What is 8.8: Interpreting differential equation solutions?

Interpret constants and initial values with their units, and examine limiting behaviour to identify equilibrium values or long-term predictions.

How many marks are 8.8 questions worth?

8.8 practice drills on this page are worth 3–8 marks each; marks in a real exam vary by question.

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We have not yet indexed a verified past-paper appearance for 8.8 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

8.8

Interpret the solution of a differential equation in the context of solving a problem, including identifying limitations of the solution; includes links to kinematics.

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Interpreting differential equation solutions

Worked answers and methods for 8.8 on Edexcel A-level Maths 9MA0.

Explanation

Interpret constants and initial values with their units, and examine limiting behaviour to identify equilibrium values or long-term predictions.
In kinematics, $v=\frac{\mathrm ds}{\mathrm dt}$ and $a=\frac{\mathrm dv}{\mathrm dt}$ ; the sign of velocity gives direction, while a change of sign marks a reversal of motion.
Check whether the mathematical solution remains meaningful on the time interval and within the physical range assumed by the model.
A common error is to state only that a model is 'unrealistic'; name a specific assumption, such as constant environmental conditions or neglect of resistance, and explain its effect.

Worked example

The velocity of a particle is modelled by $v=20(1-e^{-0.5t})\,\text{m s}^{-1}$ for $t\geq0$ . Find the limiting velocity and the time when $v=15\,\text{m s}^{-1}$ . State one limitation of the model.

1.As $t\to\infty$ , $e^{-0.5t}\to0$ , so the limiting velocity is $20\,\text{m s}^{-1}$ .
2.For $v=15$ , $15=20(1-e^{-0.5t})$ , so $e^{-0.5t}=1/4$ and $t=2\ln4\approx2.77\,\text{s}$ .
3.A real resistance law or driving force may change with conditions, so the same constants need not remain valid indefinitely.

Answer: $20\,\text{m s}^{-1}$ ; $t=2\ln4\approx2.77\,\text{s}$ ; One valid limitation, such as a constant resistance law being assumed

Common mistakes

Don't extend a population solution beyond the time at which the model predicts an impossible negative value.
Don't report a limiting value algebraically and fail to interpret its units or challenge the long-term model.

Exam tip

For a contextual differential model, give numerical conclusions with units and state a limitation tied to its assumptions.

Worked practice

Tier 1 · Easy

A population model has solution

N=500e^{0.12t}

, where

t

is measured in years. Interpret the constants

500

and

0.12

, and state one limitation of the model.

(4)

(Total for Question 1 is 4 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Initial population $500$ ; continuous relative growth rate $0.12$ per year One valid limitation, such as finite resources eventually preventing unlimited exponential growth	4
Notes
At $t=0$ , $N=500$ , so $500$ is the initial population. Also $\frac1N\frac{\mathrm dN}{\mathrm dt}=0.12$ , so $0.12$ is the continuous proportional growth rate per year. The model predicts unbounded growth and therefore ignores a limiting factor such as finite food, space or changing birth and death rates.

(4 marks)

Tier 2 · Standard

The mass of a substance is modelled by

M=50e^{-0.2t}

grams, where

t

is in hours. Find the time at which the mass reaches

5

grams, giving the time exactly and to

3

significant figures; hence state when the mass is below

5

grams. Interpret the limiting value of

M

and state one limitation of the model.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Exact crossing time: $t=5\ln10$ hours Crossing time to $3$ significant figures: $t=11.5$ hours The mass is below $5$ grams for $t>5\ln10$ hours $M$ tends to $0$ grams as $t\to\infty$ . For example, the decay constant may not remain fixed.	5
Notes
At the crossing, $50e^{-0.2t}=5$ , so $e^{-0.2t}=0.1$ and $t=5\ln10=11.512\ldots$ hours, which is $11.5$ hours to $3$ significant figures. Since the model decreases continuously, $M<5$ for $t>5\ln10$ . As $t\to\infty$ , the exponential tends to zero, so the model predicts $M\to0$ . A limitation is that it assumes one constant proportional decay rate indefinitely.

(5 marks)

Tier 3 · Hard

A particle has velocity

v=18-24e^{-0.3t}\,\text{m s}^{-1}

for

t\geq0

, with displacement

s=0

t=0

. Find when the particle changes direction, its displacement then, and its limiting velocity. Give one limitation of using this model for arbitrarily large

t

(8)

(Total for Question 3 is 8 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$t=\frac{10}{3}\ln\frac43\approx0.959\,\text{s}$ $s\approx-2.74\,\text{m}$ Limiting velocity $18\,\text{m s}^{-1}$ One valid limitation tied to the assumed force or resistance law	8
Notes
Solve $v=0$ : $18=24e^{-0.3t}$ , so $t=\frac{10}{3}\ln(4/3)\approx0.959\,\text{s}$ . Since $v(0)=-6$ and $v$ tends to $18>0$ , the continuous velocity changes sign there, proving a reversal. Integrating and using $s(0)=0$ gives $s=18t+80e^{-0.3t}-80$ , which is approximately $-2.74\,\text{m}$ at the turn. The limiting velocity is $18\,\text{m s}^{-1}$ . Over long times, the assumed force or resistance relationship and constant conditions may cease to apply.

