| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Substitution reducing to first order linear ODE |
| Difficulty | Challenging +1.3 This is a structured multi-part question where part (a) guides students through a given substitution, part (b) involves solving a separable differential equation (standard Further Maths technique), and parts (c-d) require back-substitution and curve sketching. While it requires multiple techniques and careful algebra, the substitution is provided and each step is scaffolded, making it more accessible than questions requiring independent insight into solution methods. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dv}{dx} = \frac{dy}{dx} - 2\) | B1 | Correct differentiation of transformation; allow any correct connecting derivative |
| \(\frac{dy}{dx} + 2yx(y-4x) = 2-8x^3 \Rightarrow \frac{dv}{dx}+2+2(v+2x)x(v+2x-4x) = 2-8x^3\) | M1 | Complete substitution into equation |
| \(\Rightarrow \frac{dv}{dx} + 2 + 2x(v^2-4x^2) = 2-8x^3\) | M1 | Applies difference of squares or expands |
| \(\Rightarrow \frac{dv}{dx} = -2xv^2\) * | A1* | Reaches given answer with no errors |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{1}{v^2}\frac{dv}{dx} = -2x \Rightarrow \int v^{-2}\,dv = -2\int x\,dx\) | B1 | Correct separation of variables |
| \(\frac{v^{-1}}{-1} = -2\frac{x^2}{2}(+c)\) | M1 | Power increased by 1 on at least one term |
| \(\frac{1}{v} = x^2 + c\) | A1 | Correct integration including \(+c\) |
| \(v = \frac{1}{x^2+c}\) | A1 | Correct expression for \(v\) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y = 2x + \frac{1}{x^2+c}\) | B1ft | Follow through their \(v\) from (b) |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(-1 = 2(-1) + \frac{1}{1+c} \Rightarrow c = \ldots\) | M1 | Uses point \((-1,-1)\); must have had constant of integration |
| \(y = 2x + \frac{1}{x^2}\) | A1 | Correct equation; withhold if \(y = 2x + \frac{1}{x^2} + c\) led here |
| Attempts sketch with at least one of: one branch correct / vertical asymptote / long-term behaviour to infinity / minimum in quadrant 1 | M1 | At least one key feature shown |
| Fully correct shape: two branches tending to infinity as \(x \to \pm\infty\), minimum in first quadrant | A1 | Not a follow-through; must be correct curve |
| \(y\)-axis labelled as vertical asymptote | B1ft | Follow through; need not label if clearly \(y\)-axis |
| (5) | ||
| (14 marks) |
# Question 8:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dv}{dx} = \frac{dy}{dx} - 2$ | B1 | Correct differentiation of transformation; allow any correct connecting derivative |
| $\frac{dy}{dx} + 2yx(y-4x) = 2-8x^3 \Rightarrow \frac{dv}{dx}+2+2(v+2x)x(v+2x-4x) = 2-8x^3$ | M1 | Complete substitution into equation |
| $\Rightarrow \frac{dv}{dx} + 2 + 2x(v^2-4x^2) = 2-8x^3$ | M1 | Applies difference of squares or expands |
| $\Rightarrow \frac{dv}{dx} = -2xv^2$ * | A1* | Reaches given answer with no errors |
| | **(4)** | |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{1}{v^2}\frac{dv}{dx} = -2x \Rightarrow \int v^{-2}\,dv = -2\int x\,dx$ | B1 | Correct separation of variables |
| $\frac{v^{-1}}{-1} = -2\frac{x^2}{2}(+c)$ | M1 | Power increased by 1 on at least one term |
| $\frac{1}{v} = x^2 + c$ | A1 | Correct integration including $+c$ |
| $v = \frac{1}{x^2+c}$ | A1 | Correct expression for $v$ |
| | **(4)** | |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $y = 2x + \frac{1}{x^2+c}$ | B1ft | Follow through their $v$ from (b) |
| | **(1)** | |
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $-1 = 2(-1) + \frac{1}{1+c} \Rightarrow c = \ldots$ | M1 | Uses point $(-1,-1)$; must have had constant of integration |
| $y = 2x + \frac{1}{x^2}$ | A1 | Correct equation; withhold if $y = 2x + \frac{1}{x^2} + c$ led here |
| Attempts sketch with at least one of: one branch correct / vertical asymptote / long-term behaviour to infinity / minimum in quadrant 1 | M1 | At least one key feature shown |
| Fully correct shape: two branches tending to infinity as $x \to \pm\infty$, minimum in first quadrant | A1 | Not a follow-through; must be correct curve |
| $y$-axis labelled as vertical asymptote | B1ft | Follow through; need not label if clearly $y$-axis |
| | **(5)** | |
| | **(14 marks)** | |\begin{enumerate}
\item (a) Show that the transformation $v = y - 2 x$ transforms the differential equation
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y x ( y - 4 x ) = 2 - 8 x ^ { 3 }$$
into the differential equation
$$\frac { \mathrm { d } v } { \mathrm {~d} x } = - 2 x v ^ { 2 }$$
(b) Solve the differential equation (II) to determine $v$ as a function of $x$\\
(c) Hence obtain the general solution of the differential equation (I).\\
(d) Sketch the solution curve that passes through the point $( - 1 , - 1 )$.
On your sketch show clearly the equation of any horizontal or vertical asymptotes.\\
You do not need to find the coordinates of any intercepts with the coordinate axes or the coordinates of any stationary points.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0d458344-42cb-48d1-90b3-e071df8ea7bb-32_2817_1962_105_105}
\end{center}
\hfill \mbox{\textit{Edexcel F2 2022 Q8 [14]}}