Differential Equations (Edexcel A Level Maths: Pure): Exam Questions

4 hours28 questions
1
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4 marks

Find the general solution to the differential equation

fraction numerator straight d y over denominator straight d x end fraction equals 2 x y

where y greater than 0.

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2
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3 marks

Find the general solution to the differential equation

fraction numerator d y over denominator d x end fraction equals 3 x squared y

where y not equal to 0.

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3a
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1 mark

The differential equation

fraction numerator straight d V over denominator straight d t space end fraction equals negative k V

is used to model the rate at which water is leaking from a container, where

  • V is the volume of water in the container

  • t is the time in seconds

  • k is a positive constant

Explain, in context, the significance of the negative sign in the model.

3b
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3 marks

Find the general solution to the differential equation.

3c
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2 marks

Given that

  • k equals 0.02

  • the initial volume of the container is 300 litres

find a complete equation linking V and t.

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4a
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4 marks

Given that y greater than 1, find the general solution to the differential equation

fraction numerator d y over denominator d x end fraction equals 6 x squared open parentheses y minus 1 close parentheses

writing your answer in the form

y equals A straight e to the power of straight f left parenthesis x right parenthesis end exponent plus 1

where A is a constant and straight f open parentheses x close parentheses is a function of x which you should find.

4b
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4 marks

Given that y greater than negative 2, find the general solution to the differential equation

fraction numerator d y over denominator d x end fraction equals 9 left parenthesis y plus 2 right parenthesis square root of x

writing your answer in the form

y equals A straight e to the power of straight f open parentheses x close parentheses end exponent minus 2

where A is a constant and straight f open parentheses x close parentheses is a function of x which you should find.

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5a
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2 marks

The volume of water in a sink, V, decreases with time t, measured from the point at which the plug is removed.

It is known that Vdecreases at a rate proportional to its volume.

Use this information to write down a suitable differential equation for V and t, using a constant of proportionality k where k greater than 0.

5b
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2 marks

The general solution to the differential equation in part (a) can be written in the form

V equals A straight e to the power of negative k t end exponent

where k greater than 0.

(i) State, in the context of the question, what the constant A represents.

(ii) Briefly explain the significance of the negative sign in the solution.

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6a
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3 marks

A differential equation is given by

fraction numerator d y over denominator d x end fraction equals sec squared x

where y equals 2 square root of 3 when x equals pi over 3.

Show that

y equals a plus tan space x

where a is a constant to be found.

6b
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5 marks

A differential equation is given by

sec space x fraction numerator d y over denominator d x end fraction equals cosec space y

where y equals 0 when x equals pi over 2.

Show that

cos space y equals b minus sin space x

where b is a constant to be found.

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1a3 marks

A large spherical balloon is deflating.

At time t seconds the balloon has radius r cm and volume V cm3.

The volume of the balloon is modelled as decreasing at a constant rate.

Using this model, show that

fraction numerator straight d r over denominator straight d t end fraction equals negative k over r squared

where k is a positive constant.

1b
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5 marks

Given that

  • the initial radius of the balloon is 40 cm

  • after 5 seconds the radius of the balloon is 20 cm

  • the volume of the balloon continues to decrease at a constant rate until the balloon is empty

solve a differential equation to find a complete equation linking r and t.

1c
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2 marks

Find the limitation on the values of t for which the equation in part (b) is valid.

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2
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4 marks

By solving the differential equation, show that the general solution to

2 x fraction numerator d y over denominator d x end fraction equals 3 k x cubed y

where y not equal to 0 is

y equals A straight e to the power of 1 half k x cubed end exponent

where A and k are constants.

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3
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4 marks

Find the general solution to the differential equation

fraction numerator d y over denominator d x end fraction equals sin squared open parentheses 2 y close parentheses

giving your answer in the form x equals straight f left parenthesis y right parenthesis plus c where c is a constant and straight f open parentheses y close parentheses is a function to be found.

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4a
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2 marks

Find the general solution to the differential equation

9 t squared minus 4 plus fraction numerator d x over denominator d t end fraction equals 0

4b
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3 marks

Find the particular solution to the differential equation

fraction numerator d V over denominator d x end fraction minus 4 equals 2 straight e to the power of x

given that the graph of V against x passes through the point with coordinates open parentheses 0 comma space 3 close parentheses.

