| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Bernoulli equation |
| Difficulty | Standard +0.8 This is a structured Bernoulli equation problem from Further Maths that guides students through the transformation and solution process. While it requires knowledge of Bernoulli equations and integrating factors (Further Maths topics), the question scaffolds the approach by providing the transformation and breaking it into clear steps. The algebraic manipulation and integrating factor application are standard techniques, making this moderately challenging but not requiring novel insight. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -\frac{1}{z^2}\frac{dz}{dx}\) | B1 | Correct differentiation of \(y = \frac{1}{z}\) |
| \(-\frac{x^2}{z^2}\frac{dz}{dx} + \frac{x}{z} = \frac{2}{z^2}\) | M1 | Substitutes into given differential equation |
| \(\frac{dz}{dx} - \frac{z}{x} = -\frac{2}{x^2}\) | A1* | Achieves printed answer with no errors; allow written down from correct substitution with no intermediate step |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = \frac{1}{x}\) | B1 | Correct integrating factor of \(\frac{1}{x}\) |
| \(\frac{z}{x} = -\int \frac{2}{x^3}\,dx\) | M1 | For \(Iz = -\int\frac{2I}{x^2}\,dx\); condone missing "dx" |
| \(\frac{z}{x} = \frac{1}{x^2} + c\) | A1 | Correct equation including constant |
| \(z = \frac{1}{x} + cx\) | A1 | Correct equation in required form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{y} = \frac{1}{x} + cx \Rightarrow -\frac{8}{3} = \frac{1}{3} + 3c \Rightarrow c = -1\) | M1 | Reverses substitution and uses given conditions to find constant |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{y} = \frac{1}{x} - x \Rightarrow y = \frac{x}{1-x^2}\) | A1 | Correct equation for \(y\) in terms of \(x\). Allow any correct equivalents e.g. \(y = \frac{1}{x^{-1}-x}\), \(y = \frac{1}{\frac{1}{x}-x}\) |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -\frac{1}{z^2}\frac{dz}{dx}$ | B1 | Correct differentiation of $y = \frac{1}{z}$ |
| $-\frac{x^2}{z^2}\frac{dz}{dx} + \frac{x}{z} = \frac{2}{z^2}$ | M1 | Substitutes into given differential equation |
| $\frac{dz}{dx} - \frac{z}{x} = -\frac{2}{x^2}$ | A1* | Achieves printed answer with no errors; allow written down from correct substitution with no intermediate step |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = \frac{1}{x}$ | B1 | Correct integrating factor of $\frac{1}{x}$ |
| $\frac{z}{x} = -\int \frac{2}{x^3}\,dx$ | M1 | For $Iz = -\int\frac{2I}{x^2}\,dx$; condone missing "dx" |
| $\frac{z}{x} = \frac{1}{x^2} + c$ | A1 | Correct equation including constant |
| $z = \frac{1}{x} + cx$ | A1 | Correct equation in required form |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{y} = \frac{1}{x} + cx \Rightarrow -\frac{8}{3} = \frac{1}{3} + 3c \Rightarrow c = -1$ | M1 | Reverses substitution and uses given conditions to find constant |
## Question (Inverse Function - first page):
| $\frac{1}{y} = \frac{1}{x} - x \Rightarrow y = \frac{x}{1-x^2}$ | A1 | Correct equation for $y$ in terms of $x$. Allow any correct equivalents e.g. $y = \frac{1}{x^{-1}-x}$, $y = \frac{1}{\frac{1}{x}-x}$ |
**Total: 2 marks**
---\begin{enumerate}
\item (a) Show that the transformation $y = \frac { 1 } { z }$ transforms the differential equation
\end{enumerate}
$$x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + x y = 2 y ^ { 2 }$$
into the differential equation
$$\frac { \mathrm { d } z } { \mathrm {~d} x } - \frac { z } { x } = - \frac { 2 } { x ^ { 2 } }$$
(b) Solve differential equation (II) to determine $z$ in terms of $x$.\\
(c) Hence determine the particular solution of differential equation (I) for which $y = - \frac { 3 } { 8 }$ at $x = 3$
Give your answer in the form $y = \mathrm { f } ( x )$.
\hfill \mbox{\textit{Edexcel F2 2023 Q3 [9]}}