| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 20 |
| 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 structured multi-part question with clear signposting through each step. Part (a) involves routine verification of a given solution and finding max/min values of a simple function. Part (b) guides students through separable variables with partial fractions (a standard C4 technique), with the partial fractions decomposition explicitly requested and the target answer provided. While it requires multiple techniques, each step is scaffolded and represents standard textbook exercises rather than requiring problem-solving insight. |
| Spec | 1.02y Partial fractions: decompose rational functions1.06f Laws of logarithms: addition, subtraction, power rules1.07b Gradient as rate of change: dy/dx notation1.07l Derivative of ln(x): and related functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P_{\max} = \frac{2}{2-1} = 2\) | B1 | |
| \(P_{\min} = \frac{2}{2+1} = 2/3\) | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = \frac{2}{2-\sin t} = 2(2-\sin t)^{-1}\) | ||
| \(\Rightarrow \frac{dP}{dt} = -2(2-\sin t)^{-2} \cdot -\cos t\) | M1 | chain rule |
| \(= \frac{2\cos t}{(2-\sin t)^2}\) | B1 | \(-1(\ldots)^{-2}\) soi |
| A1 | (or quotient rule M1, numerator A1, denominator A1) | |
| \(\frac{1}{2}P^2\cos t = \frac{1}{2}\cdot\frac{4}{(2-\sin t)^2}\cos t\) | DM1 | attempt to verify |
| \(= \frac{2\cos t}{(2-\sin t)^2} = \frac{dP}{dt}\) | E1 [5] | or by integration as in (b)(ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{P(2P-1)} = \frac{A}{P} + \frac{B}{2P-1}\) | M1 | correct partial fractions |
| \(= \frac{A(2P-1)+BP}{P(2P-1)}\) | ||
| \(1 \equiv A(2P-1)+BP\) | M1 | substituting values, equating coeffs or cover up rule |
| \(P=0 \Rightarrow 1 = -A \Rightarrow A = -1\) | A1 | \(A = -1\) |
| \(P=\frac{1}{2} \Rightarrow 1 = A\cdot 0 + \frac{1}{2}B \Rightarrow B = 2\) | A1 | \(B = 2\) |
| So \(\frac{1}{P(2P-1)} = -\frac{1}{P} + \frac{2}{2P-1}\) | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dP}{dt} = \frac{1}{2}(2P-P^2)\cos t\) | ||
| \(\Rightarrow \int\frac{1}{2P^2-P}\,dP = \int\frac{1}{2}\cos t\,dt\) | M1 | separating variables |
| \(\Rightarrow \int\left(\frac{2}{2P-1} - \frac{1}{P}\right)dP = \int\frac{1}{2}\cos t\,dt\) | ||
| \(\Rightarrow \ln(2P-1) - \ln P = \frac{1}{2}\sin t + c\) | A1 | \(\ln(2P-1) - \ln P\) ft their A,B from (i) |
| A1 | \(\frac{1}{2}\sin t\) | |
| When \(t=0\), \(P=1\): \(\ln 1 - \ln 1 = \frac{1}{2}\sin 0 + c \Rightarrow c = 0\) | B1 | finding constant \(= 0\) |
| \(\Rightarrow \ln\left(\frac{2P-1}{P}\right) = \frac{1}{2}\sin t\) * | E1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P_{\max} = \frac{1}{2-e^{1/2}} = 2.847\) | M1A1 | www |
| \(P_{\min} = \frac{1}{2-e^{-1/2}} = 0.718\) | M1A1 [4] | www |
## Question 5:
### Part 5(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P_{\max} = \frac{2}{2-1} = 2$ | B1 | |
| $P_{\min} = \frac{2}{2+1} = 2/3$ | B1 [2] | |
### Part 5(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = \frac{2}{2-\sin t} = 2(2-\sin t)^{-1}$ | | |
| $\Rightarrow \frac{dP}{dt} = -2(2-\sin t)^{-2} \cdot -\cos t$ | M1 | chain rule |
| $= \frac{2\cos t}{(2-\sin t)^2}$ | B1 | $-1(\ldots)^{-2}$ soi |
| | A1 | (or quotient rule M1, numerator A1, denominator A1) |
| $\frac{1}{2}P^2\cos t = \frac{1}{2}\cdot\frac{4}{(2-\sin t)^2}\cos t$ | DM1 | attempt to verify |
| $= \frac{2\cos t}{(2-\sin t)^2} = \frac{dP}{dt}$ | E1 [5] | or by integration as in (b)(ii) |
### Part 5(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{P(2P-1)} = \frac{A}{P} + \frac{B}{2P-1}$ | M1 | correct partial fractions |
| $= \frac{A(2P-1)+BP}{P(2P-1)}$ | | |
| $1 \equiv A(2P-1)+BP$ | M1 | substituting values, equating coeffs or cover up rule |
| $P=0 \Rightarrow 1 = -A \Rightarrow A = -1$ | A1 | $A = -1$ |
| $P=\frac{1}{2} \Rightarrow 1 = A\cdot 0 + \frac{1}{2}B \Rightarrow B = 2$ | A1 | $B = 2$ |
| So $\frac{1}{P(2P-1)} = -\frac{1}{P} + \frac{2}{2P-1}$ | [4] | |
### Part 5(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dt} = \frac{1}{2}(2P-P^2)\cos t$ | | |
| $\Rightarrow \int\frac{1}{2P^2-P}\,dP = \int\frac{1}{2}\cos t\,dt$ | M1 | separating variables |
| $\Rightarrow \int\left(\frac{2}{2P-1} - \frac{1}{P}\right)dP = \int\frac{1}{2}\cos t\,dt$ | | |
| $\Rightarrow \ln(2P-1) - \ln P = \frac{1}{2}\sin t + c$ | A1 | $\ln(2P-1) - \ln P$ ft their A,B from (i) |
| | A1 | $\frac{1}{2}\sin t$ |
| When $t=0$, $P=1$: $\ln 1 - \ln 1 = \frac{1}{2}\sin 0 + c \Rightarrow c = 0$ | B1 | finding constant $= 0$ |
| $\Rightarrow \ln\left(\frac{2P-1}{P}\right) = \frac{1}{2}\sin t$ * | E1 [5] | |
### Part 5(b)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P_{\max} = \frac{1}{2-e^{1/2}} = 2.847$ | M1A1 | www |
| $P_{\min} = \frac{1}{2-e^{-1/2}} = 0.718$ | M1A1 [4] | www |
---5 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.
\begin{enumerate}[label=(\alph*)]
\item Suppose that the number of cases, $P$ thousand, after time $t$ months is modelled by the equation $P = \frac { 2 } { 2 - \sin t }$. Thus, when $t = 0 , P = 1$.
\begin{enumerate}[label=(\roman*)]
\item By considering the greatest and least values of $\sin t$, write down the greatest and least values of $P$ predicted by this model.
\item Verify that $P$ satisfies the differential equation $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t$.
\end{enumerate}\item An alternative model is proposed, with differential equation
$$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$
As before, $P = 1$ when $t = 0$.
\begin{enumerate}[label=(\roman*)]
\item Express $\frac { 1 } { P ( 2 P - 1 ) }$ in partial fractions.
\item Solve the differential equation (*) to show that
$$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$
This equation can be rearranged to give $P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }$.
\item Find the greatest and least values of $P$ predicted by this model.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 Q5 [20]}}