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AP®MathsCollege BoardCalculus BCStudy GuidesUnit 7: Differential EquationsFirst-Order Differential EquationsApproximating Solutions Using Euler's Method

Approximating Solutions Using Euler's Method (College Board AP® Calculus BC)

Study Guide

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

Updated on

Euler's method

What is Euler’s method?

  • Euler’s method is a numerical method for finding approximate solutions to first order differential equations

  • It treats the derivatives in the equation as being constant over short ‘steps’

    • At each step you use a linear approximation to approximate the value at the next point

      • See the 'Approximating Values of a Function' study guide in the Linearization topic

  • The differential equation must be in the form fraction numerator d y over denominator d x end fraction equals f to the power of apostrophe open parentheses x comma blank y close parentheses

    • i.e. fraction numerator d y over denominator d x end fraction is expressed in terms of x and y

  • Start with a known value y subscript 0 corresponding to a value x subscript 0

    • Then y subscript 1 equals y subscript 0 plus increment x times f to the power of apostrophe open parentheses x subscript 0 comma space y subscript 0 close parentheses

      • increment x is the step size

    • y subscript 2 equals y subscript 1 plus increment x times f to the power of apostrophe open parentheses x subscript 1 comma space y subscript 1 close parentheses

      • x subscript 1 equals x subscript 0 plus increment x

    • y subscript 3 equals y subscript 2 plus increment x times f to the power of apostrophe open parentheses x subscript 2 comma space y subscript 2 close parentheses

      • x subscript 2 equals x subscript 1 plus increment x

    • etc.

Graph showing Euler's method with points (x0, y0) to (x4, y4). Black line for approximation, grey for exact solution, and slopes labelled.
Example of an Euler's method approximation

  • An Euler’s method approximation tends to wander away from the exact solution

    • The accuracy can be improved by making the step size smaller

Examiner Tips and Tricks

If an exam question asks you how to improve an Euler’s method approximation, the answer will almost always involve decreasing the step size increment x.

How do I use Euler’s method with a first order differential equation?

  • STEP 1: Make sure your differential equation is in fraction numerator d y over denominator d x end fraction equals f to the power of apostrophe open parentheses x comma blank y close parentheses form

  • STEP 2: Write down the recursion equations y subscript n plus 1 end subscript equals y subscript n plus increment x times f to the power of apostrophe open parentheses x subscript n comma blank y subscript n close parentheses and x subscript n plus 1 end subscript equals x subscript n plus increment x

    • An exam question may tell you the correct value of increment x to use 

      • or you may need to calculate it from other info provided

      • See the Worked Example

  • STEP 3: Use the recursion feature on your calculator to calculate the Euler’s method approximation over the correct number of steps

    • The values for x subscript 0 and y subscript 0 will come from the initial conditions given in the question

    • Your calculator will output the answer as a table of values

Examiner Tips and Tricks

Be careful with letters. In the exam (and in your calculator’s recursion app) the variables may not be x and y.

Worked Example

Consider the differential equation fraction numerator d y over denominator d x end fraction plus y equals x plus 1 with the initial condition space y left parenthesis 0 right parenthesis equals 0.5.

(a) Apply Euler’s method with five steps to approximate the solution to the differential equation at x equals 1.

First rearrange into fraction numerator d y over denominator d x end fraction equals f to the power of apostrophe open parentheses x comma blank y close parentheses form

fraction numerator d y over denominator d x end fraction equals x minus y plus 1

Next calculate the step size increment x; you need to go from x equals 0 to x equals 1 in five steps so

increment x equals fraction numerator 1 minus 0 over denominator 5 end fraction equals 0.2

Put that information into the recursion equations y subscript n plus 1 end subscript equals y subscript n plus increment x times f to the power of apostrophe open parentheses x subscript n comma blank y subscript n close parentheses and x subscript n plus 1 end subscript equals x subscript n plus increment x

y subscript n plus 1 end subscript equals y subscript n plus 0.2 times open parentheses x subscript n minus y subscript n plus 1 close parentheses

x subscript n plus 1 end subscript equals x subscript n plus 0.2

Put those equations into the recursion app on your calculator, and specify the required number of steps (n equals 5)

Your calculator will output the answers in a table

n

x subscript n

y subscript n

0

0

0.5

1

0.2

0.6

2

0.4

0.72

3

0.6

0.856

4

0.8

1.0048

5

1

1.1638

Note that the last row in the table gives the result you're looking for (i.e. the Euler approximation for y when x equals 1)

Round to 3 decimal places

y open parentheses 1 close parentheses almost equal to 1.164 space open parentheses 3 space straight d. straight p. close parentheses

(b) Explain how the accuracy of the approximation in part (a) could be improved.

Decrease the step size by using a greater number of steps to get from 0 to 1

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Mark Curtis

    Author: Mark Curtis

    Expertise: Maths

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.