SPS 2281 - Mathematical Methods
Lecture #8 - Applications of First-order Differential Equations
1. Linear Models
(a) Growth and Decay
(b) Half-life of Radioactive
(c) Carbon Dating
(d) Newton’s Law of Cooling / Warming
(e) Chemical Mixtures
(f) Series Circuit
2. Non-Linear Equations
(a) Logistics Equation
(b) Chemical Reactions
3. Systems of Differential Equations
(a) Radioactive Series
(b) Mixtures
(c) Predator-Prey Models
(d) Competition Models
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Non-Linear Models - Logistic Equation
A non-linear equation
dP
= P (a − bP )
dt
is known as logistic equation studied by P.F. Verhulst to predict the human populations for various countries.
If P (0) = P0, P0 6= a/b, we find that
c1 = P0/(a − bP0) so the final solution
will be
aP0
P (t) =
bP0 + (a − bP0) e−at
The solution for this equation is


b/a 
 1/a

+

 dP = dt
P
a − bP
1
1
ln |P | − ln |a − bP | = t + c
a
a P
= at + ac
ln a − bP P
= c1 eat
a − bP
ac1
P (t) =
bc1 + e−at
2
Non-Linear Models - Logistic Equation
Example - Logistic Growth
Suppose a student carrying a flu virus returns to an isolated college campus of 1000
students. If it is assumed that the rate at which the virus spreads is proportional
not only to the number x of infected students but also to the number of students
not infected, determine the number of infected students after 6 days if it is further
observed that after 4 days x(4) = 50.
Finding k from x(4) = 50, we get
Answer
Assumin no one leave the campus, the initial value
1000
50
=
problem is
1 + 999 e−4000k
1 19
−1000k = ln
4 999
dx
= −0.9906
= kx(1000 − x), x(0) = 1.
dt
so the equation becomes
Since a = 1000k and b = k, the solution from
previous slide
x(t) =
1000
1 + 999 e−0.9906t
for 6 days
x(t) =
1000
1 + 999 e−1000kt
x(6) =
3
1000
= 276 students
1 + 999 e−5.9436
Non-Linear Models - Chemical Reactions
Suppose that a grams of chemical A are combined with b grams of chemical B. If
there are M parts of A and N parts of B formed in the compound and X(t) is the
number of grams of chemical C formed, then the number of grams of chemical A and
the number of grams of chemical B remaining at time t are, respectively,
M
N
a−
X and b −
X.
M +N
M +N
The law of mass action states that when no temperature change is involved, the rate
at which the two substances react is proportional to the product of the amounts of A
and B that are untransformed (remaining) at time t:



dX 
M
N
∝ a −
X  b −
X 
dt
M +N
M +N
If k is constant of proportionality,
dX
= k(α − X)(β − X)
dt
where
α = a(M + N )/M
β = b(M + N )/N
4
Non-Linear Models - Chemical Reactions
Example - Second-Order Chemical Reaction
A compound C is formed when two chemicals A and B are combined. The resulting reaction between the two
chemicals is such that for each gram of A, 4 grams of B is used. It is observed that 30 grams of the compound C
is formed in 10 minutes.
Determine the amount of C at time t if the rate of the reaction is proportional to the amounts of A and B
remaining and if initially there are 50 grams of A and 32 grams of B.
How much of the compound C is present at 15 minutes?
Interpret the solution as t → ∞.
Integration gives
Answer
The amounts of A and B remaining at time t are
1
4
50 − X and 32 − X.
5
5
Compound C satisfy
!
!
1
4
dX
∝ 50 − X 32 − X
dt
5
5
using k,
dX
= k(250 − X)(40 − X), X(0) = 0, X(10) = 30.
dt
By separation of variables
1/210
1/210
−
dX +
dX = kdt
250 − X
40 − X
ln
250 − X
= 210kt + c
40 − X
250 − X
= c2 e210kt
40 − X
Using initial value, we get
1 − e−0.1258t
.
X(t) = 1000
25 − 4 e−0.1258t
When t → ∞, X → 40, leaving
1
4
50 − (40) = 42 g of A and 32 − (40) = 0 g of B
5
5
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Exercises
1. The number N (t) of supermarkets throughout the country that are using a computerized checkout system is described by the initial-value problem
dN
= N (1 − 0.0005N ), N (0) = 1.
dt
(a) Solve the initial value problem.
(b) How many companies are expected to adopt the new technology when t = 10?
2. Two chemicals A and B are combined to form a chemical C. The rate, or velocity,
of the reaction is proportional to the product of the instantaneous amounts of A
and B not converted to chemical C. Initially, there are 40 grams of A and 50
grams of B, and for each gram of B, 2 grams of A is used. It is observed that 10
grams of C is formed in 5 minutes.
(a) How much is formed in 20 minutes?
(b) What is the limiting amount of C after a long time?
(c) How much of chemicals A and B remains after a long time?
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Answers for Exercises
2000 et
1999+et
1.(a) N (t) =
(b) N (10) = 1834
2.(a) 29.3 g
(b) X → 60 as t → ∞
(c) 0 g of A and 30 g of B.
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