Chabot Mathematics
§9.3 ODE
Applications
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Review §
9.2
 Any QUESTIONS About
• §9.2 First Order, Linear, Ordinary
Differential Equations
 Any QUESTIONS
About HomeWork
• §9.2 → HW-14
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
§9.3 Learning Goals
 Use differential equations to model
applications involving public health,
orthogonal trajectories, and finance.
 Explore the predator-prey model
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 Consider a population of individuals
amidst an outbreak of some disease,
with fractions of the total population S
susceptible, I immune, and D diseased.
 One model for the spread of an
epidemic is that the rate of change in
the fraction of diseased individuals is
jointly proportional to the number
susceptible and diseased individuals.
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 (a) Supposing that the size of the population
and value of I are fixed, write a differential
equation modeling the change in the fraction
of diseased individuals over time.
 (b) If a constant 10% of the total population is
immune, initially 0.04% of the population has
the disease, and after one week 0.1% of the
population has the disease, find an equation
giving the fraction of the population that is
diseased after t days.
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 SOLUTION:
 (a) As with many modeling problems,
start by carefully translating the English
Words into mathematics:
 “the rate of change in the fraction of
diseased individuals is jointly
proportional to the fraction of susceptible
and diseased individuals”
dD
 k S  D
dt
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 Since EveryOne is Either Sick, Immune,
or Diseased then S, I, and D are
Percentages that must add up to 100%:
S  I  D  100%  1.00
 Then the fraction of susceptible
individuals “leftover” after immune and
diseased individuals are accounted for.
S  1.00  I  D  1  I  D
 Thus the
revised ODE
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dD
= k × (1- I - D)× D
dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 (b) With a constant 10% Immune, solve
an initial value problem for the
differential equation
dD
= k × (1- I - D)× D
dt
dD
 k  (1  0.1  D )  D
dt
 Note that this eqn is Variable Separable
dt
 dD k  0.9  D   D 



1
 dt
 0.9  D   D
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dD
 k  dt
0.9  D   D
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 Integrating the
1
dD   k  dt

Separated eqn: 0.9  D   D
 The complex AntiDerivative can be
accomplished using the Table of
Integrals #6 from Section 6.1:
1
D
ln
+ C2 = kt + C1
0.9 0.9 - D
 Use Algebra on the above Equation to
solve for D(t)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 Working to Isolate D(t)
1
D
ln
 C2  kt  C1 
0.9 0.9  D
1
D
ln
 kt  C1  C2 
0.9 0.9  D
 Letting C1  C2   C
 Find
D
ln
0.9  D
 0.9kt  C  e
ln
D
0.9  D
 e 0.9 kt C
 Remove ABS bars as expect D<0.9 (90%)
 
D
 e 0.9 kt C  eC  e 0.9 kt  eC  e 0.9 kt   A  e 0.9 kt  Ae0.9 kt
0.9  D
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 Working to D  0.9  D Ae0.9kt
Isolate D(t) D + DAe0.9kt = 0.9Ae0.9kt


0.9 kt
0
.
9
Ae
D 1  Ae 0.9 kt  0.9 Ae 0.9 kt  Dt  
1  Ae 0.9 kt
 Now Use Initial Condition: Dt  0  0.0004
 In the
0.9 Ae 0.9 k 0 0.9 Ae 0 0.9 A
D0  


