| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Find higher derivatives from equation |
| Difficulty | Challenging +1.2 This is a Further Maths FP1 question requiring systematic differentiation of a differential equation and series solution methods. Part (a) involves repeated differentiation using the product rule (3-4 steps), which is mechanical but requires care. Part (b) applies standard series solution techniques with given initial conditions. While this is harder than typical A-level due to being Further Maths content, it follows a well-established algorithmic procedure without requiring novel insight or complex problem-solving. |
| Spec | 4.08a Maclaurin series: find series for function4.10a General/particular solutions: of differential equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y'' = 2xy' - y \Rightarrow y''' = 2xy'' + 2y' - y'\) | M1, A1 | Attempts differentiation with product rule; all terms on one side |
| \(y''' = 2xy'' + y' \Rightarrow y'''' = 2xy''' + 2y'' + y''\) | M1 | Continues differentiating |
| \(y'''' = 2xy''' + 3y'' \Rightarrow y''''' = 2xy'''' + 5y'''\) | A1 | Completes to fifth derivative in correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=0, y=0, y'=1 \Rightarrow y''(0) = 0\) from equation | B1 | Deduces correct value for \(y''(0)\) |
| \(y'''(0) = 2\times0\times y''(0)+1 = 1\); \(y''''(0) = 2\times0\times1+3\times0 = 0\) | M1, A1 | Finds values of derivatives at given point; all correct |
| \(x=0\), \(y'''(0)=1\), \(y''''(0)=0 \Rightarrow y'''''(0) = 5\) | A1 | |
| \(y = y(0) + y'(0)x + \frac{y''(0)}{2}x^2 + \frac{y'''(0)}{6}x^3 + \frac{y''''(0)}{24}x^4 + \frac{y'''''(0)}{120}x^5 + \ldots\) | M1 | Correct mathematical language, given denominators |
| \(y = x + \frac{1}{6}x^3 + \frac{1}{24}x^5 + \ldots\) | A1ft | Correct series; must start \(y = \ldots\) |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y'' = 2xy' - y \Rightarrow y''' = 2xy'' + 2y' - y'$ | M1, A1 | Attempts differentiation with product rule; all terms on one side |
| $y''' = 2xy'' + y' \Rightarrow y'''' = 2xy''' + 2y'' + y''$ | M1 | Continues differentiating |
| $y'''' = 2xy''' + 3y'' \Rightarrow y''''' = 2xy'''' + 5y'''$ | A1 | Completes to fifth derivative in correct form |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=0, y=0, y'=1 \Rightarrow y''(0) = 0$ from equation | B1 | Deduces correct value for $y''(0)$ |
| $y'''(0) = 2\times0\times y''(0)+1 = 1$; $y''''(0) = 2\times0\times1+3\times0 = 0$ | M1, A1 | Finds values of derivatives at given point; all correct |
| $x=0$, $y'''(0)=1$, $y''''(0)=0 \Rightarrow y'''''(0) = 5$ | A1 | |
| $y = y(0) + y'(0)x + \frac{y''(0)}{2}x^2 + \frac{y'''(0)}{6}x^3 + \frac{y''''(0)}{24}x^4 + \frac{y'''''(0)}{120}x^5 + \ldots$ | M1 | Correct mathematical language, given denominators |
| $y = x + \frac{1}{6}x^3 + \frac{1}{24}x^5 + \ldots$ | A1ft | Correct series; must start $y = \ldots$ |
---4.
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 5 } y } { \mathrm {~d} x ^ { 5 } } = a x \frac { \mathrm {~d} ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } + b \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$$
where $a$ and $b$ are integers to be found.
\item Hence find a series solution, in ascending powers of $x$, as far as the term in $x ^ { 5 }$, of the differential equation (I) where $y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ at $x = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q4 [9]}}