| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question involving coupled differential equations, requiring elimination to form a second-order ODE, finding complementary functions, and justifying why certain particular integral forms fail. While the individual techniques are standard for FM students, the context of coupled equations and the need to explain the failure of a PI form adds conceptual depth beyond routine exercises. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral4.10h Coupled systems: simultaneous first order DEs |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | If C – P> 0(or more children are playing in the |
| Answer | Marks | Guidance |
|---|---|---|
| will be negative(and so must be negative). | B1 | |
| [1] | 2.4 | Properly reasoned argument. |
| Answer | Marks | Guidance |
|---|---|---|
| of derivative | Ignore reference to dP/dt. | |
| (c) | (i) | Auxiliary equation is m2 + 4m = 0 so m = 0 or |
| m = –4 so CF is ( P = ) A + B e − 4 t | B1 | |
| [1] | 1.1 | Condone missing “P =” and/or x |
| Answer | Marks |
|---|---|
| Condone Ae0t | m2 –2m = 0 so m = 0 or m = 2 |
| Answer | Marks |
|---|---|
| (ii) | Because the function P = b is one of the |
| Answer | Marks | Guidance |
|---|---|---|
| when plugged in to the LHS). | B1 | |
| [1] | 2.4 | or other correct reason eg this |
| Answer | Marks |
|---|---|
| in the trial function. | “The complementary function |
| Answer | Marks |
|---|---|
| (iii) | P = at2 + bt =>P = 2at + b and P = 2a |
| Answer | Marks |
|---|---|
| P = A + B e − 4 t + 8 t 2 + 4 t | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | Correctly differentiating to find |
| Answer | Marks |
|---|---|
| Must be “P =”. Must be P(t). | 2a−2(2at+b)=−t+ |
| Answer | Marks |
|---|---|
| (d) | t = 0, P = 55 =>A + B = 55 |
| Answer | Marks |
|---|---|
| so 63 children cao www | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | Using first initial condition. |
| Answer | Marks | Guidance |
|---|---|---|
| (e) | The number of children cannot increase | |
| indefinitely. | B1 | |
| [1] | 3.5b | or other reasonable limitation of |
| Answer | Marks |
|---|---|
| not realistic. | Not just “Children are being added |
Question 8:
8 | (a) | If C – P> 0(or more children are playing in the
bouncy castle than in the ball pits so) children
will leave the bouncy castle and hence dC/dt
will be negative(and so must be negative). | B1
[1] | 2.4 | Properly reasoned argument.
Could be related to C – P< 0.
ie
• some conditional
• consequence in terms of
movement of children
• consequence in terms of sign
of derivative | Ignore reference to dP/dt.
(c) | (i) | Auxiliary equation is m2 + 4m = 0 so m = 0 or
m = –4 so CF is ( P = ) A + B e − 4 t | B1
[1] | 1.1 | Condone missing “P =” and/or x
instead of t / y instead of P here.
Condone Ae0t | m2 –2m = 0 so m = 0 or m = 2
so (P=)A+Be2t
(ii) | Because the function P = b is one of the
functions that form the CF (so will come to 0
when plugged in to the LHS). | B1
[1] | 2.4 | or other correct reason eg this
trial function doesn’t produce a
term in t to compare with 64t.
If considering the CF, must
address the fact that there is a
constant term in the CF and so
there cannot be a constant term
in the trial function. | “The complementary function
already has a constant in it”.
“Part of the trial function is
included in the complementary
function”.
Not “There is no P term in the
differential equation”.
Not “It matches the form of the
complementary function”.
(iii) | P = at2 + bt =>P = 2at + b and P = 2a
2a+4(2at+b)=...
8at+4b+2a=64t+32
8a=64a=8,
4b+2a=324b=32−2(8)=16b=4
P = A + B e − 4 t + 8 t 2 + 4 t | M1
A1
A1FT
[3] | 3.4
3.4
1.1 | Correctly differentiating to find
the first and second derivatives
and substituting into the LHS of
the DE in (b).
Finding a and b correctly.
FT their CF + their numerical
PI.
Must be “P =”. Must be P(t). | 2a−2(2at+b)=−t+
−4a=−a=
4
2a−2b=2b= −=−
2 2
−
b=
4
P A B e 2 t t 2 t = + + −
4 4
(d) | t = 0, P = 55 =>A + B = 55
P ' = − 4 B e − 4 t + 1 6 t + 4
t = 0, P = 24 => –4B + 4 = 24 =>B = –5
=>A = 60
1 − 4 1 1 2 1
t = P = 6 0 − 5 e + 2 8 + 4
2 2 2
= 6 0 − 5 e − 2 + 2 + 2 = 6 3 .3 2 3 ...
so 63 children cao www | M1
M1
M1
A1
[4] | 3.3
3.4
3.3
3.4 | Using first initial condition.
Correctly differentiating their P
to find P (can be implied by
next line)
Substituting t=0 correctly in P
to obtain the second initial
condition and using the latter to
solve for A and B.
Must be from correct P.
(e) | The number of children cannot increase
indefinitely. | B1
[1] | 3.5b | or other reasonable limitation of
the model in the long-term
which could be environmental.
eg the room has a finite size
the centre will close.
P ~ 8t2 as t gets large which is
not realistic. | Not just “Children are being added
constantly”.
Not “The number of children grows
exponentially”.A children's play centre has two rooms, a room full of bouncy castles and a room full of ball pits. At any given instant, each child in the centre is playing either on the bouncy castles or in the ball pits. Each child can see one room from the other room and can decide to change freely between the two rooms. It is assumed that such changes happen instantaneously.
The number of children playing on the bouncy castles at time $t$ hours, is denoted by $C$ and the corresponding number of children playing in the ball pits is $P$. Because the number of children is large for most of the time, $C$ and $P$ are modelled as being continuous.
When there is a different number of children in each room, some children will move from the room with more children to the room with fewer children. A researcher therefore decides to model $C$ and $P$ with the following coupled differential equations.
$$\frac{dP}{dt} = \alpha(P-C) + \gamma t$$
$$\frac{dC}{dt} = \alpha(C-P)$$
\begin{enumerate}[label=(\alph*)]
\item Explain why $\alpha$ must be negative. [1]
\end{enumerate}
After examining data, the researcher chooses $\alpha = -2$ and $\gamma = 32$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $P$ satisfies the second order differential equation $\frac{d^2P}{dt^2} + 4\frac{dP}{dt} = 64t + 32$. [2]
\item \begin{enumerate}[label=(\roman*)]
\item Find the complementary function for the differential equation from part (b). [1]
\item Explain why a particular integral of the form $P = at + b$ will not work in this situation. [1]
\item Using a particular integral of the form $P = at^2 + bt$, find the general solution of the differential equation from part (b). [3]
\end{enumerate}
\end{enumerate}
At a certain time there are 55 children playing in the ball pits and 24 children per hour are arriving at the ball pits.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Use the model, starting from this time, to estimate the number of children in the ball pits 30 minutes later. [4]
\item Explain why the model becomes unreliable as $t$ gets very large. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2024 Q8 [13]}}