0.0004
0.9 k 0
0
D(t) eqn
1  Ae
1  Ae
1 A
 Next, Solve the Above Eqn for Constant
Exponential PreFactor, A
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 Solving for A 0.9A = 0.0004
1+ A
0.9A - 0.0004A = 0.0004
0.0004
A=
» 0.000445
0.8996
 Use A in D(t)
0.9(0.000445)e0.9kt
0.0004e0.9kt
D(t) =
»
0.9kt
1+ (0.000445)e
1+ 0.000445e0.9kt
 Now Use the 2nd Temporal Condition
D7  0.1%  0.001
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 By 2nd Time-Based Condition
0.0004e0.9k(7)
D(7) =
= 0.001
0.9k(7)
1+ 0.000445e
0.0004e6.3k = 0.001(1+ 0.000445e6.3k )
0.0004e
6.3k
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- 0.000000445e
= 0.001
0.001
6.3k
e =
0.0003996
e6.3k » 2.5
1
k»
ln 2.5 » 0.1454
6.3
6.3k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Model Epidemic
 Use the Values of A & k to construct the
completed Function, D(t)
0.0004e0.1309t
Dt  
1  0.000445e 0.1309t
Fraction Diseased People (%)
MTH16 • Bruce Mayer, PE
80
70
60
50
40
30
20
10
0
MTH15 Quick Plot BlueGreenBkGnd 130911.m
0
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20
40
60
t (days)
80
100
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 The price of gasoline and the number of
purchased electric cars depend on one
another. Assume that the rate of change
in price of gasoline is a decreasing linear
function of the price of electric cars.
Similarly, the rate of change in the number
of electric cars purchased is an increasing
linear function of the price of gasoline.
 (a) Model the relationships in rates of
change as linked differential equations
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 (b) Say that gasoline increases by $1 per
year in the absence of electric cars and
the rate decreases by 5 cents for each
additional thousand electric cars that are
produced. Also, say that if gas were
(magically) priced at $0/gal, there would
be a growth rate of 3k ElecCars, with the
rate increasing by 4k ElecCars for each $1
increase in the gallon price of gas. Solve
the differential equations implicitly.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 SOLUTION:
 (a) Again Very CareFully Translate the
Word-Statement to Math Relations
 “the rate of change in price of gasoline
is a decreasing linear function of the
price of electric cars”
dG
 a  bC
dt
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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 Now construct the differential equation
for the change in the car price:
 “the rate of change in the number
of electric cars purchased is an
increasing linear function of the price of
gasoline”
dC
 d  fG
dt
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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 Thus have constructed TWO ODE’s in
for G(t) and C(t) with 4 unknown
constants: a, b, d, and f
dG
= a - bC
dt
dC
= d + fG
dt
 More translation is in order to find
values of the constants in the two
ODEs:
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 “gasoline increases by $1 per year in the
absence of electric cars and the rate
decreases by 5 cents for each additional
thousand NonGasoline cars produced”
dG
=1 - 0.05C
dt
dC
= 3 + 4G
dt
 Also
 “if gas were priced at $0/gal, there would
be a growth rate of 3k ElecCars, with the
rate increasing by 4k ElecCars for each
$1 increase in the gallon of gas price”
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 Instead of Finding G(t) and C(t)
determine an Implicit relation between
the two dependent variables
 Note that G depends on C, and C
depends on G; i.e. the Equations are
COUPLED
 Try one of dG 
dC
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dG
dC
dt
dt
dC
or
dt  dC
dG
dG
dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 Find dG/dC: dG  dG dt  1  0.05C
dC
3  4G
dC dt
 Separating the Variables
 dG 1  0.05C  3  4G   dC




3  4G 
 dC
1

3  4G   dG  1  0.05C   dC
 Finding the AntiDerivatives
 3  4G  dG   1  0.05C  dC
3G + 2G 2 + K1 = C - 0.025C 2 + K2
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Linked ODE’s
 The Final
3G  2G 2  K1  C  0.025C 2  K 2
G & C Relation:
 This relationship does not define a
function as it has Double-Values
• Can NOT isolate G or C
 However the eqn nevertheless is a
predictable relation between gasoline
price and electric car sales
• The graph in the CG plane is an ellipse
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
WhiteBoard Work
 Problems From §9.3
• P16: Atmospheric Pressure
• P28: Predator-Prey
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
All Done for Today
Predator
vs
Prey
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Wolves vs. Elk
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
2a
–
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BMayer@ChabotCollege.edu
2b
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
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BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx