| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| 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 differential equation and series expansion. Part (a) involves careful application of the product rule and quotient rule to differentiate the given equation, which is algebraically demanding but methodical. Part (b) requires substituting initial conditions to find coefficients in a Taylor series, which is a standard technique. While the algebra is somewhat involved and the topic is Further Maths level, the question follows a well-established procedure without requiring novel insight or particularly deep conceptual understanding. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = \frac{4}{y}\left(\frac{dy}{dx}\right)^2 - 3\) | M1 | Divide through by \(y\) |
| \(\frac{d^3y}{dx^3} = -\frac{4}{y^2}\left(\frac{dy}{dx}\right)^3 + \frac{8}{y}\times\frac{d^2y}{dx^2}\times\frac{dy}{dx}\) | M1A1A1 | Attempt differentiation using product rule and chain rule; A1 either RHS term correct, A1 second RHS term correct with no extras |
| \(\frac{d^3y}{dx^3} = -\frac{4}{y^2}\left(\frac{dy}{dx}\right)^3 + \frac{8}{y}\left(\frac{4}{y}\left(\frac{dy}{dx}\right)^2 - 3\right)\left(\frac{dy}{dx}\right)\) | ||
| \(\frac{d^3y}{dx^3} = \frac{28}{y^2}\left(\frac{dy}{dx}\right)^3 - \frac{24}{y}\left(\frac{dy}{dx}\right)\) | A1* (5) | Eliminate \(\frac{d^2y}{dx^2}\) and obtain the given result |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx}\frac{d^2y}{dx^2} + y\frac{d^3y}{dx^3} - 8\frac{dy}{dx}\times\frac{d^2y}{dx^2} + 3\frac{dy}{dx} = 0\) | M1A1A1 | Re-arrange equation; attempt differentiation using product rule and chain rule; A1 two terms correct, A1 all correct |
| \(\frac{d^3y}{dx^3} = \frac{1}{y}\left(7\frac{dy}{dx}\right)\left(\frac{4}{y}\left(\frac{dy}{dx}\right)^2 - 3\right) - \frac{3}{y}\frac{dy}{dx}\) | M1 | |
| \(\frac{d^3y}{dx^3} = \frac{28}{y^2}\left(\frac{dy}{dx}\right)^3 - \frac{24}{y}\left(\frac{dy}{dx}\right)\) | A1* (5) | Eliminate \(\frac{d^2y}{dx^2}\) and obtain correct result |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| At \(x=0\): \(\frac{d^2y}{dx^2} = \frac{4}{8}(1)^2 - 3 = -\frac{5}{2}\) oe | B1 | Correct value for \(\frac{d^2y}{dx^2}\) |
| \(\frac{d^3y}{dx^3} = \frac{28}{64}\times 1^3 - \frac{24}{8}\times 1 = -\frac{41}{16}\) | M1 | Use the given expression from (a) to obtain value for \(\frac{d^3y}{dx^3}\); award if correct value seen |
| \(y = 8 + x - \frac{5}{2}\times\frac{x^2}{2!} - \frac{41}{16}\times\frac{x^3}{3!} + \ldots\) | M1 | Taylor's series formed using their values for derivatives (\(2!\) or \(2\), \(3!\) or \(6\)) |
| \(y = 8 + x - \frac{5}{4}x^2 - \frac{41}{96}x^3 + \ldots\) | A1 (4) | Correct series; must start (or end) \(y = \ldots\); correct terms must be seen |
## Question 4(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = \frac{4}{y}\left(\frac{dy}{dx}\right)^2 - 3$ | M1 | Divide through by $y$ |
| $\frac{d^3y}{dx^3} = -\frac{4}{y^2}\left(\frac{dy}{dx}\right)^3 + \frac{8}{y}\times\frac{d^2y}{dx^2}\times\frac{dy}{dx}$ | M1A1A1 | Attempt differentiation using product rule and chain rule; A1 either RHS term correct, A1 second RHS term correct with no extras |
| $\frac{d^3y}{dx^3} = -\frac{4}{y^2}\left(\frac{dy}{dx}\right)^3 + \frac{8}{y}\left(\frac{4}{y}\left(\frac{dy}{dx}\right)^2 - 3\right)\left(\frac{dy}{dx}\right)$ | | |
| $\frac{d^3y}{dx^3} = \frac{28}{y^2}\left(\frac{dy}{dx}\right)^3 - \frac{24}{y}\left(\frac{dy}{dx}\right)$ | A1* (5) | Eliminate $\frac{d^2y}{dx^2}$ and obtain the given result |
**ALT:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx}\frac{d^2y}{dx^2} + y\frac{d^3y}{dx^3} - 8\frac{dy}{dx}\times\frac{d^2y}{dx^2} + 3\frac{dy}{dx} = 0$ | M1A1A1 | Re-arrange equation; attempt differentiation using product rule and chain rule; A1 two terms correct, A1 all correct |
| $\frac{d^3y}{dx^3} = \frac{1}{y}\left(7\frac{dy}{dx}\right)\left(\frac{4}{y}\left(\frac{dy}{dx}\right)^2 - 3\right) - \frac{3}{y}\frac{dy}{dx}$ | M1 | |
| $\frac{d^3y}{dx^3} = \frac{28}{y^2}\left(\frac{dy}{dx}\right)^3 - \frac{24}{y}\left(\frac{dy}{dx}\right)$ | A1* (5) | Eliminate $\frac{d^2y}{dx^2}$ and obtain correct result |
## Question 4(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| At $x=0$: $\frac{d^2y}{dx^2} = \frac{4}{8}(1)^2 - 3 = -\frac{5}{2}$ oe | B1 | Correct value for $\frac{d^2y}{dx^2}$ |
| $\frac{d^3y}{dx^3} = \frac{28}{64}\times 1^3 - \frac{24}{8}\times 1 = -\frac{41}{16}$ | M1 | Use the given expression from (a) to obtain value for $\frac{d^3y}{dx^3}$; award if correct value seen |
| $y = 8 + x - \frac{5}{2}\times\frac{x^2}{2!} - \frac{41}{16}\times\frac{x^3}{3!} + \ldots$ | M1 | Taylor's series formed using their values for derivatives ($2!$ or $2$, $3!$ or $6$) |
| $y = 8 + x - \frac{5}{4}x^2 - \frac{41}{96}x^3 + \ldots$ | A1 (4) | Correct series; must start (or end) $y = \ldots$; correct terms must be seen |
---4. Given that
$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y = 0$$
\begin{enumerate}[label=(\alph*)]
\item show that
$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 28 } { y ^ { 2 } } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 3 } - \frac { 24 } { y } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$
Given also that $y = 8$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ at $x = 0$
\item find a series solution for $y$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, simplifying the coefficients where possible.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2021 Q4 [9]}}