d) The prey population ($112$) is now well above $75$, so $-0.3 + 0.004x \gt 0$ and $\dfrac{dy}{dt} \gt 0$: with plenty of food, the predator population will start to grow quickly.

7. (Core) A damped spring satisfies $\dfrac{d^2x}{dt^2} + 0.4\dfrac{dx}{dt} + 4x = 0$, with $x(0) = 2$ and $\dfrac{dx}{dt}(0) = 0$.

• (a) Write the equation as a coupled first-order system.
• (b) Use technology to estimate $x(1)$ with $h = 0.1$ and with $h = 0.01$. Comment on your results.

Solution

(a) Let $y = \dfrac{dx}{dt}$. Rearranging, $\dfrac{d^2x}{dt^2} = -4x - 0.4\dfrac{dx}{dt}$, so

$$\frac{dx}{dt} = y, \qquad \frac{dy}{dt} = -4x - 0.4y, \qquad x(0) = 2,\ y(0) = 0$$

(b) With $h = 0.1$ ($10$ steps): $x(1) \approx -0.656$. With $h = 0.01$ ($100$ steps): $x(1) \approx -0.532$ (3 s.f.).

The two estimates differ by more than $0.1$, so the larger step isn’t reliable. The smaller step is more accurate. (A further check with $h = 0.001$ gives about $-0.518$, close to the true value of about $-0.516$.) Either way, after $1$ second the mass is on the other side of its rest position.

8. (Challenge) Consider $\dfrac{dx}{dt} = -y$, $\dfrac{dy}{dt} = x$, whose exact solutions move around circles centred at the origin.

• (a) Show that one step of Euler’s method with step $h$ multiplies $x^2 + y^2$ by $1 + h^2$.
• (b) Starting at $(1, 0)$ with $h = 0.1$, find the distance from the origin after $10$ steps, to 4 s.f. What does this tell you about Euler’s method for this system?

Solution

(a) One step gives $x_{n+1} = x_n - hy_n$ and $y_{n+1} = y_n + hx_n$. Then

$$\begin{aligned} x_{n+1}^2 + y_{n+1}^2 &= x_n^2 - 2hx_ny_n + h^2y_n^2 + y_n^2 + 2hx_ny_n + h^2x_n^2 \\ &= (1 + h^2)\left(x_n^2 + y_n^2\right) \end{aligned}$$

(b) The squared distance starts at $1$ and is multiplied by $1.01$ each step, so after $10$ steps it is $1.01^{10}$. The distance is $\sqrt{1.01^{10}} = 1.01^5 \approx 1.051$.

The true path stays on the unit circle, but every Euler step moves outward, so the Euler path spirals outward. Euler’s method always overestimates the size of the orbit here; smaller steps reduce, but never remove, the drift. (This is the same drift you can see in the phase plane in Example 2.)

9. (Challenge) A pendulum of length $1$ m swings so that its angle $\theta$ (in radians) from the vertical after $t$ seconds satisfies

$$\frac{d^2\theta}{dt^2} = -9.8\sin\theta, \qquad \theta(0) = 0.2, \quad \frac{d\theta}{dt}(0) = 0$$

• (a) Use Euler’s method with $h = 0.05$ to estimate $\theta$ and $\dfrac{d\theta}{dt}$ at $t = 0.1$, to 3 s.f.
• (b) Use technology with $h = 0.01$ to estimate the time when the pendulum first passes through the vertical.

Solution

(a) Let $\omega = \dfrac{d\theta}{dt}$, so $\dfrac{d\theta}{dt} = \omega$ and $\dfrac{d\omega}{dt} = -9.8\sin\theta$.

$t$	$\theta$	$\omega$	$\dfrac{d\theta}{dt} = \omega$	$\dfrac{d\omega}{dt} = -9.8\sin\theta$
$0$	$0.2$	$0$	$0$	$-1.94696$
$0.05$	$0.2$	$-0.097348$	$-0.097348$	$-1.94696$
$0.1$	$0.195133$	$-0.194696$

$\theta(0.1) \approx 0.195$ rad and $\dfrac{d\theta}{dt}(0.1) \approx -0.195$ rad/s (3 s.f.).

(b) The pendulum passes through the vertical when $\theta = 0$. In the spreadsheet with $h = 0.01$, $\theta$ is still positive at $t = 0.50$ ($\theta \approx 0.0021$) and first becomes negative at $t = 0.51$ ($\theta \approx -0.0043$). So it first passes through the vertical at about $t = 0.50$ s (a quarter of a full swing).