(8 marks)

Tier 1 · Easy

The temperature

T

°C of a drink satisfies

\dfrac{\mathrm dT}{\mathrm dt}=-0.25(T-18)

, where

t

is measured in minutes. Its solution is

T=18+62e^{-0.25t}

. State the initial temperature and interpret the limiting value of

T

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	Initial temperature $80$ °C The temperature tends to the surrounding temperature of $18$ °C	3
Notes
At $t=0$ , $T=18+62=80$ °C. As $t\to\infty$ , $e^{-0.25t}\to0$ , so $T\to18$ °C. In the differential equation, $\mathrm dT/\mathrm dt=0$ when $T=18$ , so this is the equilibrium temperature of the surroundings.

(3 marks)

Tier 2 · Standard

The mass

M

grams of dissolved salt in a mixing vessel satisfies

\dfrac{\mathrm dM}{\mathrm dt}=6-0.03M

, where

t

is measured in minutes and

M(0)=40

. The solution is

M=200-160e^{-0.03t}

. (a) Find the time when

M=120

, giving your answer exactly and to

3

significant figures. (b) Explain the significance of

M=200

in both the differential equation and the model.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$t=\dfrac{100}{3}\ln2$ minutes $=23.1$ minutes to $3$ significant figures $M=200$ g is the equilibrium mass and the limiting mass predicted by the model	5
Notes
Set $120=200-160e^{-0.03t}$ . Then $e^{-0.03t}=1/2$ , so $t=\ln2/0.03=(100/3)\ln2=23.104\ldots$ minutes, which is $23.1$ minutes to $3$ significant figures. In the differential equation, $\mathrm dM/\mathrm dt=0$ when $M=200$ . Also $e^{-0.03t}\to0$ , so the solution approaches $200$ g; it is the model's equilibrium and limiting mass.

(5 marks)

Tier 3 · Hard

A particle moves in a straight line with velocity

v>0

. Its motion is modelled by

\dfrac{\mathrm dv}{\mathrm dt}=\dfrac4v

, where

v=2\,\text{m s}^{-1}

and displacement

s=0

when

t=0

. Find

v

in terms of

t

. Hence find the displacement when

t=3

seconds, giving your answer exactly and to

3

significant figures. Explain why the model cannot be valid for arbitrarily large

t

(7)

(Total for Question 6 is 7 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$v=\sqrt{8t+4}\,\text{m s}^{-1}$ $s=\dfrac{14\sqrt7-2}{3}\,\text{m}=11.7\,\text{m}$ to $3$ significant figures The model predicts unbounded speed, so the inverse-speed acceleration law cannot remain valid indefinitely	7
Notes
Separating gives $v\,\mathrm dv=4\,\mathrm dt$ , so $v^2/2=4t+C$ . The initial condition gives $v^2=8t+4$ , and $v>0$ selects $v=\sqrt{8t+4}$ . Since $v=\mathrm ds/\mathrm dt$ , $s(3)=\int_0^3\sqrt{8t+4}\,\mathrm dt=[(8t+4)^{3/2}/12]_0^3=(14\sqrt7-2)/3=11.680\ldots$ , which is $11.7$ m to $3$ significant figures. The formula gives $v\to\infty$ , so the assumed acceleration law is not credible indefinitely.

(7 marks)

Tier 2 · Standard

A population is modelled by

P=900-100e^{0.06t}

, where

t

is the number of years after observations begin. State the initial population. Find the time when the model first predicts no animals, giving your answer exactly and to

3

significant figures. Hence give the time interval over which the model gives a non-negative population, and state one limitation of the model.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Initial population $800$ $t=\dfrac{50}{3}\ln9$ years $=36.6$ years to $3$ significant figures $0\leq t\leq\dfrac{50}{3}\ln9$ One valid limitation, such as the rate of population loss not remaining unchanged as the population and conditions change	5
Notes
At $t=0$ , $P=900-100=800$ . Setting $P=0$ gives $e^{0.06t}=9$ , so $t=\ln9/0.06=(50/3)\ln9=36.620\ldots$ years, which is $36.6$ years to $3$ significant figures. The model gives $P\geq0$ only from $t=0$ to this time. It assumes the same exponential loss pattern and unchanged conditions throughout; beyond the zero-population time it becomes physically impossible.

(5 marks)

Tier 3 · Hard

The area

A

hectares of a site that has been restored is modelled by

\dfrac{\mathrm dA}{\mathrm dt}=0.1A\left(1-\dfrac{A}{50}\right)

, where

t

is measured in years. A solution of this differential equation is

A=\dfrac{50}{1+4e^{-0.1t}}

for

t\geq0

. Interpret the constants

50

4

and

0.1

in the model. Find when the restored area first reaches

40

hectares, giving the time exactly and to

3

significant figures. Describe the limiting behaviour of both

A

and

\dfrac{\mathrm dA}{\mathrm dt}

t\to\infty

, and state one limitation of the model.