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5a
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5 marks

A differential equation is given by

straight e to the power of negative 3 x end exponent fraction numerator d y over denominator d x end fraction equals 2 straight e to the power of y

It is known that y equals 0 when x equals 0.

Solve the differential equation, giving your answer in the form

p straight e to the power of 3 x end exponent plus straight e to the power of negative y end exponent equals q

where p and q are rational numbers to be found.

5b
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6 marks

A differential equation is given by

sin squared x fraction numerator d y over denominator d x end fraction equals cos squared y

where y equals 0 when x equals pi over 4.

Solve the differential equation, giving your answer in the form

tan space y equals straight f open parentheses x close parentheses

where straight f open parentheses x close parentheses is a function to be found.

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6a
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3 marks

A weather balloon of volume V m3 is being inflated, where t is the time in minutes after inflation begins.

  • The rate of change of its volume is inversely proportional to its volume

  • When the rate of inflation of the balloon is 10 m3 min-1, the volume of the balloon is 20 m3

Use this information to write down a suitable differential equation for V and t.

6b
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3 marks

Show that the general solution to the differential equation is

V squared equals 400 t plus c

where c is a constant.

6c
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3 marks

Initially, the balloon is flat with a volume of 0 m3.

Find the volume of the balloon after 25 minutes.

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7a
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7 marks

A disease affecting trees is spreading throughout a large forested area. Let N be the number of infected trees t days after the disease was first discovered.

A model for N and t is given by

space fraction numerator d N over denominator d t end fraction equals k N t

where k is a positive constant.

It is known that

  • When the disease was first discovered, 3trees were infected

  • Ten days after the disease was first discovered, 10 trees were infected

Solve the differential equation to show that

N equals 3 straight e to the power of a t squared end exponent

where

a equals 1 over 100 ln open parentheses 10 over 3 close parentheses

7b
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3 marks

Scientists believe the majority of the forest can be saved from infection if action is taken before 30 trees are infected.

Find the number of days (since first discovering the disease) that the model predicts scientists have in order to take action.

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8a
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7 marks

A bar of soap in the shape of a cuboid is placed in a bowl of warm water and its volume is recorded at regular intervals.  The water is maintained at a constant temperature.

  • Before being placed in the water, the soap measures 3 cm by 6 cm by 10 cm

  • Two minutes later, the bar of soap measures 2.85 cm by 5.7 cm by 9.5 cm

  • The rate of decrease in volume of the bar of soap is modelled as being directly proportional to its volume

Defining any variables you use, find and solve a differential equation linking the volume of the bar of soap and time.

8b
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2 marks

By considering the volume of soap as time increases, suggest a limitation of the model.

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1a3 marks

Express fraction numerator 3 over denominator open parentheses 2 x minus 1 close parentheses open parentheses x plus 1 close parentheses end fraction in partial fractions.

1b5 marks

When chemical A and chemical B are mixed, oxygen is produced.

A scientist mixed these two chemicals and measured the total volume of oxygen produced over a period of time.

The total volume of oxygen produced, V m3, t hours after the chemicals were mixed, is modelled by the differential equation

fraction numerator straight d V over denominator straight d t end fraction equals fraction numerator 3 V over denominator open parentheses 2 t minus 1 close parentheses open parentheses t plus 1 close parentheses end fraction space space space space space space space space space space space space space V greater or equal than 0 space space space space space space space t greater or equal than k

where k is a constant.

Given that exactly 2 hours after the chemicals were mixed, a total volume of 3 m3 of oxygen had been produced, solve the differential equation to show that

V equals fraction numerator 3 open parentheses 2 t minus 1 close parentheses over denominator open parentheses t plus 1 close parentheses end fraction

1c2 marks

The scientist noticed that

  • there was a time delay between the chemicals being mixed and oxygen being produced

  • there was a limit to the total volume of oxygen produced

Deduce from the model

(i) the time delay giving your answer in minutes,

(ii) the limit giving your answer in m3

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2a3 marks
Diagram of a rectangular prism with dimensions labelled: length 20 m, width 10 m, height 5 m. The depth of water is labelled as h m.
Figure 1

A tank in the shape of a cuboid is being filled with water.

The base of the tank measures 20 m by 10 m and the height of the tank is 5 m, as shown in Figure 1.

At time t minutes after water started flowing into the tank the height of the water was h m and the volume of the water in the tank was V m3.

