| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| 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 (part a) and then solving a linear first-order ODE with integrating factor (part b). While it involves multiple steps and is from Further Maths content, the substitution is provided and the techniques are routine applications of the syllabus material, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. \(2y\frac{dy}{dx}=-\frac{1}{t^2}\frac{dt}{dx}\) or \(y^2=\frac{1}{t}\Rightarrow t=y^{-2}\Rightarrow\frac{dt}{dx}=-2y^{-3}\frac{dy}{dx}\) | B1 | Any appropriate correct first derivative statement connecting \(\frac{dy}{dx}\) and \(\frac{dt}{dx}\). May be implied by use of chain rule |
| \(\frac{dy}{dx}+y=xy^3 \Rightarrow -\frac{t^{-\frac{3}{2}}}{2}\frac{dt}{dx}+\frac{1}{\sqrt{t}}=xt^{-\frac{3}{2}}\) | M1 | Substitutes completely into given DE to obtain equation in \(x\) and \(t\) only |
| \(\frac{dt}{dx}-2t=-2x\) | A1* | Achieves printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I=e^{\int-2\,dx}=e^{-2x}\) | B1 | Correct integrating factor. May be implied by working |
| \(te^{-2x}=-2\int xe^{-2x}\,dx\) | M1 | For their \(IF\times t = {-2}\int x\times\) their \(IF\,(dx)\) provided IF is a function in \(x\) |
| \(=-2\left(xe^{-2x}-\int e^{-2x}\,dx\right)\) | M1 | Integrates RHS by parts to obtain \(\alpha xe^{\pm kx}+\beta\int e^{\pm kx}\,(dx)\) |
| \(=xe^{-2x}+\frac{1}{2}e^{-2x}(+c)\) | A1 | Correct RHS (not follow through). The \(+c\) not needed for this mark |
| \(t=\frac{1}{y^2}=x+\frac{1}{2}+ce^{2x}\) \(\Rightarrow y^2=\frac{1}{x+\frac{1}{2}+ce^{2x}}\) | M1A1 | Reverses substitution making \(y^2\) subject (constant of integration must be introduced correctly); correct answer in required form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m-2=0\Rightarrow m=2\), so CF is \(t=Ae^{2x}\) | B1 | Solves auxiliary equation and forms correct CF |
| For PI try \(t=ax+b\) | M1 | Selects appropriate form for PI |
| \(\frac{dt}{dx}-2t=-2x \Rightarrow a-2(ax+b)=-2x \Rightarrow a=\ldots, b=\ldots\) | M1 | Differentiates and substitutes into DE to find PI constants |
| \(a=1,\ b=\frac{1}{2} \Rightarrow t=Ae^{2x}+x+\frac{1}{2}\) | A1 | Correct values, CF and PI added (\(t=\) may be missing) |
| \(t=\frac{1}{y^2}=x+\frac{1}{2}+Ae^{2x}\) \(\Rightarrow y^2=\frac{1}{x+\frac{1}{2}+Ae^{2x}}\) | M1A1 | Reverses substitution making \(y^2\) subject. Must have used both PI and CF. Correct answer in required form |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $2y\frac{dy}{dx}=-\frac{1}{t^2}\frac{dt}{dx}$ or $y^2=\frac{1}{t}\Rightarrow t=y^{-2}\Rightarrow\frac{dt}{dx}=-2y^{-3}\frac{dy}{dx}$ | B1 | Any appropriate correct first derivative statement connecting $\frac{dy}{dx}$ and $\frac{dt}{dx}$. May be implied by use of chain rule |
| $\frac{dy}{dx}+y=xy^3 \Rightarrow -\frac{t^{-\frac{3}{2}}}{2}\frac{dt}{dx}+\frac{1}{\sqrt{t}}=xt^{-\frac{3}{2}}$ | M1 | Substitutes completely into given DE to obtain equation in $x$ and $t$ only |
| $\frac{dt}{dx}-2t=-2x$ | A1* | Achieves printed answer with no errors |
## Part (b) - Main scheme (integrating factor):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I=e^{\int-2\,dx}=e^{-2x}$ | B1 | Correct integrating factor. May be implied by working |
| $te^{-2x}=-2\int xe^{-2x}\,dx$ | M1 | For their $IF\times t = {-2}\int x\times$ their $IF\,(dx)$ provided IF is a function in $x$ |
| $=-2\left(xe^{-2x}-\int e^{-2x}\,dx\right)$ | M1 | Integrates RHS by parts to obtain $\alpha xe^{\pm kx}+\beta\int e^{\pm kx}\,(dx)$ |
| $=xe^{-2x}+\frac{1}{2}e^{-2x}(+c)$ | A1 | Correct RHS (not follow through). The $+c$ not needed for this mark |
| $t=\frac{1}{y^2}=x+\frac{1}{2}+ce^{2x}$ $\Rightarrow y^2=\frac{1}{x+\frac{1}{2}+ce^{2x}}$ | M1A1 | Reverses substitution making $y^2$ subject (constant of integration must be introduced correctly); correct answer in required form |
## Part (b) - ALT (complementary function + particular integral):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m-2=0\Rightarrow m=2$, so CF is $t=Ae^{2x}$ | B1 | Solves auxiliary equation and forms correct CF |
| For PI try $t=ax+b$ | M1 | Selects appropriate form for PI |
| $\frac{dt}{dx}-2t=-2x \Rightarrow a-2(ax+b)=-2x \Rightarrow a=\ldots, b=\ldots$ | M1 | Differentiates and substitutes into DE to find PI constants |
| $a=1,\ b=\frac{1}{2} \Rightarrow t=Ae^{2x}+x+\frac{1}{2}$ | A1 | Correct values, CF and PI added ($t=$ may be missing) |
| $t=\frac{1}{y^2}=x+\frac{1}{2}+Ae^{2x}$ $\Rightarrow y^2=\frac{1}{x+\frac{1}{2}+Ae^{2x}}$ | M1A1 | Reverses substitution making $y^2$ subject. Must have used both PI and CF. Correct answer in required form |\begin{enumerate}
\item (a) Show that the substitution $y ^ { 2 } = \frac { 1 } { t }$ transforms the differential equation
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + y = x y ^ { 3 }$$
into the differential equation
$$\frac { \mathrm { d } t } { \mathrm {~d} x } - 2 t = - 2 x$$
(b) Solve differential equation (II) and determine $y ^ { 2 }$ in terms of $x$.
\hfill \mbox{\textit{Edexcel F2 2024 Q4 [9]}}