Standard +0.8 This is a numerical methods question requiring careful application of two different techniques (Euler's method and midpoint formula) to a non-linear differential equation. While the methods themselves are standard, the question demands precision with multiple steps, correct sequencing (Euler first to get y₁, then midpoint to get y₂), and accurate arithmetic with the given function. The 5 significant figure requirement and small step size increase computational care needed. This is moderately challenging but within reach for well-prepared Further Maths students.
A curve passes through the point \((-2, 4.73)\) and satisfies the differential equation
$$\frac{dy}{dx} = \frac{y^2 - x^2}{2x + 3y}$$
Use Euler's step by step method once, and then the midpoint formula
$$y_{r+1} = y_{r-1} + 2hf(x_r, y_r), \quad x_{r+1} = x_r + h$$
once, each with a step length of \(0.02\), to estimate the value of \(y\) when \(x = -1.96\)
Give your answer to five significant figures.
[4 marks]
Question 9:
9 | Uses Euler’s
method once to
calculate an
estimate of y
when x = −1.98
Condone one
error. | 1.1a | M1 | x = –2
0
y = 4.73
0
h = 0.02
4.732 –(–2)2
y = 4.73+0.02
1 2(–2)+3(4.73)
= 4.766060648
x = –1.98
1
4.7660606482 –(–1.98)2
y = 4.73+0.04
2 2(–1.98)+3(4.766060648)
= 4.8027
Obtains AWRT
4.766
PI by final
AWRT 4.8027 | 1.1b | A1
Uses midpoint
formula once with
their y
1
Condone one
error. | 3.1a | M1
Obtains AWRT
4.8027 | 1.1b | A1
Question total | 4
Q | Marking instructions | AO | Marks | Typical solution
A curve passes through the point $(-2, 4.73)$ and satisfies the differential equation
$$\frac{dy}{dx} = \frac{y^2 - x^2}{2x + 3y}$$
Use Euler's step by step method once, and then the midpoint formula
$$y_{r+1} = y_{r-1} + 2hf(x_r, y_r), \quad x_{r+1} = x_r + h$$
once, each with a step length of $0.02$, to estimate the value of $y$ when $x = -1.96$
Give your answer to five significant figures.
[4 marks]
\hfill \mbox{\textit{AQA Further Paper 2 2024 Q9 [4]}}