| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Series solution from differential equation |
| Difficulty | Challenging +1.3 This is a Further Maths F2 question requiring differentiation of an implicit equation and series solution construction. While it involves second-order DEs and requires careful algebraic manipulation through multiple derivatives, the method is systematic and procedural: differentiate the given equation to find the third derivative, then use initial conditions to build the Maclaurin series term-by-term. The techniques are standard for F2, though the implicit differentiation and bookkeeping across multiple derivatives elevates it slightly above average difficulty. |
| Spec | 4.08a Maclaurin series: find series for function4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}\) seen in differentiation | B1 | Product of these terms seen. |
| \(\frac{d^3y}{dx^3} = -\frac{4}{y}\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + \frac{2}{y^2}\left(\frac{dy}{dx}\right)^3\) | M1A1A1 (4) | Divide by \(y\) and differentiate using chain and product rules; either RHS term correct; both correct. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \(x=0\): \(\frac{d^2y}{dx^2}=\frac{1}{2}(-2\times1^2+4)=1\) | B1 | Correct evaluation. |
| \(\frac{d^3y}{dx^3}=\frac{1}{2}(-5\times1+2)\times1=-\frac{3}{2}\) | M1 | Substitute to find \(\frac{d^3y}{dx^3}\). |
| \((y=)\ 2+x+(1)\frac{x^2}{2!}+\left(-\frac{3}{2}\right)\frac{x^3}{3!}+\ldots\) | M1 | Form Taylor series with their derivatives. |
| \(y=2+x+\frac{1}{2}x^2-\frac{1}{4}x^3+\ldots\) | A1 (4) | All terms correct. |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}$ seen in differentiation | B1 | Product of these terms seen. |
| $\frac{d^3y}{dx^3} = -\frac{4}{y}\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + \frac{2}{y^2}\left(\frac{dy}{dx}\right)^3$ | M1A1A1 (4) | Divide by $y$ and differentiate using chain and product rules; either RHS term correct; both correct. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $x=0$: $\frac{d^2y}{dx^2}=\frac{1}{2}(-2\times1^2+4)=1$ | B1 | Correct evaluation. |
| $\frac{d^3y}{dx^3}=\frac{1}{2}(-5\times1+2)\times1=-\frac{3}{2}$ | M1 | Substitute to find $\frac{d^3y}{dx^3}$. |
| $(y=)\ 2+x+(1)\frac{x^2}{2!}+\left(-\frac{3}{2}\right)\frac{x^3}{3!}+\ldots$ | M1 | Form Taylor series with their derivatives. |
| $y=2+x+\frac{1}{2}x^2-\frac{1}{4}x^3+\ldots$ | A1 (4) | All terms correct. |\begin{enumerate}
\item Given that
\end{enumerate}
$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 2 y = 0 \quad y > 0$$
(a) determine $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ in terms of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x }$ and $y$
Given that $y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ at $x = 0$\\
(b) determine a series solution for $y$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, giving each coefficient in its simplest form.
\hfill \mbox{\textit{Edexcel F2 2022 Q5 [8]}}