| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Bernoulli equation |
| Difficulty | Challenging +1.2 This is a standard Bernoulli equation problem from Further Maths requiring a given substitution to linearize the DE, then solving using integrating factor. While it involves multiple steps (substitution verification, finding integrating factor, integration by parts), the substitution is provided and the techniques are routine for F2 students. Slightly above average difficulty due to the algebraic manipulation and integration by parts required, but follows a well-established template. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z=y^{-2} \Rightarrow \frac{dz}{dy}=-2y^{-3}\) or equivalent | B1 | Any correct equation following differentiation of substitution. Could be implied |
| \(\frac{dy}{dx}=\frac{dz}{dx}\cdot\frac{dy}{dz}\) e.g. \(\frac{dy}{dx}=-\frac{1}{2}y^3\frac{dz}{dx}\) | M1 A1 | M1: Correct chain rule linking \(\frac{dy}{dx}\) and \(\frac{dz}{dx}\); A1: Correct equation |
| \(x\frac{dy}{dx}+y+4x^2y^3\ln x=0 \Rightarrow -\frac{1}{2}xy^3\frac{dz}{dx}+y+4x^2y^3\ln x=0\) | dM1 | Dependent on M1. Substitutes \(\frac{dy}{dx}\) into differential equation |
| \(\frac{dz}{dx}-\frac{2}{xy^2}-8x\ln x=0 \Rightarrow \frac{dz}{dx}-\frac{2z}{x}=8x\ln x\)* | A1* | Correct given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{IF} = e^{\int-\frac{2}{x}dx}\) | M1 | Attempt integrating factor |
| \(= e^{-2\ln x} = x^{-2}\) | A1 | Correct integrating factor |
| \(x^{-2}z = \int x^{-2}(8x\ln x)\,dx\) | M1 | Multiply through by IF and integrate RHS |
| \(\int x^{-1}\ln x\,dx\): by parts \((u=\ln x,\,u'=x^{-1},\,v'=x^{-1},\,v=\ln x)\) \(\Rightarrow I=(\ln x)^2-kI \Rightarrow I=p(\ln x)^2\) | M1 | Integration by parts or substitution for \(\int x^{-1}\ln x\,dx\) |
| \(\int x^{-1}\ln x\,dx = \frac{1}{2}(\ln x)^2 [+c]\) | A1 | Correct integral |
| \(x^{-2}z=4(\ln x)^2+k \Rightarrow z=4x^2(\ln x)^2+kx^2\) \(\Rightarrow y^2=\frac{1}{4x^2(\ln x)^2+kx^2}\) oe | A1 | Correct expression for \(y^2\) |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z=y^{-2} \Rightarrow \frac{dz}{dy}=-2y^{-3}$ or equivalent | B1 | Any correct equation following differentiation of substitution. Could be implied |
| $\frac{dy}{dx}=\frac{dz}{dx}\cdot\frac{dy}{dz}$ e.g. $\frac{dy}{dx}=-\frac{1}{2}y^3\frac{dz}{dx}$ | M1 A1 | M1: Correct chain rule linking $\frac{dy}{dx}$ and $\frac{dz}{dx}$; A1: Correct equation |
| $x\frac{dy}{dx}+y+4x^2y^3\ln x=0 \Rightarrow -\frac{1}{2}xy^3\frac{dz}{dx}+y+4x^2y^3\ln x=0$ | dM1 | Dependent on M1. Substitutes $\frac{dy}{dx}$ into differential equation |
| $\frac{dz}{dx}-\frac{2}{xy^2}-8x\ln x=0 \Rightarrow \frac{dz}{dx}-\frac{2z}{x}=8x\ln x$* | A1* | Correct given answer |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{IF} = e^{\int-\frac{2}{x}dx}$ | M1 | Attempt integrating factor |
| $= e^{-2\ln x} = x^{-2}$ | A1 | Correct integrating factor |
| $x^{-2}z = \int x^{-2}(8x\ln x)\,dx$ | M1 | Multiply through by IF and integrate RHS |
| $\int x^{-1}\ln x\,dx$: by parts $(u=\ln x,\,u'=x^{-1},\,v'=x^{-1},\,v=\ln x)$ $\Rightarrow I=(\ln x)^2-kI \Rightarrow I=p(\ln x)^2$ | M1 | Integration by parts or substitution for $\int x^{-1}\ln x\,dx$ |
| $\int x^{-1}\ln x\,dx = \frac{1}{2}(\ln x)^2 [+c]$ | A1 | Correct integral |
| $x^{-2}z=4(\ln x)^2+k \Rightarrow z=4x^2(\ln x)^2+kx^2$ $\Rightarrow y^2=\frac{1}{4x^2(\ln x)^2+kx^2}$ oe | A1 | Correct expression for $y^2$ |\begin{enumerate}
\item (a) Show that the substitution $z = y ^ { - 2 }$ transforms the differential equation
\end{enumerate}
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y + 4 x ^ { 2 } y ^ { 3 } \ln x = 0 \quad x > 0$$
into the differential equation
$$\frac { \mathrm { d } z } { \mathrm {~d} x } - \frac { 2 z } { x } = 8 x \ln x \quad x > 0$$
(b) By solving differential equation (II), determine the general solution of differential equation (I), giving your answer in the form $y ^ { 2 } = \mathrm { f } ( x )$
\hfill \mbox{\textit{Edexcel F2 2023 Q7 [11]}}