| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Applied/modelling contexts |
| Difficulty | Standard +0.3 This is a straightforward integrating factor question with standard steps: find integrating factor e^(-0.1t), multiply through, integrate, and apply initial condition. Part (b) is simple verification by substitution. While it requires multiple steps, the method is entirely routine for FP2 students with no conceptual challenges or novel insights needed. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Integrating factor is \(e^{-\int 0.1\,dt} = e^{-0.1t}\) | B1 | |
| \(Pe^{-0.1t} = \int 0.05te^{-0.1t}\,dt\) | M1 | |
| \(= \frac{-0.05te^{-0.1t}}{0.1} + \int \frac{0.05e^{-0.1t}}{0.1}\,dt\) | M1 | |
| \(= -0.5te^{-0.1t} - 5e^{-0.1t} + c\) | A1 | |
| \(\therefore P = -\frac{1}{2}t - 5 + ce^{0.1t}\) | A1 | |
| At \(t=0\), \(P=10000\): \(c = 10005\), \(\therefore P = -\frac{1}{2}t - 5 + 10005e^{0.1t}\) | M1 A1 (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| When \(t=6\), \(P = 18222 < 20000\) | ||
| When \(t=7\), \(P = 20139 > 20000\) | M1 | |
| So \(P\) reaches 20000 during the seventh year | A1 (2) |
# Question 7:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| Integrating factor is $e^{-\int 0.1\,dt} = e^{-0.1t}$ | B1 | |
| $Pe^{-0.1t} = \int 0.05te^{-0.1t}\,dt$ | M1 | |
| $= \frac{-0.05te^{-0.1t}}{0.1} + \int \frac{0.05e^{-0.1t}}{0.1}\,dt$ | M1 | |
| $= -0.5te^{-0.1t} - 5e^{-0.1t} + c$ | A1 | |
| $\therefore P = -\frac{1}{2}t - 5 + ce^{0.1t}$ | A1 | |
| At $t=0$, $P=10000$: $c = 10005$, $\therefore P = -\frac{1}{2}t - 5 + 10005e^{0.1t}$ | M1 A1 (7) | |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| When $t=6$, $P = 18222 < 20000$ | | |
| When $t=7$, $P = 20139 > 20000$ | M1 | |
| So $P$ reaches 20000 during the seventh year | A1 (2) | |
---7. A population $P$ is growing at a rate which is modelled by the differential equation
$$\frac { d P } { d t } - 0.1 P = 0.05 t$$
where $t$ years is the time that has elapsed from the start of observations.\\
It is given that the population is 10000 at the start of the observations.
\begin{enumerate}[label=(\alph*)]
\item Solve the differential equation to obtain an expression for $P$ in terms of $t$.
\item Show that the population doubles between the sixth and seventh year after the observations began.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 Q7 [9]}}