Modeling First Order O.D.E’s
1- Rate of Growth and Decay
The I.V.P. (1)
! dx
= kx
#
" dt
#$x(t 0 ) = x 0
where k is a constant, occurs in many physical sciences models that involve growth or
decay.
For example, in biology, it is often observed that the rate at which certain bacteria grow is
proportional to the number of bacteria present at any time.
Over short intervals of time the population of small animals, such as rodents, can be
predicted fairly accurately by the solution of the I.P.V. (1).
The constant k can be determined from the solution of the differential equation by using a
subsequent measurement of the population at time t > t0.
In physics I.V.P.(1) provides a model for approximating the remaining amount of a
substance which is disintegrating through radioactivity.
Also I.V.P. (1) determines the temperature in a cooling body. In chemistry the amount of
a substance remaining during certain reaction is also described by I.V.P. (1).
Example:
1) A culture initially has N0 number of bacteria. At time t = 1 hour the number of bacteria
is measured to be 3/2 N0.
If the rate of growth is proportional to the number of bacteria present at any time,
determine the time necessary for the number of bacteria to triple.
Solution:
Let’s write the I.V.P. that models the problem:
! dN
= kN
#
" dt
#$N(0) = N 0
First solve the first order equation
dN
= kN
dt
it is separable, and can be expressed as
dN
= kt
N
integrating, we get
ln N = kt + c, where N > 0
or
N(t) = cekt
Using the initial condition: N(0) = N0 = ce0 = c
then
N(t) = N0ekt
Since at time t = 1 hour the population is 3/2N0,
then
N(1) = 3/2 N0 = N0ek
or
3/2 = ek
and solving for k,
k = ln(3/2) = 0.4055
Therefore,
N(t) = N 0 e 0.4055t
Now, we must find the time t when the number of bacteria triple the original number.
3N 0 = N 0 e 0.4055t
!!!!!3 = e 0.4055t
taking natural logarithm both side
ln 3 = 0.4055t
ln 3
t=
= 2.71!hours
0.4055
Remark: The function ekt increases as t increases if k > 0 and decreases as t increases if
k < 0. Thus, problems describing growth, such as population, bacteria, or even capital
are characterized by a positive value of k. Meanwhile, problems involving decay, as in
radioactive disintegration will yield a negative k value.
2) A breeder reactor converts the relatively stable uranium 238 into isotope plutonium
239. After 15 years it is determined that 0.043% of the initial amount A0 has
disintegrated.
Find the time t it takes for one-half of the atoms in the initial amount A0 to disintegrate
(half-life).
Solution:
A(t) is the amount of plutonium remaining at time t.
So we can create the I.V.P.
! dA
= kA
#
" dt
#$ A(0) = A 0
The solution of the separable equation is
A(t) = A0ekt,
If 0.043% of the atoms of A0 have disintegrated, then 99.957% of A0 remains.
To find k, we solve
0.99957A0 = A0 e15k
15k = ln (0.99957)
ln(0.99957)
k=
= !0.0000286
15
Hence,
A(t) = A 0 e !0.0000286t
Half-life
1
A 0 = A 0 e !0.0000287t
2
1
= e !0.0000287t
2
" 1%
!0.0000287t = ln $ ' = ! ln 2
# 2&
t=
!0.693
= 24,180!years
!0.0000287
2- Cooling:
Newton’s law of cooling states that the rate at which the temperature T(t) changes in a
cooling body is proportional to the difference between the temperature in the body and
the constant temperature T0 of the surrounding medium.
dT
= k ( T ! T0 )
dt
where k is the constant of proportionality.
Example:
When a cake is removed from a baking oven its temperature is measured at 300ºF. Three
minutes later its temperature is 200ºF. How long will it take to cool off to room
temperature of 70ºF?
Solution:
The problem can be described by the I.V.P.
" dT
$ = k ( T ! 70 )
# dt
$% T(0) = 300
We solve the separable equation
dT
= kdt
T ! 70
ln T ! 70 = kt + d
T ! 70 =!ce kt
T(t) = 70 + ce kt
Using the initial condition when t = 0, T = 300, we get
300 = 70 + c or c = 230
then
T(t) = 300 + 230 ekt
Since
then
e 3k
T(3) = 200,
200 ! 70 130
1 " 13 %
=
=
!!or!!k = ln $ ' = !0.19018
230
230
3 # 23 &
T(t) = 70 + 230e !0.19018t
Therefore,
We want to determine how long it will take to reach room temperature.
Notice that lim T(t) = 70 , so the equation 70 = T(t) does not have a solution.
t!"
Intuitively we expect that the cake will reach room temperature after a reasonably period
of time. How long?
Let’s create the following table
t (min) T(t)
20.1
75º
21.3
74º
22.8
73º
24.9
72º
28.6
71º
32.3
70º
The table shows it will reach room temperature in about 30 minutes.
3- A Mixture Problem
At time t = 0 a tank contains Q0 lb of salt dissolved in 100 gal of water; see figure.
