Transcript File

Setting up and Solving
Differential Equations
Setting up and Solving Differential Equations
We 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
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?
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
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?
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.
dx
 x 
dt
dx
 kx , where k is a constant.
dt
Setting up and Solving Differential Equations
You may remember the solution to this equation but, if not,
we can separate the variables to find it.


dx
1
 kx 
dx  k dt
dt
x
 ln x  kt  C
kt
x

Ae
    (1 )

We were given “ A solution initially contains 200 bacteria . . .
“ and “. . . the rate of increase of the number is initially 100
per hour”
There are 2 pairs of conditions here which enable us to solve
for 2 unknowns.
Setting up and Solving Differential Equations
You may remember the solution to this equation but, if not,
we can separate the variables to find it.


dx
1
 kx 
dx  k dt
dt
x
 ln x  kt  C
kt
x

Ae
    (1 )

We were given “ A solution initially contains 200 bacteria . . .
“ and “. . . the rate of increase of the number is initially 100
per hour”
t  0 , x  200 :
200  Ae 0 
A  200
Setting up and Solving Differential Equations
You may remember the solution to this equation but, if not,
we can separate the variables to find it.


dx
1
 kx 
dx  k dt
dt
x
 ln x  kt  C
kt
x

Ae
    (1 )

We were given “ A solution initially contains 200 bacteria . . .
“ and “. . . the rate of increase of the number is initially 100
per hour”
t  0 , x  200 :
dx
t  0,
 100 :
dt
200  Ae 0  A  200
dx
 kx  100  k ( 200 )  k  0  5
dt
Setting up and Solving Differential Equations
You may remember the solution to this equation but, if not,
we can separate the variables to find it.


dx
1
 kx 
dx  k dt
dt
x
 ln x  kt  C
kt
x

Ae
    (1 )

We were given “ A solution initially contains 200 bacteria . . .
“ and “. . . the rate of increase of the number is initially 100
per hour”
t  0 , x  200 :
dx
t  0,
 100 :
dt
Substituting in (1):
200  Ae 0 
A  200
dx
 kx  100  k ( 200 )  k  0  5
dt
Setting up and Solving Differential Equations
You may remember the solution to this equation but, if not,
we can separate the variables to find it.


dx
1
 kx 
dx  k dt
dt
x
 ln x  kt  C
kt
x

Ae
    (1 )

We were given “ A solution initially contains 200 bacteria . . .
“ and “. . . the rate of increase of the number is initially 100
per hour”
t  0 , x  200 :
dx
t  0,
 100 :
dt
Substituting in (1):
t2

200  Ae 0 
A  200
dx
 kx  100  k ( 200 )  k  0  5
dt
05 t
x  200 e
x  544 ( nearest integer )
Setting up and Solving Differential Equations
The graph showing the growth function is
Number after 2
hours
Number at start of
measurements
x  200 e
05 t
Setting up and Solving Differential Equations
e.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 Equations
e.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.
dm
dm
m 
  km , where k is a positive
dt
dt
constant.
dm
We could write
 km where k is negative, but most
dt
people prefer to emphasise the negative gradient.
Setting up and Solving Differential Equations
e.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.
dm
dm
m 
  km ,
dt
dt
We can quote the solution:
where k is a positive
constant.
Setting up and Solving Differential Equations
e.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.
dm
dm
m 
  km , where k is a positive
dt
dt
constant.
kt
We can quote the solution:
mAe
Make sure you write t on the r.h.s. !
Setting up and Solving Differential Equations
e.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.
dm
dm
m 
  km , where k is a positive
dt
dt
constant.
kt
We can quote the solution:
mAe
0

10

Ae
 A  10
t  0 , m  10
Setting up and Solving Differential Equations
e.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.
dm
dm
m 
  km , where k is a positive
dt
dt
constant.
kt
We can quote the solution:
mAe
0

10

Ae
 A  10
t  0 , m  10
 20 k
 k ( 2 0)

0

5

e
t  20 , m  5 
5  10 e
A log is just an index !
Setting up and Solving Differential Equations
e.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.
dm
dm
m 
  km , where k is a positive
dt
dt
constant.
kt
We can quote the solution:
mAe
0

10

Ae
 A  10
t  0 , m  10
 20 k
 k ( 2 0)

