| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Series solution from differential equation |
| Difficulty | Standard +0.3 This is a standard FP2 series solution question requiring differentiation of the given differential equation, substitution of initial conditions, and construction of a Maclaurin series. The steps are routine and methodical with no novel insight required, making it slightly easier than average for A-level but typical for Further Maths material. |
| Spec | 1.07q Product and quotient rules: differentiation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y''' + xy'' + y' = -2\sin x\) so \(y''' = -xy'' - y' - 2\sin x\) | M1, A1, A1 | |
| Subtotal | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y''' + 3 = 0\) so \(y''' = -3\) | B1 | |
| Subtotal | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute to give \(\frac{d^2y}{dx^2} = 2\) | B1 | |
| Use Taylor Expansion to give \(y = 1 + 3x + x^2 - \frac{x^3}{2}\ldots\) | M1 A1 | |
| Subtotal | (3) | |
| Total | (7) |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y''' + xy'' + y' = -2\sin x$ so $y''' = -xy'' - y' - 2\sin x$ | M1, A1, A1 | |
| **Subtotal** | **(3)** | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y''' + 3 = 0$ so $y''' = -3$ | B1 | |
| **Subtotal** | **(1)** | |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute to give $\frac{d^2y}{dx^2} = 2$ | B1 | |
| Use Taylor Expansion to give $y = 1 + 3x + x^2 - \frac{x^3}{2}\ldots$ | M1 A1 | |
| **Subtotal** | **(3)** | |
| **Total** | **(7)** | |
---1.
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2 \cos x$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ in terms of $x , \frac { \mathrm {~d} y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
At $x = 0 , y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$
\item Find the value of $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ at $x = 0$
\item Express $y$ as a series in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2013 Q1 [7]}}