What is 7.6: Constructing differential equations?

Translate a rate statement into derivative notation after defining the dependent and independent variables, including their units where relevant.

How many marks are 7.6 questions worth?

7.6 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 7.6 appeared in?

We have not yet indexed a verified past-paper appearance for 7.6 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

7.6

Construct simple differential equations in pure mathematics and in context (contexts may include kinematics, population growth and modelling the relationship between price and demand).

Draft — not yet indexed

Constructing differential equations

Worked answers and methods for 7.6 on Edexcel A-level Maths 9MA0.

Explanation

Translate a rate statement into derivative notation after defining the dependent and independent variables, including their units where relevant.
Phrases such as 'proportional to' introduce a positive constant $k$ ; words such as 'decreases' or 'decays' determine whether a minus sign is needed.
A limiting or equilibrium value often appears as a difference, for example growth towards a capacity $K$ can be modelled by $\frac{\mathrm dP}{\mathrm dt}=k(K-P)$ .
A common error is to solve the differential equation when only its construction is requested, while failing to state or determine the proportionality constant.

Worked example

A population $P$ grows at a rate proportional to the difference between $1200$ and the current population. When $P=800$ , the population is increasing at $50$ individuals per year. Construct the differential equation, including the value of the constant of proportionality.

1.Write $\frac{\mathrm dP}{\mathrm dt}=k(1200-P)$ .
2.Using $P=800$ and rate $50$ gives $50=k(400)$ , so $k=\frac18$ .
3.Therefore $\frac{\mathrm dP}{\mathrm dt}=\frac18(1200-P)$ .

Answer: $\frac{\mathrm dP}{\mathrm dt}=\frac18(1200-P)$

Common mistakes

Don't assign the wrong sign to a decay or resistance term, predicting growth when the quantity should fall.
Don't write a proportional relationship without defining the constant or using the supplied rate to determine it.

Exam tip

Translate each verbal rate statement into a signed differential equation, then substitute the calibration data to find the constant.

Worked practice

Tier 1 · Easy

The acceleration of a particle is proportional to its speed

v

and acts opposite to the motion. Write down a differential equation for

v

in terms of time

t

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\frac{\mathrm dv}{\mathrm dt}=-kv$ , where $k>0$	2
Notes
Acceleration is $\frac{\mathrm dv}{\mathrm dt}$ . Proportionality to $v$ gives magnitude $kv$ with $k>0$ , and opposition to the motion supplies the minus sign: $\frac{\mathrm dv}{\mathrm dt}=-kv$ .

(2 marks)

Tier 2 · Standard

A population

P

grows at a rate proportional to the product

P(1000-P)

. When

P=200

, the population is increasing at

32

individuals per day. Construct the differential equation, including the constant of proportionality.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\dfrac{\mathrm dP}{\mathrm dt}=\dfrac{1}{5000}P(1000-P)$	4
Notes
The statement gives $dP/dt=kP(1000-P)$ with $k>0$ . Using $P=200$ and $dP/dt=32$ gives $32=k(200)(800)=160000k$ , so $k=1/5000$ . Hence $dP/dt=P(1000-P)/5000$ .

(4 marks)

Tier 3 · Hard

Demand

D

is modelled as a function of price

p

. The rate of decrease of demand with respect to price is proportional to

D

and inversely proportional to

p^2

. When

p=5

and

D=800

\frac{\mathrm dD}{\mathrm dp}=-64

. Construct the differential equation, determining its constant.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\frac{\mathrm dD}{\mathrm dp}=-\frac{2D}{p^2}$	5
Notes
The description gives $\frac{\mathrm dD}{\mathrm dp}=-\frac{kD}{p^2}$ with $k>0$ . Substitute the data: $-64=-\frac{k(800)}{25}=-32k$ , so $k=2$ . Hence $\frac{\mathrm dD}{\mathrm dp}=-\frac{2D}{p^2}$ .

(5 marks)

Tier 1 · Easy

The rate of change of

y

with respect to

x

is proportional to

x^2

. Construct a differential equation relating

y

and

x

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac{\mathrm dy}{\mathrm dx}=kx^2$ , where $k$ is a constant	2
Notes
The derivative $\dfrac{\mathrm dy}{\mathrm dx}$ represents the stated rate of change. Proportionality to $x^2$ therefore gives $\dfrac{\mathrm dy}{\mathrm dx}=kx^2$ for a constant $k$ .

(2 marks)

Tier 2 · Standard

A particle moves in a straight line with speed

v

. Its acceleration is

6

minus a quantity proportional to

v^2

. When

v=2

, its acceleration is

2

. Construct a differential equation for

v

in terms of time

t

, including the value of the constant.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$\dfrac{\mathrm dv}{\mathrm dt}=6-v^2$	4
Notes
Write $\dfrac{\mathrm dv}{\mathrm dt}=6-kv^2$ . Using $v=2$ and acceleration $2$ gives $2=6-4k$ , so $k=1$ . Hence $\dfrac{\mathrm dv}{\mathrm dt}=6-v^2$ .