In a model of this situation

  • the sides of the tank have negligible thickness

  • the rate of change of V is inversely proportional to the square root of h

Show that

fraction numerator d h over denominator d t end fraction equals fraction numerator lambda over denominator square root of h end fraction

where lambda is a constant.

2b
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5 marks

Given that

  • initially the height of the water in the tank was 1.44 m

  • exactly 8 minutes after water started flowing into the tank the height of the water was 3.24 m

use the model to find an equation linking h with t, giving your answer in the form

h to the power of 3 over 2 end exponent equals A t plus B

where A and B are constants to be found.

2c
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2 marks

Hence find the time taken, from when water started flowing into the tank, for the tank to be completely full.

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3a
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4 marks

Find the general solution to the differential equation

fraction numerator 2 y minus 1 over denominator 3 end fraction fraction numerator d y over denominator d x end fraction equals x squared y squared minus x squared y

where y greater than 1, giving your answer in the form

y squared minus y equals straight f open parentheses x close parentheses

3b
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4 marks

Find the general solution to the differential equation

3 fraction numerator d y over denominator d x end fraction equals fraction numerator cosec space y cubed over denominator space y to the power of space space 2 end exponent end fraction

giving your answer in the form x equals straight g left parenthesis y right parenthesis.

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4a
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5 marks

Show that the general solution to the differential equation

y cot space x space fraction numerator d y over denominator d x end fraction equals y squared plus 3

is given by

y squared plus 3 equals A sec to the power of 2 space end exponent x

where A is a constant.

4b
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6 marks

Solve the differential equation

straight e to the power of x squared end exponent space fraction numerator d y over denominator d x end fraction equals 2 x space cosec space 3 y 

given that y equals pi over 3 when x equals 0.

Give your answer in the form

cos space 3 y equals straight f open parentheses x close parentheses

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5a
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2 marks

A hot air balloon is being inflated at a rate that is inversely proportional to the square of its volume.

Defining variables for the volume of the balloon (m3) and time (seconds), write down a differential equation to describe the relationship between volume and time as the hot air balloon is inflated.

5b
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8 marks

You are given the following information:

  • Initially, the hot air balloon has a volume of zero

  • After 400 seconds of inflating, its volume is 600 m3

  • The hot air balloon is considered ready for release when its volume reaches 1250 m3

If the hot air balloon needs to be ready for release by midday, find the latest time that it can start being inflated.

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6a
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2 marks

Find the general solution to the differential equation

1 half sec squared open parentheses 3 t close parentheses plus 2 fraction numerator d x over denominator d t end fraction equals 0

6b
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6 marks

Find the particular solution to the differential equation

2 x straight e to the power of 4 x end exponent minus 3 fraction numerator d V over denominator d x end fraction equals 1

where the graph of V against x passes through the point with coordinates open parentheses 0 comma space 2 close parentheses.

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7
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8 marks

A petrol tanker is leaking petrol. The volume of petrol inside the petrol tanker is V litres, which is decreasing over time t minutes.

  • The volume of petrol inside the tanker decreases at a rate proportional to the square of its volume

  • The initial volume of petrol inside the tanker is 4000 litres

  • After 10 minutes, the volume of petrol in the tanker has dropped by 30%

By forming and solving a suitable differential equation, show that

V equals 4000 open parentheses fraction numerator a over denominator 3 t plus b end fraction close parentheses

where a and b are constants to be found.

Hence, describe what happens to the volume of petrol after a long time.

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1a
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5 marks

A spherical mint of radius 5 mm is placed in the mouth and sucked.

Four minutes later, the radius of the mint is 3 mm.

In a simple model, the rate of decrease of the radius of the mint is inversely proportional to the square of the radius.

Using this model and all the information given, find an equation linking the radius of the mint and the time.

(You should define the variables that you use.)

1b
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2 marks

Hence find the total time taken for the mint to completely dissolve. Give your answer in minutes and seconds to the nearest second.

1c
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1 mark

Suggest a limitation of the model.

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2a6 marks

Use the substitution u equals 4 minus square root of h to show that

integral fraction numerator d h over denominator 4 minus square root of h end fraction equals negative 8 ln open vertical bar 4 minus square root of h close vertical bar minus 2 square root of h plus k

where k is a constant.

2b2 marks

A team of scientists is studying a species of slow growing tree.

The rate of change in height of a tree in this species is modelled by the differential equation

fraction numerator d h over denominator d t end fraction equals fraction numerator t to the power of 0.25 end exponent open parentheses 4 minus square root of h close parentheses over denominator 20 end fraction

where h is the height in metres and t is the time, measured in years, after the tree is planted.

