| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - direct proportionality (dN/dt = kN) |
| Difficulty | Standard +0.3 This is a straightforward differential equations question testing standard separation of variables technique. Parts (a) and (b) involve the most basic exponential growth model, while parts (c) and (d) require the same technique with a slightly less routine integral (cos λt). All parts follow predictable patterns with no problem-solving insight required, making this slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int \frac{dP}{P} = \int k\, dt\) | M1 | Separates variables; integral signs not necessary |
| \(\ln P = kt\,(+c)\) | A1 | Must see \(\ln P\) and \(kt\); correct equation with/without \(+c\) |
| When \(t=0\), \(P=P_0 \Rightarrow \ln P_0 = c\) (or \(P=Ae^{kt} \Rightarrow P_0=A\)) | M1 | Use of boundary condition to find constant of integration |
| \(\ln P = kt + \ln P_0 \Rightarrow P = P_0 e^{kt}\) | A1 | \(P = P_0 e^{kt}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = 2P_0\) and \(k = 2.5 \Rightarrow 2P_0 = P_0 e^{2.5t}\) | M1 | Substitutes \(P=2P_0\) into expression involving \(P\) |
| \(e^{2.5t} = 2 \Rightarrow \ln e^{2.5t} = \ln 2\) or \(2.5t = \ln 2\) | M1 | Eliminates \(P_0\) and takes \(\ln\) of both sides |
| \(t = \frac{1}{2.5}\ln 2 = 0.277258872...\) days | ||
| \(t = 0.277258872... \times 24 \times 60 = 399.252776...\) minutes | ||
| \(t = \underline{399}\) min or \(t = 6\) hr \(39\) mins (to nearest minute) | A1 | awrt \(t=399\) or 6 hr 39 mins |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int \frac{dP}{P} = \int \lambda\cos\lambda t\, dt\) | M1 | Separates variables; integral signs not necessary |
| \(\ln P = \sin\lambda t\,(+c)\) | A1 | Must see \(\ln P\) and \(\sin\lambda t\); correct equation with/without \(+c\) |
| When \(t=0\), \(P=P_0 \Rightarrow \ln P_0 = c\) | M1 | Use of boundary condition to find constant of integration |
| \(P = P_0 e^{\sin\lambda t}\) | A1 | \(P = P_0 e^{\sin\lambda t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = 2P_0\), \(\lambda = 2.5 \Rightarrow 2P_0 = P_0 e^{\sin 2.5t}\) | ||
| \(e^{\sin 2.5t} = 2 \Rightarrow \sin 2.5t = \ln 2\) | M1 | Eliminates \(P_0\) and makes \(\sin\lambda t\) or \(\sin 2.5t\) the subject by taking \(\ln\)'s |
| \(t = \frac{1}{2.5}\sin^{-1}(\ln 2)\) | dM1 | Rearranges to make \(t\) the subject (must use \(\sin^{-1}\)) |
| \(t = 0.306338477...\) days | ||
| \(t = 0.306338477... \times 24 \times 60 = 441.1274082...\) minutes | ||
| \(t = \underline{441}\) min or \(t = 7\) hr \(21\) mins (to nearest minute) | A1 | awrt \(t=441\) or 7 hr 21 mins |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int \frac{dP}{\lambda P} = \int \cos \lambda t \, dt\) | M1 | Separates variables with \(\int \frac{dP}{\lambda P}\) and \(\int \cos \lambda t \, dt\) on either side; integral signs not necessary |
