IB Analysis and Approaches HL · Integral calculus · Section 8 of 13

Euler's method

3 worked examples · By Álvaro López

Key idea

Many differential equations cannot be solved with formulas. Euler's method solves them approximately: it follows the tangent to the solution over a small step, stops, recalculates the gradient and repeats. It is HL content, and it is usually asked as a table of a few steps on Paper 1, or with the graphic display calculator (GDC) on Paper 2.

The idea: follow the tangent

You have a differential equation $\frac{d y}{d x} = f (x, y)$ and a starting point $(x_{0}, y_{0})$ through which the solution passes. The equation gives the gradient of the solution at any point $(x, y)$ , even if you do not know the solution.

If you take a small horizontal step $h$ (the step size) and follow the tangent instead of the curve, you make a small error. From the new point you calculate the gradient again with $f$ and advance another step. In this way you build a polygonal line that imitates the solution curve.

The number of steps you need is $n = \frac{x _{end} - x _{0}}{h}$ . With $h = 0.2$ to go from $x = 0$ to $x = 1$ you need $5$ steps.

Exact solution y=e^x and Euler polygon with step 0.25 for y'=y, y(0)=1; the polygon lies below the curve

Slope field

The slope field draws, on a grid of points, a short segment with the gradient $f (x, y)$ at each point. Any solution of the equation is a curve that just touches all those segments. Each Euler step does exactly that: it relies on one segment and moves in its direction.

Slope field of y'=x-y and the solution through (0,1)

Accuracy: step size and error

Euler gives an approximation, not the exact value. Two ideas are worth being clear about.

Errors accumulate: each step starts from a point that is already approximate, not from the exact point of the solution.

Worked examples

Example 1 Basic

Three Euler steps

Let $\frac{d y}{d x} = 2 x - y$ with $y = 1$ when $x = 0$ . Use Euler's method with step $h = 0.5$ .

a) Calculate $y_{1}$ .

b) Calculate $y_{2}$ .

c) Estimate the value of $y$ for $x = 1.5$ .

Solution

Here $f (x, y) = 2 x - y$ , $x_{0} = 0$ , $y_{0} = 1$ and $h = 0.5$ .

a) $y_{1} = y_{0} + h f (x_{0}, y_{0}) = 1 + 0.5 (0 - 1) = 0.5$ , with $x_{1} = 0.5$ .

b) $y_{2} = y_{1} + h f (x_{1}, y_{1}) = 0.5 + 0.5 (1 - 0.5) = 0.75$ , with $x_{2} = 1$ .

c) For $x = 1.5$ we need $y_{3} = y_{2} + h f (x_{2}, y_{2}) = 0.75 + 0.5 (2 - 0.75) = 1.375$ .

$n$	$x_{n}$	$y_{n}$
0	0	1
1	0.5	0.5
2	1	0.75
3	1.5	1.375

Answer: $y (1.5) \approx 1.375$ .

Example 2 Medium

Euler against the exact solution

Consider $\frac{d y}{d x} = y$ with $y (0) = 1$ .

a) Apply Euler's method with $h = 0.25$ to estimate $y (1)$ .

b) Apply the method with $h = 0.5$ to estimate $y (1)$ .

c) Solve the equation exactly and find $y (1)$ .

d) Which estimate is better? Is the estimate an underestimate or an overestimate?

Solution

a) With $f (x, y) = y$ each step multiplies by $1 + h = 1.25$ . You need $4$ steps: $y_{1} = 1.25, y_{2} = \frac{25}{16}, y_{3} = \frac{125}{64}, y_{4} = \frac{625}{256} \approx 2.441.$

b) With $h = 0.5$ there are $2$ steps and each one multiplies by $1.5$ : $y_{1} = 1.5, y_{2} = \frac{9}{4} = 2.25.$

c) Separating variables, $\frac{d y}{y} = d x$ , so $ln y = x + C$ , and with $y (0) = 1$ we get $y = e^{x}$ . Hence $y (1) = e \approx 2.718$ .

d) The error with $h = 0.25$ is $2.718 - 2.441 \approx 0.277$ and with $h = 0.5$ it is $2.718 - 2.25 \approx 0.468$ . The estimate with the smaller step is better. Both are underestimates, because $e^{x}$ is convex and the tangents lie below the curve.

Answer: $y (1) \approx 2.441$ ( $h = 0.25$ ), $y (1) \approx 2.25$ ( $h = 0.5$ ), $y (1) = e$ (exact); the $h = 0.25$ estimate is better, and it is an underestimate.

Example 3 Hard

Cooling a drink

A drink at $90^{\circ} C$ is left in a room at $20^{\circ} C$ . Its temperature $T$ (in degrees) at minute $t$ satisfies $\frac{d T}{d t} = - \frac{1}{5} (T - 20), T (0) = 90.$

a) Use Euler's method with $h = 1$ to estimate $T (3)$ .

b) Solve the equation exactly and calculate $T (3)$ to three significant figures.

c) Calculate the error of the estimate in part a) to three significant figures and state whether it is an underestimate or an overestimate.

d) Estimate $T (3)$ with $h = 0.5$ (six steps). Give the result to two decimal places.

Solution

a) With $f (t, T) = - \frac{1}{5} (T - 20)$ and $h = 1$ : $T_{1} = 90 - \frac{1}{5} (70) = 76, T_{2} = 76 - \frac{1}{5} (56) = 64.8, T_{3} = 64.8 - \frac{1}{5} (44.8) = 55.84.$

b) It is separable: $\frac{d T}{T - 20} = - \frac{1}{5} d t$ , so $ln (T - 20) = - \frac{t}{5} + C$ , that is, $T = 20 + K e^{- t /5}$ . With $T (0) = 90$ we get $K = 70$ . Then $T (3) = 20 + 70 e^{- 3/5} \approx 58.4.$

c) The error is $58.417 - 55.84 \approx 2.58$ . The estimate is an underestimate: the solution is convex (it decreases more and more slowly).

d) Each step with $h = 0.5$ multiplies the difference $T - 20$ by $1 - \frac{1}{2} \cdot \frac{1}{5} = 0.9$ : $T_{6} = 20 + 70 \cdot 0. 9^{6} = 20 + 70 \cdot 0.531441 \approx 57.20.$ The error falls to about $1.2$ : when the step is halved, the error is roughly halved.

In the exam: for "estimate using Euler's method" you must write the formula and the table of steps: the final result alone does not earn the method marks. If they ask you to compare with the exact solution, make clear which is which.

Common mistakes

Summary

The equation $\frac{d y}{d x} = f (x, y)$ gives the gradient of the solution at each point.
Euler: $x_{n + 1} = x_{n} + h$ and $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ , starting from $(x_{0}, y_{0})$ .
It is organised in a table $n, x_{n}, y_{n}$ , and you need $\frac{x _{end} - x _{0}}{h}$ steps.
Smaller step, more accuracy; the error accumulates. Convex solution: Euler underestimates; concave: it overestimates.
The slope field shows the direction of the solutions at each point of the grid.

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