Euler's Method (AA HL)
Not every differential equation has a solution that can be written down in a closed form. Euler's method instead builds a numerical estimate of the solution, stepping forward from a known starting point using the gradient at each point along the way. This page covers the formula, worked examples, and the mistakes that lose the most marks. It's part of the broader Differential Equations topic.
10 questions on this sub-topic.
The step formula
Covered under IB syllabus reference AHL5.18. The formula is in the booklet, but examiners expect you to apply it in an organised iteration table, not just quote it.
Euler's method
\[y_{n+1}=y_n+h\cdot f(x_n,y_n)\]
Step forward from a known starting point using the gradient at each step - smaller \(h\) gives a more accurate estimate, at the cost of more steps.
✓ In the formula bookletx-step
\(x_{n+1}=x_n+h\)
The \(x\)-value always increases by exactly \(h\) each step - keep a running column for it alongside \(y_n\) so nothing gets muddled.
Need the full syllabus wording, separable equations, and the integrating factor too? See Differential Equations.
Worked examples
Use Euler's method with step length \(h=0.5\) to estimate \(y(1)\) for the differential equation \(\dfrac{dy}{dx}=x+y,\ y(0)=1.\)
Worked solution
\(x_{n+1}=x_n+h,\ y_{n+1}=y_n+h\cdot f(x_n,y_n)\) with \(f(x,y)=x+y.\) M1
| \(n\) | \(x_n\) | \(y_n\) | \(f(x_n,y_n)=x_n+y_n\) |
|---|---|---|---|
| 0 | 0 | 1 | 1 |
| 1 | 0.5 | \(1+0.5(1)=1.5\) | 2 |
| 2 | 1.0 | \(1.5+0.5(2)\) M1 \(=2.5\) | - |
\(y(1)\approx2.5.\) A1
Use Euler's method with step length \(h=0.1\) to estimate \(y(0.3)\) for \(\dfrac{dy}{dx}=2x-y,\ y(0)=1,\) giving your answer to 3 decimal places.
Worked solution
\(y_{n+1}=y_n+h(2x_n-y_n).\) M1
| \(n\) | \(x_n\) | \(y_n\) | \(f(x_n,y_n)=2x_n-y_n\) |
|---|---|---|---|
| 0 | 0.0 | 1 | \(-1\) |
| 1 | 0.1 | \(1+0.1(-1)=0.9\) | \(-0.7\) |
| 2 | 0.2 | \(0.9+0.1(-0.7)=0.83\) | \(-0.43\) |
| 3 | 0.3 | \(0.83+0.1(-0.43)\) M1 \(=0.787\) | - |
\(y(0.3)\approx0.787.\) A1
Casio fx-CG50 / fx-CG100: Spreadsheet mode - enter \(x,y\) columns and fill down \(y_{n+1}=y_n+0.1(2x_n-y_n)\).
TI-Nspire CX: Lists & Spreadsheet - build the recursive column formula for \(y_n\).
On the GDC: a recursive list or spreadsheet column set up as \(y_n=y_{n-1}+0.1(2x_{n-1}-y_{n-1})\) fills in every step in seconds. See the parent topic's GDC guidance for calculator-specific steps.
Common mistakes
- Using too large a step size without saying so. A bigger \(h\) means fewer steps but a less accurate estimate - if the question asks for a specific \(h,\) use exactly that value, not a bigger one to save arithmetic.
- Substituting the new \(x\)-value into \(f(x,y)\) before updating \(y\). Each step uses the gradient at the old point \((x_n,y_n)\) to find the new \(y_{n+1}\) - working out \(f\) at the wrong pair of coordinates throws off every step that follows.
- Rounding too early in a multi-step iteration. Carry full accuracy through each intermediate step and only round the final answer to the number of decimal places the question asks for - early rounding compounds into a noticeably wrong estimate by the last step.
Ready to practise properly?
11 Euler's-method questions, marked instantly like the real exam.
Quick answers
What is Euler's method?
Euler's method is a numerical way to estimate the solution of \(\dfrac{dy}{dx}=f(x,y)\): starting from a known point, it steps forward using \(y_{n+1}=y_n+h\cdot f(x_n,y_n)\), where \(h\) is the chosen step length.
Does a smaller step size in Euler's method make it more accurate?
Yes - a smaller \(h\) produces an estimate closer to the true solution curve, but requires more steps to reach the same \(x\)-value, so there is a trade-off between accuracy and workload.