Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^3} - {x^3}$ for which $y = 1$ when $x = 0$. Use Euler’s method with a step length of $0.1$ to find an approximation for the value of $y$ when $x = 0.4$.

use of $y \to y + h\frac{{{\text{d}}y}}{{{\text{d}}x}}$     (M1)

Note: After the first line, award A1 for each subsequent $y$ value, provided it is correct to 3sf.

approximate value of $y(0.4) = 1.70$     A1

Note: Accept $1.7$ or any answers that round to $1.70$.

[7 marks]