Find, according to the model, the range in heights of trees in this species.

2c
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7 marks

One of these trees is one metre high when it is first planted.

According to the model, calculate the time this tree would take to reach a height of 12 metres, giving your answer to 3 significant figures.

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3a
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4 marks
Diagram of a rectangular tank with dimensions labelled: height 5m, length 8m, width 3m. The water level is at height h. Point T is marked at the bottom left front corner.
Figure 5

Water flows at a constant rate into a large tank.

The tank is a cuboid, with all sides of negligible thickness.

The base of the tank measures 8 m by 3 m and the height of the tank is 5 m.

There is a tap at a point T at the bottom of the tank, as shown in Figure 5.

At time t minutes after the tap has been opened

  • the depth of the water in the tank is h metres

  • water is flowing into the tank at a constant rate of 0.48 m3 per minute

  • water is modelled as leaving the tank through the tap at a rate of 0.1 h m3 per minute

Show that, according to the model,

1200 space fraction numerator straight d h over denominator straight d t end fraction equals 24 minus 5 h

3b6 marks

Given that when the tap was opened, the depth of the water in the tank was 2 m, show that, according to the model,

h equals A plus B straight e to the power of negative k t end exponent

where A, B and k are constants to be found.

3c2 marks

Given that the tap remains open, determine, according to the model, whether the tank will ever become full, giving a reason for your answer.

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4
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5 marks

Find the general solution to the differential equation

fraction numerator d y over denominator d x end fraction equals 2 x y plus 2 x minus y minus 1

where y greater than negative 1, giving your answer in the form y equals straight f open parentheses x close parentheses.

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5a
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6 marks

Palm trees are being planted on an island. Let N be the total number of palm trees planted on the island after t days.

The variables N and t are modelled by the differential equation

fraction numerator straight d N over denominator straight d t end fraction equals k N open parentheses N minus 1 close parentheses comma space space space N greater than 1

where N greater than 1 and k is a positive constant.

By solving the differential equation, show that

N equals fraction numerator 1 over denominator 1 minus A straight e to the power of k t end exponent end fraction

where A is a positive constant.

5b
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3 marks

It is known that

  • Initially 2 palm trees are planted

  • After 14 days, 4 palm trees in total have been planted

Use this information to show that

k equals 1 over 14 ln space p

where p is a rational number to be found.

5c
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3 marks

By considering the form of the solution to the differential equation, suggest a range of values of t for which the model is valid.

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6a
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6 marks

The temperature of a heated object, T°C, cools over time, t minutes. The room temperature (called the ambient temperature) is constant, T subscript amb, where T greater than T subscript amb.

Newton’s Law of Cooling states that the rate of decrease in temperature of a heated object is directly proportional to the difference between the object’s temperature and the ambient temperature.

By forming and solving a differential equation in T and t (involving the constant T subscript amb and a positive constant of proportionality, k) show that

T equals T subscript amb plus A straight e to the power of negative k t end exponent

6b
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4 marks

For food safety reasons, a meat processing factory must store its products at a temperature of below -1 °C.

  • One particular product has a temperature of 7 °C

  • It is placed in one of the factory's freezers, which has a constant ambient temperature of -4 °C

  • One minute later, its temperature has dropped to 4.7 °C.

  • Any products that fail to cool to below -1 °C within 6 minutes must be discarded

Determine whether or not this product will need to be discarded.

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7a
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4 marks

Show that the solution to the differential equation

cos space x space fraction numerator d y over denominator d x end fraction equals cos space y space

where y equals 0 when x equals 0 may be written in the form

open vertical bar tan space open parentheses y over 2 plus pi over 4 close parentheses space close vertical bar equals open vertical bar tan space open parentheses x over 2 plus pi over 4 close parentheses close vertical bar

7b
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6 marks

(i) Prove that if open vertical bar tan space open parentheses y over 2 plus pi over 4 close parentheses close vertical bar space equals open vertical bar tan open parentheses space x over 2 plus pi over 4 close parentheses close vertical bar then

y equals x plus 2 n pi space space space space space space or space space space space space space y equals negative x plus open parentheses 2 n minus 1 close parentheses pi

where n is an integer.

(ii) Hence deduce that the particular solution to the differential equation in part (a) is

y equals pi minus x

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