| \(\frac{1}{\lambda} \ln P = \frac{1}{\lambda} \sin \lambda t \ (+c)\) | A1 | Must see \(\frac{1}{\lambda} \ln P\) and \(\frac{1}{\lambda} \sin \lambda t\); correct equation with/without \(+c\) |
| When \(t=0\), \(P=P_0 \Rightarrow \frac{1}{\lambda}\ln P_0 = c\) (or \(P = Ae^{\sin\lambda t} \Rightarrow P_0 = A\)) | M1 | Use of boundary condition to attempt to find constant of integration |
| Hence \(P = P_0 e^{\sin \lambda t}\) | A1 | \(P = P_0 e^{\sin \lambda t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int \frac{dP}{\lambda P} = \int \cos \lambda t \, dt\) | M1 | Separates variables with \(\int \frac{dP}{\lambda P}\) and \(\int \cos \lambda t \, dt\) on either side; integral signs not necessary |
| \(\frac{1}{\lambda}\ln(\lambda P) = \frac{1}{\lambda}\sin \lambda t \ (+c)\) | A1 | Must see \(\frac{1}{\lambda}\ln(\lambda P)\) and \(\frac{1}{\lambda}\sin \lambda t\); correct equation with/without \(+c\) |
| When \(t=0\), \(P=P_0 \Rightarrow \frac{1}{\lambda}\ln(\lambda P_0) = c\) (or \(\lambda P = Ae^{\sin\lambda t} \Rightarrow \lambda P_0 = A\)) | M1 | Use of boundary condition to attempt to find constant of integration |
| Hence \(P = P_0 e^{\sin \lambda t}\) | A1 | \(P = P_0 e^{\sin \lambda t}\) |
# Question 8:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \frac{dP}{P} = \int k\, dt$ | M1 | Separates variables; integral signs not necessary |
| $\ln P = kt\,(+c)$ | A1 | Must see $\ln P$ and $kt$; correct equation with/without $+c$ |
| When $t=0$, $P=P_0 \Rightarrow \ln P_0 = c$ (or $P=Ae^{kt} \Rightarrow P_0=A$) | M1 | Use of boundary condition to find constant of integration |
| $\ln P = kt + \ln P_0 \Rightarrow P = P_0 e^{kt}$ | A1 | $P = P_0 e^{kt}$ |
> Note: $P = P_0 e^{kt}$ written down without the first M1 scores all four marks.
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = 2P_0$ and $k = 2.5 \Rightarrow 2P_0 = P_0 e^{2.5t}$ | M1 | Substitutes $P=2P_0$ into expression involving $P$ |
| $e^{2.5t} = 2 \Rightarrow \ln e^{2.5t} = \ln 2$ or $2.5t = \ln 2$ | M1 | Eliminates $P_0$ and takes $\ln$ of both sides |
| $t = \frac{1}{2.5}\ln 2 = 0.277258872...$ days | | |
| $t = 0.277258872... \times 24 \times 60 = 399.252776...$ minutes | | |
| $t = \underline{399}$ min or $t = 6$ hr $39$ mins (to nearest minute) | A1 | awrt $t=399$ or 6 hr 39 mins |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \frac{dP}{P} = \int \lambda\cos\lambda t\, dt$ | M1 | Separates variables; integral signs not necessary |
| $\ln P = \sin\lambda t\,(+c)$ | A1 | Must see $\ln P$ and $\sin\lambda t$; correct equation with/without $+c$ |
| When $t=0$, $P=P_0 \Rightarrow \ln P_0 = c$ | M1 | Use of boundary condition to find constant of integration |
| $P = P_0 e^{\sin\lambda t}$ | A1 | $P = P_0 e^{\sin\lambda t}$ |
> Note: $P = P_0 e^{\sin\lambda t}$ written down without the first M1 scores all four marks.
