© Christine Crisp Teach A Level Maths Vol. 2: A2 Core Modules 49: Setting up and Solving...

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

49: Setting up and 49: Setting up and Solving Differential Solving Differential EquationsEquations

Setting up and Solving Differential Equations

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Module C4

Setting up and Solving Differential EquationsWe have seen how to solve differential

equations by the method of separating the variables.We have also met equations that describe situations of growth and decay.

This presentation brings the 2 topics together and we see how to set up and solve the differential equations for growth and decay.

We will also set up and solve some differential equations that describe other situations.


Solution: The description in the question is typical of exponential growth.

e.g. 1. A solution initially contains 200 bacteria. Assuming the number, x, increases at a rate proportional to the number present, write down a differential equation connecting x and the time, t. If the rate of increase of the number is initially 100 per hour, how many are there after 2 hours?

We have to set up a differential equation which describes the situation and solve it to find x when t = 2.


xdt

dx


, where k is a constant.

kxdt

dx

The description in the question is typical of exponential growth. We have to set up a differential equation which describes the situation and solve it to find x when t = 2.

Solution:


kxdt

dx

Cktx ln

dtkdxx

1

We were given “ A solution initially contains 200 bacteria . . . “ and “. . . the rate of increase of the number is initially 100 per hour”

You may remember the solution to this equation but, if not, we can separate the variables to find it.

There are 2 pairs of conditions here which enable us to solve for 2 unknowns.

)1( ktAex


:200,0 xt

kxdt

dx

Cktx ln

dtkdxx

1



)1( ktAex

200A 0200 Ae


kxdt

dx

Cktx ln

dtkdxx

1


:200,0 xt

:100,0 dt

dxt

200A


0200 Ae

50k

)1( ktAex

)200(100 kkxdt

dx


kxdt

dx

Cktx ln

dtkdxx

1


:100,0 dt

dxtSubstituting in (1):

200A


0200 Ae

)1( ktAex

50k )200(100 kkxdt

dx:200,0 xt


50k

kxdt

dx

Cktx ln

dtkdxx

1

)1( ktAexWe were given “ A solution initially contains 200 bacteria . . . “ and “. . . the rate of increase of the number is initially 100 per hour”

tex 50200 )integer nearest ( 5442 xt

200A


:100,0 dt

dxtSubstituting in (1):

0200 Ae

)200(100 kkxdt

dx:200,0 xt


tex 50200

The graph showing the growth function is

Number at start of measurements

Number after 2 hours

Setting up and Solving Differential Equationse.g. 2 A radioactive element decays at a rate

that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. Solution:

Let m be mass in mg and t time in days.


that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg.

mdt

dm

Solution:

,kmdt

dm where k is a

positive constant.


We could write where k is negative, but

most people prefer to emphasise the negative

gradient.

kmdt

dm



mdt

dm

Solution:

,kmdt

dm where k is a

positive constant.


We can quote the solution:



mdt

dm

Solution:

,kmdt

dm where k is a

positive constant.


We can quote the solution:Make sure you write t on the

r.h.s. !

Am tke



mdt

dm

Solution:

,kmdt

dm where k is a

positive constant.



10,0 mt 010 Ae 10 AAm tke



mdt

dm

Solution:

,kmdt

dm where k is a

positive constant.



)20(105 ke ke 2050 10,0 mt

5,20 mt

010 Ae 10 A

A log is just an index !

Am tke



mdt

dm

Solution:

,kmdt

dm where k is a

positive constant.



)20(105 ke ke 2050 50ln20 k

) d.p. ( 30350 k

10,0 mt

5,20 mt

010 Ae 10 AAm tke


that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. We now have

tem 035010


that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. We now have

tem 035010 1m

te 035010

10ln0350 t

0350

10ln

t 66It takes 66 days to decay to 1 mg.

te 0350101

( nearest integer )


tem 035010

The graph showing the decay function is

Mass at start of measurements

Time when mass is 1 mg


The words “ a rate proportional to . . . ” followed by the quantity the rate refers to, gives the differential equation for growth or decay.

SUMMARY

e.g. “ the number, x, increases at a rate proportional to x ” gives kx

dt

dxx

dt

dx

The solution to the above equation is

ktAex ( but if we forget it, we can easily separate

the variables in the differential equation and solve ).

either 1 pair of values of x and t and 1 pair

of values of and t,dt

dx

or 2 pairs of values of x and t.

The values of A and k are found by substituting

Setting up and Solving Differential EquationsExercis

eFor the following problems, choose suitable letters and set up the differential equations but don’t solve them.

When you have the first 2 equations, check you agree with me and then solve the complete problems.


Exercise1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred.2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute.

50

050

10

( When solving, use k correct to 3 s.f. )

nkdt

dn1.

