| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Systems of differential equations |
| Type | Solve simultaneous ODEs directly |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring interpretation of a physical system to extract parameters (parts a-b), then solving coupled linear differential equations (part c). While the setup and parameter extraction require careful reasoning about flow rates and concentrations, the actual solution of coupled DEs is a standard Further Maths technique. The 14 total marks and multi-stage nature elevate it above average, but it follows established methods without requiring novel mathematical insight. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts4.10h Coupled systems: simultaneous first order DEs |
| Answer | Marks |
|---|---|
| 15(a) | Uses the capacity of |
| Answer | Marks | Guidance |
|---|---|---|
| equations to find | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| answer | AO3.2a | A1 |
| (b) | Deduces that 12 L/min | |
| are flowing into A | AO3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Divides 36 by their 12 | AO2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| required result. | AO2.1 | R1 |
| (c) | Differentiates one | |
| equation | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑎𝑎v̇a ria𝑦𝑦ble 𝑦𝑦̇ | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| order differential equation | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| auxiliary equation | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| for their DE | AO2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑎𝑎 𝑦𝑦 | AO1.1b | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| CAO | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| constant | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑎𝑎 𝑦𝑦 | AO1.1b | A1 |
| Answer | Marks |
|---|---|
| Total | 14 |
Question 15:
--- 15(a) ---
15(a) | Uses the capacity of
either tank and the
coefficients of the
equations to find | AO3.4 | M1 | 𝑟𝑟 = 0.005× 800
𝑟𝑟 = 4
Obtains the corre𝑟𝑟ct
answer | AO3.2a | A1
(b) | Deduces that 12 L/min
are flowing into A | AO3.3 | M1 | Flow into A = flow out of A (to keep
volume of water constant)
= 16−4 = 12
12𝜇𝜇 = 36
𝜇𝜇 = 3
Divides 36 by their 12 | AO2.2a | M1
Completes a rigorous
argument to show the
required result. | AO2.1 | R1
(c) | Differentiates one
equation | AO3.1a | M1 | and
(2)…0.02𝑎𝑎 = 𝑦𝑦̇ + 0.02𝑦𝑦
Sub in (1): 𝑎𝑎 = 50𝑦𝑦̇ +𝑦𝑦
𝑎𝑎̇ = 50𝑦𝑦̈ +𝑦𝑦̇
C50F𝑦𝑦: ̈ +𝑦𝑦̇ = 36−0.02(50𝑦𝑦̇ +𝑦𝑦 )+0.005𝑦𝑦
50 𝑦𝑦̈ +2𝑦𝑦̇ +0.015𝑦𝑦 = 36
2
50𝑚𝑚 +2𝑚𝑚+0.015= 0
PI:
𝑚𝑚 = −0.03,−0.01
36
𝑦𝑦 = 0.015= 2400
−0.03𝑡𝑡 −0.01𝑡𝑡
∴ 𝑦𝑦 = 𝐴𝐴e + 𝐴𝐴 e +2 4 0 0
−0 .03 𝑡𝑡 −0 . 01 𝑡𝑡
𝑦𝑦̇ = −0.03𝐴𝐴e − 0 s.0o1 𝐴𝐴e
𝑎𝑎 = 5 0𝑦𝑦̇ +𝑦𝑦
−0 .03 𝑡𝑡 −0.01𝑡𝑡
W𝑎𝑎h=en− 0.5𝐴𝐴e , + a0n.5d𝐴𝐴 e so+ 2400
and
𝑡𝑡 = 0 𝑎𝑎 = 0 𝑦𝑦 = 0
𝐴𝐴+𝐴𝐴+2400 a=nd0
−0.5𝐴𝐴+0.5𝐴𝐴+2400 =0
⟹ 𝐴𝐴 = 1200 𝐴𝐴 = −3600
−0.03𝑡𝑡 −0.01𝑡𝑡
𝑎𝑎 = −600e −1800e +2400
−0.03𝑡𝑡 −0.01𝑡𝑡
or𝑦𝑦 = 1200e −3600e +2400
and
(1) 0.005𝑦𝑦 = 𝑎𝑎̇ +0.0 2𝑎𝑎−36
Sub in (𝑦𝑦2)=: 200𝑎𝑎̇ +4𝑎𝑎−7200
𝑦𝑦̇ = 200𝑎𝑎̈ +4𝑎𝑎̇
C2F0: 0𝑎𝑎̈ +4𝑎𝑎̇ = 0.02(200𝑎𝑎̇ +4 𝑎𝑎−7200 )
200 𝑎𝑎̈ +8𝑎𝑎̇ +0.06𝑎𝑎 = 144
2
200𝑚𝑚 +8𝑚𝑚+0.06= 0
PI:
𝑚𝑚 = −0.03,−0.01
144
𝑎𝑎 = 0.006= 2400
−0.03𝑡𝑡 −0.01𝑡𝑡
∴ 𝑎𝑎 = 𝐴𝐴e +𝐴𝐴e +2400
−0.03𝑡𝑡 −0.01𝑡𝑡
Substitutes both and
or and in the other
equation to elimin𝑎𝑎ate one
𝑎𝑎v̇a ria𝑦𝑦ble 𝑦𝑦̇ | AO3.1a | M1
Forms a correct second
order differential equation | AO1.1b | A1
Obtains roots of their
auxiliary equation | AO1.1a | M1
Uses a valid method to
find a particular integral
for their DE | AO2.2a | M1
States general solution
for either or with their
particular integral
𝑎𝑎 𝑦𝑦 | AO1.1b | A1F
States general solutions
for both and
CAO | AO1.1b | A1
Uses initi𝑎𝑎al cond𝑦𝑦itions to
find a value for each
constant | AO3.4 | M1
Writes correct solutions
for both and
𝑎𝑎 𝑦𝑦 | AO1.1b | A1
When ,− 0.03𝑡𝑡 and −0.01𝑡𝑡 so
𝑦𝑦 = −2𝐴𝐴e a+n2d𝐴𝐴 e +2400
𝑡𝑡 = 0 𝑎𝑎 = 0 𝑦𝑦 = 0
𝐴𝐴+𝐴𝐴+2400 a=nd0
−2𝐴𝐴+2𝐴𝐴+2400= 0
⟹ 𝐴𝐴 = −600 𝐴𝐴 = −1800
−0.03𝑡𝑡 −0.01𝑡𝑡
𝑎𝑎 = −600e −1800e +2400
−0.03𝑡𝑡 −0.01𝑡𝑡
𝑦𝑦 = 1200e −3600e +2400
Total | 14\includegraphics{figure_15}
Two tanks, A and B, each have a capacity of 800 litres.
At time $t = 0$ both tanks are full of pure water.
When $t > 0$, water flows in the following ways:
• Water with a salt concentration of $\mu$ grams per litre flows into tank A at a constant rate
• Water flows from tank A to tank B at a rate of 16 litres per minute
• Water flows from tank B to tank A at a rate of $r$ litres per minute
• Water flows out of tank B through a waste pipe
• The amount of water in each tank remains at 800 litres.
At time $t$ minutes ($t \geq 0$) there are $x$ grams of salt in tank A and $y$ grams of salt in tank B.
This system is represented by the coupled differential equations
\begin{align}
\frac{dx}{dt} &= 36 - 0.02x + 0.005y \tag{1}\\
\frac{dy}{dt} &= 0.02x - 0.02y \tag{2}
\end{align}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $r$.
[2 marks]
\item Show that $\mu = 3$
[3 marks]
\item Solve the coupled differential equations to find both $x$ and $y$ in terms of $t$.
[9 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2019 Q15 [14]}}