© Christine Crisp “ “ Teach A Level Maths” Teach A Level Maths” Vol. 2: A2 Core Vol. 2: A2 Core Modules Modules 49: Setting up and 49: Setting up and Solving Solving Differential Differential Equations Equations
Mar 26, 2015
© 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
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
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.
Setting up and Solving Differential Equations
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.
Setting up and Solving Differential Equations
xdt
dx
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?
, 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:
Setting up and Solving Differential Equations
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
Setting up and Solving Differential Equations
:200,0 xt
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.
)1( ktAex
200A 0200 Ae
Setting up and Solving Differential Equations
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”
:200,0 xt
:100,0 dt
dxt
200A
You may remember the solution to this equation but, if not, we can separate the variables to find it.
0200 Ae
50k
)1( ktAex
)200(100 kkxdt
dx
Setting up and Solving Differential Equations
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”
:100,0 dt
dxtSubstituting in (1):
200A
You may remember the solution to this equation but, if not, we can separate the variables to find it.
0200 Ae
)1( ktAex
50k )200(100 kkxdt
dx:200,0 xt
Setting up and Solving Differential Equations
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
You may remember the solution to this equation but, if not, we can separate the variables to find it.
:100,0 dt
dxtSubstituting in (1):
0200 Ae
)200(100 kkxdt
dx:200,0 xt
Setting up and Solving Differential Equations
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.
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.
mdt
dm
Solution:
,kmdt
dm where k is a
positive constant.
Let m be mass in mg and t time in days.
We could write where k is negative, but
most people prefer to emphasise the negative
gradient.
kmdt
dm
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.
mdt
dm
Solution:
,kmdt
dm where k is a
positive constant.
Let m be mass in mg and t time in days.
We can quote the solution:
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.
mdt
dm
Solution:
,kmdt
dm where k is a
positive constant.
Let m be mass in mg and t time in days.
We can quote the solution:Make sure you write t on the
r.h.s. !
Am tke
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.
mdt
dm
Solution:
,kmdt
dm where k is a
positive constant.
Let m be mass in mg and t time in days.
We can quote the solution:
10,0 mt 010 Ae 10 AAm tke
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.
mdt
dm
Solution:
,kmdt
dm where k is a
positive constant.
Let m be mass in mg and t time in days.
We can quote the solution:
)20(105 ke ke 2050 10,0 mt
5,20 mt
010 Ae 10 A
A log is just an index !
Am tke
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.
mdt
dm
Solution:
,kmdt
dm where k is a
positive constant.
Let m be mass in mg and t time in days.
We can quote the solution:
)20(105 ke ke 2050 50ln20 k
) d.p. ( 30350 k
10,0 mt
5,20 mt
010 Ae 10 AAm tke
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. We now have
tem 035010
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. 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 )
Setting up and Solving Differential Equations
tem 035010
The graph showing the decay function is
Mass at start of measurements
Time when mass is 1 mg
Setting up and Solving Differential Equations
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.
Setting up and Solving Differential Equations
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.
Setting up and Solving Differential Equations
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.
Setting up and Solving Differential Equations
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.
Setting up and Solving Differential Equations
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.)
Setting up and Solving Differential Equations
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
Setting up and Solving Differential Equations
You might meet differential equations that do not describe growth functions.
Setting up and Solving Differential Equations
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
Setting up and Solving Differential Equations
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
Setting up and Solving Differential Equations
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
Setting up and Solving Differential Equations
Cx
y 2
ln2
( 1, 1 ) on the curve:
2
1 C
2
12
x
ey
C2
11ln
2
1
2ln
2
x
ySo,
Setting up and Solving Differential Equations
There is one very well known situation which can be described by a differential equation.
The following is an example.
Setting up and Solving Differential Equations
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
Setting up and Solving Differential Equations
)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.
Setting up and Solving Differential Equations
Setting up and Solving Differential Equations
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.
Setting up and Solving Differential Equations
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 Equations
xdt
dx
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?
, 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.
Setting up and Solving Differential Equations
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
You may remember the solution to this equation but, if not, we can separate the variables to find it.
0200 Ae
)200(100 k
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.
mdt
dm
Solution:
,kmdt
dm where k is a
positive constant.
Let m be mass in mg and t time in days.
We can quote the solution:
tkeAm
)20(105 ke ke 2050 50ln20 k
) d.p. ( 30350 k
10,0 mt
5,20 mt
010 Ae 10 A
Setting up and Solving Differential Equations
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 )
Setting up and Solving Differential Equations
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.