( When solving, use k correct to 3 s.f. )

xkdt

dx 2.


Solution:

60000,0 nt

kteAn

60000A

63000,10 nt )10(6000063000 ke

Let t = 0 in 1990.

60000

63000ln10k

004880k

kndt

dn

1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred.


ten 00488060000

,60000A 004880k,kteAn

)20(0048806000020 ent

200,66 ( nearest hundred )

1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred.


Solution:

50,0 xt 50A

10,4 xt

xkdt

dx ktAex

)4(5010 ke

50

10ln4k 4020k

2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute.

50

050

10

(3 s.f.)


tex 402050 :050xSubstitute

te 402050050

50

050ln4020 t

7285tAns: 5 hrs 44 mins

,ktAex 4020k,50A

2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute.

50

050

10


You might meet differential equations that do not describe growth functions.


e.g. 1 The gradient of a curve at every point equals the square of the y-value at that point. Express this as a differential equation and find the particular solution which passes through ( 1, 1 ).Solution:

The equation is


2ydx

dy

e.g. 1 The gradient of a curve at every point equals the square of the y-value at that point. Express this as a differential equation and find the particular solution which passes through ( 1, 1 ).

dxdyy

2

1

Cxy

1

Solution:

The equation is

Separating the variables:

This is the general solution to the equation


Cxy

1

C 11 2 C

( 1, 1 ) lies on the curve

21

xy

So,

or, yx

2

1

The equation is 2

1

xy

Setting up and Solving Differential EquationsExercis

e1. The gradient of a curve at any point ( x, y )

is equal to the product of x and y. The curve passes through the point ( 1, 1 ). Form a differential equation and solve it to find the equation of the curve. Give your answer in the form . )(xfy

Solution: xy

dx

dy

dxxdyy

1 C

xy

2ln

2


Cx

y 2

ln2

( 1, 1 ) on the curve:

2

1 C

2

12

x

ey

C2

11ln

2

1

2ln

2

x

ySo,


There is one very well known situation which can be described by a differential equation.

The following is an example.


Solution:The equation gives the rate of decrease of the temperature of the coffee. It is proportional to the amount that the temperature is above room temperature.

)20( xkdt

dxExplain what the following equation is describing:

This is an example of Newton’s law of cooling.

The temperature of a cup of coffee is given byat time t minutes after it was poured. The temperature of the room in which the cup is placed is

C20

Cx


)20( xkdt

dx

We can solve this equation as follows:

kdtdxx

20

1 Cktx 20ln

ktAex 20ktAex 20

If we are given further information, we can complete the solution as in the other examples.


The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.


The words “ a rate proportional to . . . ” followed by the quantity the rate refers to, gives the differential equation for growth or decay.

SUMMARY

e.g. “ the number, x, increases at a rate proportional to x ” gives kx

dt

dxx

dt

dx

The solution to the above equation is

ktAex ( but if we forget it, we can easily separate

the variables in the differential equation and solve ).

either 1 pair of values of x and t and 1 pair

of values of and t,dt

dx

or 2 pairs of values of x and t.

The values of A and k are found by substituting


xdt

dx


, where k is a constant.

kxdt

dx

Solution: The description in the question is typical of exponential growth. We have to set up a differential equation which describes the situation and solve it to find x when t = 2.


kxdt

dx

Cktx ln

dtkdxx

1

)1( ktAexWe were given “ A solution initially contains 200 bacteria . . . “ and “. . . the rate of increase of the number is initially 100 per hour”

:200,0 xt

:100,0 dt

dxt 50k

Substituting in (1):

tex 50200 )integer nearest ( 5442 xt

200A


0200 Ae

)200(100 k



mdt

dm

Solution:

,kmdt

dm where k is a

positive constant.



tkeAm

)20(105 ke ke 2050 50ln20 k

) d.p. ( 30350 k

10,0 mt

5,20 mt

010 Ae 10 A


We now have

tem 035010

1mte 035010

10ln0350 t

0350

10ln

t 66

It takes 66 days for the mass to decay to 1 mg.

te 0350101

( nearest integer )


There is one very well known situation which can be described by a differential equation.

)20( xkdt

dx

Explain what the following equation is describing:

The following is an example.The temperature of a cup of coffee is given byat time t minutes after it was poured. The temperature of the room in which the cup is placed is

C20

Cx

Setting up and Solving Differential EquationsSolutio

n:The equation gives the rate of decrease of the temperature of the coffee. It is proportional to the amount that the temperature is above room temperature.

This is an example of Newton’s law of cooling.

We can solve this equation as follows:

kdtdxx

20

1 Cktx 20ln

ktAex 20 ktAex 20If we are given further information, we can complete the solution as in the other examples.

© Christine Crisp Teach A Level Maths Vol. 2: A2 Core Modules 49: Setting up and Solving...

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