0

5

e
t  20 , m  5 
5  10 e
  20 k  ln 0  5
 k  0  035 ( 3 d.p. )
Setting up and Solving Differential Equations
e.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
m  10 e
 0035 t
Setting up and Solving Differential Equations
e.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
m 1




m  10 e
1  10 e
 0035 t
 0035 t
 0035 t
0 1 e
 0  035 t  ln 0  1
ln 0  1
t
 66 ( nearest integer )
 0  035
It takes 66 days to decay to 1 mg.
Setting up and Solving Differential Equations
The graph showing the decay function is
m  10 e
Mass at start of
measurements
 0035 t
Time when mass is 1
mg
Setting up and Solving Differential Equations
SUMMARY
 The words “ a rate proportional to . . . ” followed by the
quantity the rate refers to, gives the differential
equation for growth or decay.
e.g. “ the number, x, increases at a rate proportional to x ”
dx
dx
gives
dt
 x

dt
 kx
kt
 The solution to the above equation is
x  Ae
( but if we forget it, we can easily separate the variables
in the differential equation and solve ).
 The values of A and k are found by substituting
either 1 pair of values of x and t and 1 pair of values of
and t,
dx
dt
or 2 pairs of values of x and t.
Setting up and Solving Differential Equations
Exercise
For 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
Exercise
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
dn to the nearest
your answer
solving, use k
1.
 k n ( Whenhundred.
correct to 3 s.f. )
dt
2. A patient is receiving drug treatment. When first
measured, there is 0  5 mg of the drug per litre of
blood. After 4 hours, there is only 0  1 mg per litre.
Assuming the amount in the blood at time t is
decreasing in proportion to the amount present at time
dx it takes for
t, find how2.long
there solving,
to be only
use0k 05 mg.
  k x ( When
Give the answer
minute.
correct
to 3 s.f. )
dt to the nearest
Setting up and Solving Differential Equations
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.
dn
kt
 kn  n  A e
dt
Let t = 0 in 1990.
Solution:
A  60000
t  0 , n  60000 
t  10 , n  63000  63000  60000e k (10)
 63000 

 10k  ln 
 60000 
k  0  00488

Setting up and Solving Differential Equations
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.
n  A e , A  60000 , k  0  00488
kt
000488 t
 n  60000 e
000488 ( 20 )
t  20  n  60000 e
 66 , 200 ( nearest hundred )
Setting up and Solving Differential Equations
2.
A patient is receiving drug treatment. When first
measured, there is 0  5 mg of the drug per litre of
blood. After 4 hours, there is only 0  1 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 0  05 mg.
Give the answer to the nearest minute.
Solution:
dx
 kt
  k x  x  Ae
dt
t  0, x  0  5  A  0  5
 k (4)
t  4 , x  0  1  0  1  0  5e
 0 1 
  4k  ln 
  k  0  402 (3 s.f.)
 05
Setting up and Solving Differential Equations
2.
A patient is receiving drug treatment. When first
measured, there is 0  5 mg of the drug per litre of
blood. After 4 hours, there is only 0  1 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 0  05 mg.
Give the answer to the nearest minute.
x  Ae
 kt
Substitute
 0402 t
A

0

5
,
k  0  402  x  0  5 e
,
x  0  05 : 


 0402 t
0  05  0  5 e
 0  05 
 0  402 t  ln 

 05 
t  5  728
Ans: 5 hrs 44 mins
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
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:
dy
 y2
dx
The equation is
y
1
Separating the
variables:

2
dy 

dx
1
 xC
y
This is the general solution to the equation
Setting up and Solving Differential Equations
1
 xC
y
( 1, 1 ) lies on the curve

So,
or,
The equation is
1 1  C
1
 x2
y
1

 y
x2
1
y
x2
 C  2
Setting up and Solving Differential Equations
Exercise
1. 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 y  f ( x ) .
Solution:


dy
 xy
dx
1
dy  x dx
y


x2
ln y 
C
2
Setting up and Solving Differential Equations
x2
ln y 
C
2
( 1, 1 ) on the curve:
So,
1
 ln 1   C 
2
2
x
1
ln y 

2
2

ye
x 2 1
2
1
C
2
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
The temperature of a cup of coffee is given by x  C
at time t minutes after it was poured. The temperature of
the room in which the cup is placed is 20  C
Explain what the following equation is describing:
dx
  k ( x  20 )
dt
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.
This is an example of Newton’s law of cooling.
Setting up and Solving Differential Equations
dx
  k ( x  20 )
dt
We can solve this equation as follows:

1
dx 
x  20

 kdt
ln x  20   kt  C
x  20  Ae  kt
x  20  Ae  kt
If we are given further information, we can complete the
solution as in the other examples.