(7)

(Total for Question 8 is 7 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$50$ hectares is the limiting restored area, $4$ encodes the initial area $A(0)=10$ hectares, and $0.1\,\text{year}^{-1}$ is the proportional restoration-rate parameter $t=10\ln16$ years $=27.7$ years to $3$ significant figures $A\to50$ hectares while $\dfrac{\mathrm dA}{\mathrm dt}\to0$ One valid limitation, such as assuming that the available area, funding and restoration conditions remain fixed	7
Notes
The factor $1-A/50$ makes $A=50$ hectares an equilibrium and limiting value. At $t=0$ , $A=50/(1+4)=10$ hectares, so $4$ encodes the initial shortfall-to-restored ratio $(50-10)/10$ . The constant $0.1\,\text{year}^{-1}$ is the model's proportional restoration-rate parameter; when $A$ is small, the relative rate $(1/A)\,\mathrm dA/\mathrm dt$ is approximately $0.1$ per year. Setting $A=40$ gives $1+4e^{-0.1t}=5/4$ , so $e^{-0.1t}=1/16$ and $t=10\ln16=27.7258\ldots$ years, which is $27.7$ years to $3$ significant figures. Since the differential equation gives $\mathrm dA/\mathrm dt>0$ for $0<A<50$ , this is the first such time. As $t\to\infty$ , $A\to50$ and the differential equation then gives $\mathrm dA/\mathrm dt\to0$ . The model assumes a fixed available area and fixed conditions, ignoring changes such as funding, weather or land becoming unsuitable.

(7 marks)

Tier 3 · Hard

The balance, £

B

, of an investment account is modelled by

\dfrac{\mathrm dB}{\mathrm dt}=0.04B-1\,200

, where

t

is measured in years. The initial balance is £

40\,000

, and a solution is

B=30\,000+10\,000e^{0.04t}

. Interpret the balance £

30\,000

using the differential equation, and explain why this equilibrium is unstable. Find when the balance first reaches £

50\,000

, giving your answer exactly and to

3

significant figures. State one limitation of the model.

(7)

(Total for Question 9 is 7 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	£ $30\,000$ is an unstable equilibrium balance: balances above it increase and balances below it decrease $t=25\ln2$ years $=17.3$ years to $3$ significant figures One valid limitation, such as assuming a fixed interest rate and a continuous withdrawal of £ $1\,200$ per year indefinitely	7
Notes
At a balance of £ $30\,000$ , the differential equation gives $\mathrm dB/\mathrm dt=0$ . Above £ $30\,000$ the derivative is positive, whereas below £ $30\,000$ it is negative, so balances move away from the equilibrium. Setting the supplied solution equal to £ $50\,000$ gives $e^{0.04t}=2$ , hence $t=25\ln2=17.328\ldots$ years, which is $17.3$ years to $3$ significant figures. The model assumes that both the interest rate and the continuous withdrawal of £ $1\,200$ per year remain fixed, which need not hold over such a long period.

(7 marks)

Q10

Tier 3 · Hard

10.

The mass

M

kilograms of a pollutant in a treatment tank is modelled by

\displaystyle\frac{\mathrm dM}{\mathrm dt}=\begin{cases}10-0.25M,&0\leq t<4,\\-0.25M,&t\geq4,\end{cases}

where

t

is measured in hours and

M(0)=0

. A solution is

\displaystyle M=\begin{cases}40(1-e^{-0.25t}),&0\leq t<4,\\40(1-e^{-1})e^{-0.25(t-4)},&t\geq4.\end{cases}

Interpret the terms

10

and

0.25M

in the differential equation, and explain what happens at

t=4

. Find the maximum mass of pollutant and the later time when

M=10

, giving each answer exactly and to

3

significant figures. State one limitation of the model.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$10$ kilograms per hour is the input rate and $0.25M$ kilograms per hour is the removal rate; at $t=4$ the input stops Maximum mass $=40(1-e^{-1})$ kilograms $=25.3$ kilograms to $3$ significant figures $t=4+4\ln\left(4(1-e^{-1})\right)$ hours $=7.71$ hours to $3$ significant figures One valid limitation, such as assuming perfect mixing and a constant proportional removal rate	7
Notes
Before $t=4$ , the first term represents pollutant entering at $10$ kilograms per hour and $0.25M$ is the removal rate. At $t=4$ the input is switched off, leaving removal only. While $0<t<4$ , $M<40$ and the first differential equation gives $\mathrm dM/\mathrm dt>0$ ; after the switch, the second gives $\mathrm dM/\mathrm dt<0$ . The maximum is therefore $M(4)=40(1-e^{-1})=25.284\ldots$ kilograms, or $25.3$ kilograms to $3$ significant figures. On the later branch, $10=40(1-e^{-1})e^{-0.25(t-4)}$ , so $t=4+4\ln(4(1-e^{-1}))=7.710\ldots$ hours, or $7.71$ hours to $3$ significant figures. The model assumes ideal mixing and a fixed removal coefficient, which may not hold in a real tank.

(7 marks)

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