AS October 2020 Q1
1. The variables \(x\) and \(y\) satisfy the differential equation
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2y^2 - x - 1\]where \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3\) and \(y = 0\) at \(x = 0\)
Use the approximations
\[\left(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_n \approx \frac{(y_{n+1} - 2y_n + y_{n-1})}{h^2} \quad \text{and} \quad \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_n \approx \frac{(y_{n+1} - y_{n-1})}{2h}\]with \(h = 0.1\) to find an estimate for the value of \(y\) at \(x = 0.2\)
(7)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2y^2 - x - 1 \Rightarrow \left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{0} = 0 - 0 - 1 = -1\) | B1 | 1.1b |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_{0} = 3 \Rightarrow \dfrac{(y_1 - y_{-1})}{0.2} \approx 3\) | B1 | 1.1b |
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{0} \approx \dfrac{(y_1 - 2y_0 + y_{-1})}{h^2} \Rightarrow \dfrac{y_1 - 2(0) + y_{-1}}{0.01} \approx -1\) | M1 | 1.1b |
| \(y_1 \approx \dfrac{1}{2}(0.6 - 0.01) = 0.295\) | dM1 | 2.1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2y^2 - x - 1 \Rightarrow \left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{1} = 2(0.295)^2 - 0.1 - 1 = -0.92595\) | dM1 | 1.1b |
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{1} \approx \dfrac{(y_2 - 2y_1 + y_0)}{h^2} \Rightarrow \dfrac{y_2 - 2(0.295) + 0}{0.01} \approx -0.92595 \Rightarrow y_2 = \ldots\) | dM1 | 2.1 |
| \(y_2 \approx 2(0.295) - 0.92595 \times 0.01 = 0.581\ (3\text{ s.f.})\) | A1 | 1.1b |
| (7) | ||
| (7 marks) |
Notes
B1: Correct value for the second derivative using the differential equation
B1: Correct equation in terms of \(y_1\) and \(y_{-1}\) using the first order approximation
M1: Uses the second order approximation to obtain another equation in terms of \(y_1\) and \(y_{-1}\)
M1: Uses their two equations in \(y_1\) and \(y_{-1}\) and solves together to find \(y\) at \(x = 0.1\).
M1: Uses the differential equation with their \(y\) at \(x = 0.1\) and \(x = 0.1\) to find a value for the second derivative at \(x = 0.1\)
M1: Completes the process by using the second order approximation and their second derivative to obtain a value for \(y_2\)
A1: Correct value for \(y\) at \(x = 0.2\)
Note that all method marks are dependent