| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | First-order integration |
| Difficulty | Standard +0.8 This is a numerical methods question requiring application of finite difference approximations to solve a second-order differential equation. While the formulas are provided, students must correctly set up a system using both initial conditions (y and dy/dx at x=0), handle the boundary conditions carefully, and solve algebraically for y at x=0.2. This requires more problem-solving than routine FP1 questions but is a standard examination of the syllabus content. |
| Spec | 1.09g Numerical methods in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = 2y^2 - x - 1 \Rightarrow \left(\frac{d^2y}{dx^2}\right)_0 = 0 - 0 - 1 = -1\) | B1 | Correct value for the second derivative using the differential equation |
| \(\left(\frac{dy}{dx}\right)_0 = 3 \Rightarrow \frac{(y_1 - y_{-1})}{0.2} \approx 3\) | B1 | Correct equation in terms of \(y_1\) and \(y_{-1}\) using the first order approximation |
| \(\left(\frac{d^2y}{dx^2}\right)_0 \approx \frac{(y_1 - 2y_0 + y_{-1})}{h^2} \Rightarrow \frac{y_1 - 2(0) + y_{-1}}{0.01} \approx -1\) | M1 | Uses the second order approximation to obtain another equation in terms of \(y_1\) and \(y_{-1}\) |
| \(y_1 \approx \frac{1}{2}(0.6 - 0.01) = 0.295\) | dM1 | Uses their two equations in \(y_1\) and \(y_{-1}\) and solves together to find \(y\) at \(x = 0.1\) |
| \(\frac{d^2y}{dx^2} = 2y^2 - x - 1 \Rightarrow \left(\frac{d^2y}{dx^2}\right)_1 = 2(0.295)^2 - 0.1 - 1 = -0.92595\) | dM1 | Uses the differential equation with their \(y\) at \(x=0.1\) and \(x=0.1\) to find the second derivative at \(x=0.1\) |
| \(\left(\frac{d^2y}{dx^2}\right)_1 \approx \frac{(y_2 - 2y_1 + y_0)}{h^2} \Rightarrow \frac{y_2 - 2(0.295) + 0}{0.01} \approx -0.92595 \Rightarrow y_2 = \ldots\) | dM1 | Completes the process by using the second order approximation and their second derivative to obtain a value for \(y_2\) |
| \(y_2 \approx 2(0.295) - 0.92595 \times 0.01 = 0.581\) (3 s.f.) | A1 | Correct value for \(y\) at \(x = 0.2\) |
| Total: (7) | Note that all method marks are dependent |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = 2y^2 - x - 1 \Rightarrow \left(\frac{d^2y}{dx^2}\right)_0 = 0 - 0 - 1 = -1$ | B1 | Correct value for the second derivative using the differential equation |
| $\left(\frac{dy}{dx}\right)_0 = 3 \Rightarrow \frac{(y_1 - y_{-1})}{0.2} \approx 3$ | B1 | Correct equation in terms of $y_1$ and $y_{-1}$ using the first order approximation |
| $\left(\frac{d^2y}{dx^2}\right)_0 \approx \frac{(y_1 - 2y_0 + y_{-1})}{h^2} \Rightarrow \frac{y_1 - 2(0) + y_{-1}}{0.01} \approx -1$ | M1 | Uses the second order approximation to obtain another equation in terms of $y_1$ and $y_{-1}$ |
| $y_1 \approx \frac{1}{2}(0.6 - 0.01) = 0.295$ | dM1 | Uses their two equations in $y_1$ and $y_{-1}$ and solves together to find $y$ at $x = 0.1$ |
| $\frac{d^2y}{dx^2} = 2y^2 - x - 1 \Rightarrow \left(\frac{d^2y}{dx^2}\right)_1 = 2(0.295)^2 - 0.1 - 1 = -0.92595$ | dM1 | Uses the differential equation with their $y$ at $x=0.1$ and $x=0.1$ to find the second derivative at $x=0.1$ |
| $\left(\frac{d^2y}{dx^2}\right)_1 \approx \frac{(y_2 - 2y_1 + y_0)}{h^2} \Rightarrow \frac{y_2 - 2(0.295) + 0}{0.01} \approx -0.92595 \Rightarrow y_2 = \ldots$ | dM1 | Completes the process by using the second order approximation and their second derivative to obtain a value for $y_2$ |
| $y_2 \approx 2(0.295) - 0.92595 \times 0.01 = 0.581$ (3 s.f.) | A1 | Correct value for $y$ at $x = 0.2$ |
| **Total: (7)** | | **Note that all method marks are dependent** |\begin{enumerate}
\item The variables $x$ and $y$ satisfy the differential equation
\end{enumerate}
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 y ^ { 2 } - x - 1$$
where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$ and $y = 0$ at $x = 0$\\
Use the approximations
$$\left( \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \right) _ { n } \approx \frac { \left( y _ { n + 1 } - 2 y _ { n } + y _ { n - 1 } \right) } { h ^ { 2 } } \text { and } \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { \left( y _ { n + 1 } - y _ { n - 1 } \right) } { 2 h }$$
with $h = 0.1$ to find an estimate for the value of $y$ at $x = 0.2$
\hfill \mbox{\textit{Edexcel FP1 AS 2020 Q1 [7]}}