(4 marks)

Tier 3 · Hard

Water enters a reservoir at a constant rate of

30\,\text{m}^3\text{h}^{-1}

. Water leaves at a rate proportional to

\sqrt V

, where

V

is the volume in

\text{m}^3

. When

V=100

, the volume is decreasing at

10\,\text{m}^3\text{h}^{-1}

. Construct a differential equation for

V

in terms of time

t

, determining the constant of proportionality.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$\dfrac{\mathrm dV}{\mathrm dt}=30-4\sqrt V$	5
Notes
The net rate is inflow minus outflow, so $\dfrac{\mathrm dV}{\mathrm dt}=30-k\sqrt V$ . At $V=100$ , the net rate is $-10$ , giving $-10=30-10k$ and hence $k=4$ . Therefore $\dfrac{\mathrm dV}{\mathrm dt}=30-4\sqrt V$ .

(5 marks)

Tier 2 · Standard

The radius

r

metres of a circular oil patch increases at a rate inversely proportional to

r

. When

r=6

, the radius is increasing at

0.5

metres per minute. Construct a differential equation for

r

in terms of time

t

minutes, including the constant of proportionality.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\dfrac{\mathrm dr}{\mathrm dt}=\dfrac3r$	4
Notes
Inverse proportionality gives $\dfrac{\mathrm dr}{\mathrm dt}=k/r$ for a positive constant $k$ . Using $r=6$ and $\dfrac{\mathrm dr}{\mathrm dt}=0.5$ gives $0.5=k/6$ , so $k=3$ . Therefore $\dfrac{\mathrm dr}{\mathrm dt}=3/r$ .

(4 marks)

Tier 3 · Hard

A heater supplies energy so that the temperature

T

would increase at a constant rate

a

. Cooling occurs at a rate proportional to the amount by which

T

exceeds the room temperature of

10^\circ\text{C}

. When

T=20

\dfrac{\mathrm dT}{\mathrm dt}=8

; when

T=50

\dfrac{\mathrm dT}{\mathrm dt}=-4

. Construct the differential equation, determining both constants, and find the equilibrium temperature.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$a=12$ and the cooling constant is $\dfrac25$ $\dfrac{\mathrm dT}{\mathrm dt}=12-\dfrac25(T-10)$ Equilibrium temperature $40^\circ\text{C}$	6
Notes
Write $\dfrac{\mathrm dT}{\mathrm dt}=a-k(T-10)$ with $k>0$ . The two observations give $8=a-10k$ and $-4=a-40k$ . Subtracting gives $12=30k$ , so $k=2/5$ , and then $a=8+10(2/5)=12$ . Thus $\dfrac{\mathrm dT}{\mathrm dt}=12-(2/5)(T-10)$ . At equilibrium the derivative is zero, so $12=(2/5)(T-10)$ and $T=40^\circ\text{C}$ .

(6 marks)

Tier 3 · Hard

A tank contains

500

litres of well-stirred brine. Brine containing

0.02

kg of salt per litre enters at

3

litres per minute, and the mixture leaves at the same volume rate. Let

S

be the mass of salt in the tank, in kg, after

t

minutes. Construct a differential equation for

S

, and find the equilibrium mass of salt.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\dfrac{\mathrm dS}{\mathrm dt}=0.06-\dfrac{3S}{500}$ Equilibrium mass $S=10$ kg	5
Notes
Salt enters at $3(0.02)=0.06$ kg per minute. Because the mixture is well stirred and the volume stays at $500$ litres, its salt concentration is $S/500$ kg per litre, so salt leaves at $3S/500$ kg per minute. Therefore the net rate is $\mathrm dS/\mathrm dt=0.06-3S/500$ . At equilibrium this rate is zero, giving $S=10$ kg.

(5 marks)

Q10

Tier 3 · Hard

10.

A drug is infused at a rate of

6t

mg per hour, where

t

is the time in hours. The body removes the drug at a rate proportional to the current mass

M

mg. At

t=2

M=20

and

\dfrac{\mathrm dM}{\mathrm dt}=7

. Construct the differential equation, determining the positive proportionality constant. Find also the relation between

M

and

t

when the input and removal rates are instantaneously equal.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\dfrac{\mathrm dM}{\mathrm dt}=6t-\dfrac14M$ The rates are instantaneously equal when $M=24t$	5
Notes
Write $\mathrm dM/\mathrm dt=6t-kM$ with $k>0$ . The data give $7=6(2)-20k$ , so $20k=5$ and $k=1/4$ . Hence $\mathrm dM/\mathrm dt=6t-M/4$ . Equality of the input and removal rates requires $6t=M/4$ , which is equivalent to $M=24t$ .

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 7.6. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 7 Differentiation

Want help turning this into marks?

Bring 7.6 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call