More Applications of Definite Integrals 329Exponentiation produces
P
K−P
=e(kt+C^1 ) ⇒P
K−P
=ekt·eC^1 ⇒P
K−P
=C 2 ekt.Solving for P yields P = C 2 ekt(K − P) ⇒ P = C 2 ektK − C 2 ektP ⇒ P +
C 2 ektP =C 2 ektK ⇒ P(1+C 2 ekt)=C 2 ektK ⇒ P =
C 2 ekK
1 +C 2 ekt
. Dividing numer-
ator and denominator byC 2 ekt, P(t)=
C 2 ektK
1 +C 2 ekt
=
K
1
C 2 ekt+ 1
=
( K
1
C 2)
e−kt+ 1.Att=0, P 0 =
K
1
C 2
+ 1
. Solving forC 2 yields P 0
(
1
C 2+ 1
)
=K ⇒1 +C 2
C 2
=
K
P 0
⇒
P 0 +P 0 C 2 =Kc 2 ⇒ P 0 =Kc 2 −P 0 C 2 orC 2 =P 0
K−P 0
. LetA=
1
C 2
, and the solutionof this logistic differential equation with initial conditionP(0)=P 0 isP(t)=
K
Ae−kt+ 1,
whereK is the carrying capacity andA=
K−P 0
P 0
.
Example 1
The population of the United Kingdom was 57.1 million in 2001 and 60.6 million in 2006.
Find a logistic model for the growth of the population, assuming a carrying capacity of 100
million. Use the model to predict the population in 2020.
Step 1: Since the carrying capacity isK=100,
dP
dt
=kP(
1 −P
100
)
.Step 2: The solution of the differential equation, ifA=
k−P 0
P 0
=
100 − 57. 1
57. 1
≈.7513, isP(t)=100
Ae−kt+ 1orP(t)=100
. 7513 e−kt+ 1
.
Step 3: Take 2006 ast=5, P(5)= 60 .6. Then 60. 6 =
100
. 7513 e−k(5)+ 1
. Solving gives
k≈ 0 .0289 soP(t)=100
. 7513 e−^0.^0289 t+ 1
.
Step 4: Since the year 2020 corresponds tot =19, Substitute and evaluate P(19)=
100
. 7513 e−^0 .0289(19)+ 1
≈ 69 .742. The population of the United Kingdom in 2020 is
predicted to be approximately 69.742 million.
Example 2
The spread of an infectious disease can often be modeled by a logistic equation, with the
total exposed population as the carrying capacity. In a community of 2000 individuals, the
first case of a new virus is diagnosed on March 31, and by April 10, there are 500 individuals
infected. Write a differential equation that models the rate at which the virus spread through
the community, and determine when 98% of the population will have contracted the
virus.