| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable with partial fractions |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard techniques: substituting initial conditions, finding limits, partial fractions decomposition, and solving a separable differential equation. All steps are routine applications of well-practiced methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.02z Models in context: use functions in modelling1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitutes \(x=3\) and \(t=0\) into \(x=\dfrac{k(9t+5)}{4t+3}\) | M1 | Also allow: substitutes \(k=1.8\) and \(t=0\) to verify \(x=3\) |
| \(3=\dfrac{5k}{3}\Rightarrow k=1.8\) | A1* | Shows \(k=1.8\) (oe e.g. \(\frac{9}{5}\)) with no errors and at least one correct line of form \(ak=b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4050\) | B1 | cao. Allow 4.05 thousand |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{3}{x(9-2x)}\equiv\dfrac{A}{x}+\dfrac{B}{9-2x}\) | M1 | Sets up correct partial fractions form, or equivalent e.g. \(3\equiv A(9-2x)+Bx\) |
| Either \(A=\dfrac{1}{3}\) or \(B=\dfrac{2}{3}\) | A1 | One correct value for \(A\) or \(B\), or one correct fraction |
| \(\dfrac{3}{9x-2x^2}\equiv\dfrac{1}{3x}+\dfrac{2}{3(9-2x)}\) | A1 | Correct fractions. Mark for correct partial fractions, not values of constants. Award once correct fractions seen; allow if seen in (d) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3\dfrac{dx}{dt}=x(9-2x)\Rightarrow\int\dfrac{3}{x(9-2x)}dx=\int dt\) or \(\int\dfrac{1}{x(9-2x)}dx=\int\dfrac{1}{3}dt\) | M1 A1ft | M1: Separates variables and integrates to obtain \(kt\) term and one of \(\alpha\ln\beta x\) or \(\gamma\ln\delta(9-2x)\). Condone missing brackets e.g. \(\ln9-2x\). ln terms must come from partial fraction of form \(\frac{A}{x}\) or \(\frac{B}{9-2x}\). A1ft: ft on their \(A\) and \(B\) |
| \(\dfrac{1}{3}\ln x-\dfrac{1}{3}\ln(9-2x)=t(+c)\) | Brackets must be present unless implied by later work | |
| Substitutes \(t=0, x=3\Rightarrow c=0\) | M1 | Sets \(t=0, x=3\) to find \(c\). If step not attempted, only first two marks available. Can "state" e.g. \(c=0\) provided it follows correct work with "\(+c\)" |
| \(\dfrac{1}{3}\ln x-\dfrac{1}{3}\ln(9-2x)=t\Rightarrow\left(\dfrac{x}{9-2x}\right)=e^{3t}\) | ddM1 | Uses correct work to remove ln's having found constant of integration. Dependent on both previous M marks |
| \(\left(\dfrac{9-2x}{x}\right)=e^{-3t}\Rightarrow\dfrac{9}{x}-2=e^{-3t}\Rightarrow x=\dfrac{9}{2+e^{-3t}}\) | A1* | Correct work showing all necessary steps to reach given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4500\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 4500 cao or 4.5 thousand | B1 |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitutes $x=3$ and $t=0$ into $x=\dfrac{k(9t+5)}{4t+3}$ | M1 | Also allow: substitutes $k=1.8$ and $t=0$ to verify $x=3$ |
| $3=\dfrac{5k}{3}\Rightarrow k=1.8$ | A1* | Shows $k=1.8$ (oe e.g. $\frac{9}{5}$) with no errors and at least one correct line of form $ak=b$ |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4050$ | B1 | cao. Allow 4.05 thousand |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{3}{x(9-2x)}\equiv\dfrac{A}{x}+\dfrac{B}{9-2x}$ | M1 | Sets up correct partial fractions form, or equivalent e.g. $3\equiv A(9-2x)+Bx$ |
| Either $A=\dfrac{1}{3}$ or $B=\dfrac{2}{3}$ | A1 | One correct value for $A$ or $B$, or one correct fraction |
| $\dfrac{3}{9x-2x^2}\equiv\dfrac{1}{3x}+\dfrac{2}{3(9-2x)}$ | A1 | Correct fractions. Mark for correct partial fractions, not values of constants. Award once correct fractions seen; allow if seen in (d) |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $3\dfrac{dx}{dt}=x(9-2x)\Rightarrow\int\dfrac{3}{x(9-2x)}dx=\int dt$ or $\int\dfrac{1}{x(9-2x)}dx=\int\dfrac{1}{3}dt$ | M1 A1ft | M1: Separates variables and integrates to obtain $kt$ term and one of $\alpha\ln\beta x$ or $\gamma\ln\delta(9-2x)$. Condone missing brackets e.g. $\ln9-2x$. ln terms must come from partial fraction of form $\frac{A}{x}$ or $\frac{B}{9-2x}$. A1ft: ft on their $A$ and $B$ |
| $\dfrac{1}{3}\ln x-\dfrac{1}{3}\ln(9-2x)=t(+c)$ | | Brackets must be present unless implied by later work |
| Substitutes $t=0, x=3\Rightarrow c=0$ | M1 | Sets $t=0, x=3$ to find $c$. If step not attempted, only first two marks available. Can "state" e.g. $c=0$ provided it follows correct work with "$+c$" |
| $\dfrac{1}{3}\ln x-\dfrac{1}{3}\ln(9-2x)=t\Rightarrow\left(\dfrac{x}{9-2x}\right)=e^{3t}$ | ddM1 | Uses correct work to remove ln's having found constant of integration. Dependent on both previous M marks |
| $\left(\dfrac{9-2x}{x}\right)=e^{-3t}\Rightarrow\dfrac{9}{x}-2=e^{-3t}\Rightarrow x=\dfrac{9}{2+e^{-3t}}$ | A1* | Correct work showing all necessary steps to reach given answer |
## Part (e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4500$ | B1 | |
# Question 7 (continued):
**(e)**
| Answer | Mark | Guidance |
|--------|------|----------|
| 4500 cao or 4.5 thousand | B1 | |
---\begin{enumerate}
\item The number of goats on an island is being monitored.
\end{enumerate}
When monitoring began there were 3000 goats on the island.\\
In a simple model, the number of goats, $x$, in thousands, is modelled by the equation
$$x = \frac { k ( 9 t + 5 ) } { 4 t + 3 }$$
where $k$ is a constant and $t$ is the number of years after monitoring began.\\
(a) Show that $k = 1.8$\\
(b) Hence calculate the long-term population of goats predicted by this model.
In a second model, the number of goats, $x$, in thousands, is modelled by the differential equation
$$3 \frac { \mathrm {~d} x } { \mathrm {~d} t } = x ( 9 - 2 x )$$
(c) Write $\frac { 3 } { x ( 9 - 2 x ) }$ in partial fraction form.\\
(d) Solve the differential equation with the initial condition to show that
$$x = \frac { 9 } { 2 + \mathrm { e } ^ { - 3 t } }$$
(e) Find the long-term population of goats predicted by this second model.
\hfill \mbox{\textit{Edexcel P4 2023 Q7 [12]}}