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = 2P_0$, $\lambda = 2.5 \Rightarrow 2P_0 = P_0 e^{\sin 2.5t}$ | | |
| $e^{\sin 2.5t} = 2 \Rightarrow \sin 2.5t = \ln 2$ | M1 | Eliminates $P_0$ and makes $\sin\lambda t$ or $\sin 2.5t$ the subject by taking $\ln$'s |
| $t = \frac{1}{2.5}\sin^{-1}(\ln 2)$ | dM1 | Rearranges to make $t$ the subject (must use $\sin^{-1}$) |
| $t = 0.306338477...$ days | | |
| $t = 0.306338477... \times 24 \times 60 = 441.1274082...$ minutes | | |
| $t = \underline{441}$ min or $t = 7$ hr $21$ mins (to nearest minute) | A1 | awrt $t=441$ or 7 hr 21 mins |
## Question 8(c):
Given: $\frac{dP}{dt} = \lambda P \cos \lambda t$ and $t = 0$, $P = P_0$
---
**Aliter Way 2:**
| Working | Mark | Guidance |
|---------|------|----------|
| $\int \frac{dP}{\lambda P} = \int \cos \lambda t \, dt$ | M1 | Separates variables with $\int \frac{dP}{\lambda P}$ and $\int \cos \lambda t \, dt$ on either side; integral signs not necessary |
| $\frac{1}{\lambda} \ln P = \frac{1}{\lambda} \sin \lambda t \ (+c)$ | A1 | Must see $\frac{1}{\lambda} \ln P$ and $\frac{1}{\lambda} \sin \lambda t$; correct equation with/without $+c$ |
| When $t=0$, $P=P_0 \Rightarrow \frac{1}{\lambda}\ln P_0 = c$ (or $P = Ae^{\sin\lambda t} \Rightarrow P_0 = A$) | M1 | Use of boundary condition to attempt to find constant of integration |
| Hence $P = P_0 e^{\sin \lambda t}$ | A1 | $P = P_0 e^{\sin \lambda t}$ |
**[4 marks total]**
---
**Aliter Way 3:**
| Working | Mark | Guidance |
|---------|------|----------|
| $\int \frac{dP}{\lambda P} = \int \cos \lambda t \, dt$ | M1 | Separates variables with $\int \frac{dP}{\lambda P}$ and $\int \cos \lambda t \, dt$ on either side; integral signs not necessary |
| $\frac{1}{\lambda}\ln(\lambda P) = \frac{1}{\lambda}\sin \lambda t \ (+c)$ | A1 | Must see $\frac{1}{\lambda}\ln(\lambda P)$ and $\frac{1}{\lambda}\sin \lambda t$; correct equation with/without $+c$ |
| When $t=0$, $P=P_0 \Rightarrow \frac{1}{\lambda}\ln(\lambda P_0) = c$ (or $\lambda P = Ae^{\sin\lambda t} \Rightarrow \lambda P_0 = A$) | M1 | Use of boundary condition to attempt to find constant of integration |
| Hence $P = P_0 e^{\sin \lambda t}$ | A1 | $P = P_0 e^{\sin \lambda t}$ |
**[4 marks total]**
> **Note:** $P = P_0 e^{\sin \lambda t}$ written down without the first M1 mark scores all four marks in part (c).8. A population growth is modelled by the differential equation
$$\frac { \mathrm { d } P } { \mathrm {~d} t } = k P ,$$
where $P$ is the population, $t$ is the time measured in days and $k$ is a positive constant.\\
Given that the initial population is $P _ { 0 }$,
\begin{enumerate}[label=(\alph*)]
\item solve the differential equation, giving $P$ in terms of $P _ { 0 } , k$ and $t$.
Given also that $k = 2.5$,
\item find the time taken, to the nearest minute, for the population to reach $2 P _ { 0 }$.
In an improved model the differential equation is given as
$$\frac { \mathrm { d } P } { \mathrm {~d} t } = \lambda P \cos \lambda t$$
where $P$ is the population, $t$ is the time measured in days and $\lambda$ is a positive constant.\\
Given, again, that the initial population is $P _ { 0 }$ and that time is measured in days,
\item solve the second differential equation, giving $P$ in terms of $P _ { 0 } , \lambda$ and $t$.
Given also that $\lambda = 2.5$,
\item find the time taken, to the nearest minute, for the population to reach $2 P _ { 0 }$ for the first time, using the improved model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2007 Q8 [14]}}