Assume that water containing ¼ lb of salt /gal is entering the tank at a rate of r gal/min
and that the well-stirred mixture is draining from the tank at the same rate.
Set up an I.V.P. that describes this flow process. Find the amount of salt Q(t) in the tank
at any time. Find the limiting amount QL that is present after a long period of time.
Solution:
We assume that salt is neither created nor destroyed in the tank. Therefore variations in
the amount of salt are due solely to the flows in and out of the tank. More precisely the
rate of change of salt in the tank dQ/dt, is equal to the rate at which salt is flowing in
minus the rate at which it is flowing out. In symbols,
dQ
= rate!in!!!rate!out
dt
The rate at which salt enters the tank is the concentration ¼ lb/gal times the flow rate
r gal/min or r/4 lb/min.
To find the rate at which salt leaves the tank, we need to multiply the concentration of
salt in the tank by the rate of outflow r gal/min.
Since the rate of flow in and out are equal, the volume of water in the tank remains
constant at 100 gal, and since the mixture is “well-stirred,” the concentration throughout
the tank is the same, namely, [Q(t)/ 100] lb/gal.
Therefore, the rate at which salt leaves the tank is [r Q(t)/100] lb/min. Thus the
differential equation governing the process is
dQ r rQ
= !
dt 4 100
the initial condition is
Q(0) = Q0
The equation can be written as
dQ
r
r
+
Q=
dt 100
4
it is a linear equation in the dependent variable Q and independent variable t, where
r
P(t) = r/100. The integrating factor is e
So
r
! 100 dt
r
t
= e 100 .
r
r
t rQ
t r
dQ
! e 100
= e 100
dt
100
4
r
r
t r
d " 100 t %
$ e Q ' = e 100
dt $#
4
'&
int egrating
e 100
t
r
t
100
e Q
=
(
r
t
100
e
!
r
t
100
Q(t) = 25 + ce
Using the initial condition, we get
C = Q0 – 25
Then
Q(t) = 25 + ( Q 0 ! 25 ) e
!
r
t
r
dt = 25e 100 + c
4
r
t
100
r %
r
"
!
t
!
t
100
100
Q(t) = 25 $ 1 ! e
' + Q0e
#
&
From this equation, we see than when t increases without bound, then Q(t)  25, so the
limiting value QL = 25.
Assume Q0 = 50, and r = 3, find the time t after which salt level is 60% of Q0.
Then,
Q(t) = 25 + 25e !0.03t
Since 60% of 50 is 30 , we want to find t such that Q(t) = 30
30 = 25 + 25e !0.03t
5 = 25e !0.03t
1
= e !0.03t
5
" 1%
ln $ ' = !0.003t
# 5&
" 1%
ln $ '
# 5&
t=
= 53.6!min
!0.03
Remark: If the mixed brine solution is pumped out of the tank at a rate faster or slower
than the rate at which is pumped in, then the equation is different from the one in the
previous example.
Example:
Initially 50 lb of salt is dissolved in a large tank holding 300 gal of water. A brine
solution is pumped into the tank at a rate of 5 gal/min, and a well-stirred solution is then
pumped out at a slower rate of 3 gal/min. If the concentration of the solution entering the
tank is 2 lb/gal, determine the amount of salt in the tank at any time. How much salt is at
t = 50 min?
Solution:
The rate at which the salt enters the tank is: IN = (5 gal/min)(2 lb/gal) = 10 lb/min
The solution is accumulating at a rate of (5 – 3) gal/min = 2 gal/min.
After t min, there are 300 + 2t gal of brine in the tank.
The rate at which salt is leaving the tank is:
3Q(t)
! Q(t)
$
lb / gal& =
lb / min
OUT = (3 gal/min) #
" 300 + 2t
% 300 + 2t
The differential equation governing the process is:
dQ
3Q(t)
= IN ! OUT = 10 !
dt
300 + 2t
dQ
3
+
Q = 10
dt 300 + 2t
It is a linear equation with P(t) = 3/(300 + 2t), so the integrating factor is
3
e
! 300+2t dt
= e 2 ln( 300+2t ) = ( 300 + 2t ) 2
3
3
3
(300 + 2t) 2
3
3
dQ
2
+ (300 + 2t) 2
Q = (300 + 2t) 2 10
dt
300 + 2t
3
3
d!
2
# = (300 + 2t) 2 10
(300
+
2t)
Q(t)
$
dt "
int egrating
5
3
2
(300 + 2t) Q(t) = % 10(300 + 2t) dt = 5
3
2
(300 + 2t) 2
5
2
+c
Q(t) = 2(300 + 2t) + c(300 + 2t)& 2
Using the initial condition Q(0) = 50, 50 = 2(300) + c (300)-3/2, c = -2.8x106.
3
Q(t) = 2(300 + 2t) + (!2.8 " 10 6 )(300 + 2t)! 2
When t = 50, Q(50) = 2(300+100) – 2800000(300+100)-3/2 = 450 lb
3