| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Iterative/numerical methods |
| Difficulty | Challenging +1.8 This question requires multiple differentiations of an implicit equation to find higher derivatives, then careful substitution into a Taylor series. While the techniques are standard for Further Maths, the algebraic manipulation is substantial and error-prone, and connecting differential equations to Taylor series requires conceptual maturity beyond typical A-level questions. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08a Maclaurin series: find series for function4.10a General/particular solutions: of differential equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\frac{dy}{dx}-y^3=4 \Rightarrow \frac{dy}{dx}+x\frac{d^2y}{dx^2}-3y^2\frac{dy}{dx}=0\) | M1A1 | \(x\frac{dy}{dx} \rightarrow \frac{dy}{dx}+x\frac{d^2y}{dx^2}\) or \(y^3 \rightarrow ky^2\frac{dy}{dx}\). Any fully correct second derivative expression. Coefficient slips condoned. |
| \(\frac{d^2y}{dx^2}+x\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}-6y\left(\frac{dy}{dx}\right)^2-3y^2\frac{d^2y}{dx^2}=0\) | dM1 | Differentiates again using product rule correctly at least once. Depends on first M mark. |
| \(\Rightarrow x\frac{d^3y}{dx^3}=6y\left(\frac{dy}{dx}\right)^2+(3y^2-2)\frac{d^2y}{dx^2}\) | A1 | Correct answer from correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=2,\ y=1 \Rightarrow \left(\frac{dy}{dx}\right)_{x=2}=\frac{4+1}{2}=\frac{5}{2},\quad \left(\frac{d^2y}{dx^2}\right)_{x=2}=\frac{3(1)\left(\frac{5}{2}\right)-\frac{5}{2}}{2}=\frac{5}{2}\) \(\left(\frac{d^3y}{dx^3}\right)_{x=2}=\frac{6(1)\left(\frac{25}{4}\right)+\frac{5}{2}}{2}=20\) | M1 | Attempts to find values for first 3 derivatives at \(x=2\) |
| \((y=)f(2)+(x-2)f'(2)+\frac{(x-2)^2}{2}f''(2)+\frac{(x-2)^3}{6}f'''(2)+\ldots\) \(=1+\frac{5}{2}(x-2)+\frac{5}{2}\cdot\frac{(x-2)^2}{2}+20\cdot\frac{(x-2)^3}{6}+\ldots\) | M1 | Correct application of Taylor's theorem up to at least \((x-2)^2\) term |
| \((y=)1+\frac{5}{2}(x-2)+\frac{5}{4}(x-2)^2+\frac{10}{3}(x-2)^3+\ldots\) | A1 | Correct simplified expression. May be called \(y\) or \(f(x)\) or unlabelled |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x\frac{dy}{dx}-y^3=4 \Rightarrow \frac{dy}{dx}+x\frac{d^2y}{dx^2}-3y^2\frac{dy}{dx}=0$ | M1A1 | $x\frac{dy}{dx} \rightarrow \frac{dy}{dx}+x\frac{d^2y}{dx^2}$ or $y^3 \rightarrow ky^2\frac{dy}{dx}$. Any fully correct second derivative expression. Coefficient slips condoned. |
| $\frac{d^2y}{dx^2}+x\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}-6y\left(\frac{dy}{dx}\right)^2-3y^2\frac{d^2y}{dx^2}=0$ | dM1 | Differentiates again using product rule correctly at least once. Depends on first M mark. |
| $\Rightarrow x\frac{d^3y}{dx^3}=6y\left(\frac{dy}{dx}\right)^2+(3y^2-2)\frac{d^2y}{dx^2}$ | A1 | Correct answer from correct work |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=2,\ y=1 \Rightarrow \left(\frac{dy}{dx}\right)_{x=2}=\frac{4+1}{2}=\frac{5}{2},\quad \left(\frac{d^2y}{dx^2}\right)_{x=2}=\frac{3(1)\left(\frac{5}{2}\right)-\frac{5}{2}}{2}=\frac{5}{2}$ $\left(\frac{d^3y}{dx^3}\right)_{x=2}=\frac{6(1)\left(\frac{25}{4}\right)+\frac{5}{2}}{2}=20$ | M1 | Attempts to find values for first 3 derivatives at $x=2$ |
| $(y=)f(2)+(x-2)f'(2)+\frac{(x-2)^2}{2}f''(2)+\frac{(x-2)^3}{6}f'''(2)+\ldots$ $=1+\frac{5}{2}(x-2)+\frac{5}{2}\cdot\frac{(x-2)^2}{2}+20\cdot\frac{(x-2)^3}{6}+\ldots$ | M1 | Correct application of Taylor's theorem up to at least $(x-2)^2$ term |
| $(y=)1+\frac{5}{2}(x-2)+\frac{5}{4}(x-2)^2+\frac{10}{3}(x-2)^3+\ldots$ | A1 | Correct simplified expression. May be called $y$ or $f(x)$ or unlabelled |
---2.
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y ^ { 3 } = 4$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$x \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = a y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + \left( b y ^ { 2 } + c \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$
where $a$, $b$ and $c$ are integers to be determined.
Given that $y = 1$ at $x = 2$
\item determine the Taylor series expansion for $y$ in ascending powers of $( x - 2 )$, up to and including the term in $( x - 2 ) ^ { 3 }$, giving each coefficient in simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2024 Q2 [7]}}