Chabot Mathematics
§9.1 ODE
Models
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Review §
8.3
 Any QUESTIONS About
• §8.3 → TrigonoMetric Applications
 Any QUESTIONS
About
HomeWork
• §8.3 → HW-12
“TriAnguLation
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
§9.1 Learning Goals
 Solve “variable separable” differential
equations and initial value problems
 Construct and use mathematical models
involving differential equations
 Explore learning
and population
models, including
exponential and
logistic growth
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
ReCall Mathematical Modeling
1. DEVELOP MATH EQUATIONS that
represent some RealWorld Process
•
Almost always involves some simplifying
ASSUMPTIONS
2. SOLVE the Math Equations for the
quanty/quantities of Interest
3. INTERPRET the Solution
– Does it MATCH the
RealWorld Results?
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Differential Equations
 A DIFFERENTIAL EQUATION is ANY
equation that includes at least ONE
calculus-type derivative
• ReCall that Derivatives are themselves the
ratio “differentials” such as dy/dx or dy/dt
 TWO Types of Differential Equations
• ORDINARY (ODE) → Exactly ONE-Each
INdependent & Dependent Variable
• PARTIAL (PDE) → Multiple Independent
Variables
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Differential Equation
 ODE Examples
dy

 y  t2
dt


d 2z
2 dz
x

4
1

z

z

e
dx
dx 2
d2y
dy
m 2  c  ky  cost 
dt
dt
d
L
 g sin   at  cos   0
dt
• ODEs Covered in MTH16
 PDE’s
1   2 Pv  1 Pv
r

2
r r  r  Dv t
 u   v   w




x
y
z
t
• PDEs NOT covered in MTH16
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Terms of the (ODE) Trade
 a SOLUTION to an ODE is a
FUNCTION that makes BOTH SIDES of
the Original ODE TRUE at same time
 A GENERAL Solution is a
Characterization of a Family of
Solutions
• Sometimes called the Complementary
Solution
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Terms of the (ODE) Trade
 ODEs coupled with side conditions are
called
• Initial Value Problems (IVP) for a temporal
(time-based) independent variable
• Boundary Value Problems (BVP) for a
spatial (distance-based) independent
variable
 a Solution that the satisfies the
complementary eqn and side-condition
is called the Particular Solution
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Develop Model
 After being implanted in a mouse, the
growth rate in volume of a human colon
cell over time is proportional to the
difference between a maximum size M
and the cell’s current volume V
 Write a differential equation in terms of
V, M, t, and/or a constant of
proportionality that expresses this rate
of change mathematically.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Develop Model
 SOLUTION:
 Translate the Problem Statement
Phrase-by-Phrase
“…the growth rate in volume of a human
colon cell over time is proportional to
the difference between a maximum size
M and the cell’s current volume V…”
 Build the ODE
dV
= k×(M - V )
Math Model
dt
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Separation of Variables
 The form of a “Variable Separable”
Ordinary Differential Equation
dy h x 

or
dx g  y 
dy u t 

dt v y 
 Find The General Solution by
SEPARATING THE VARIABLES and
Integrating
 g  y dy   hx dx
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or
 v y dy   ut dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Solve Mouse ODE
 Consider the differential equation for
cell growth constructed previously.
• The colon cell’s maximum volume is 14
cubic millimeters
• The cell’sits current volume is 0.5 cubic
millimeters
• Six days later the cell has volume
increases 4 cubic millimeters.
 Find the Particular Solution matching
the above criteria.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Solve Mouse ODE
 SOLUTION:
 ReCall the ODE dV  k M  V 
dt
Math Model
 From the Problem Statement,
3
the Maximum Volume  M  14 mm 
 Using M = 14 in the ODE State the
Initial Value Problem as
dV
= k (14 -V )
dt
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With
3


V t  0  0.5 mm
TimeBased
3


V
t

6

4
mm
Values
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Solve Mouse ODE
 The ODE is separable, so isolate
factors that can be integrated with
respect to V and those that can be
integrated with respect to t
dV
 k 14  V 
dt
dt
 dV k 14  V 

 dt 

1
 14  V 
 Then the Variable-Separated Equation
dV
= k dt
14 -V
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Solve Mouse ODE
 Integrate
Both Sides
ans Solve
dV
ò 14 -V =
òk
dt
-ln 14 -V + C1 = kt +C2
ln (14 -V ) = -kt +C, where C = -(C2 - C1 )
14 -V = e
-kt+C
V t   14  Ae  kt , where A  eC
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Solve Mouse ODE
 At this Point have 2 Unknowns: A & k
 Use the Given Time-Points (initial
values) to Generate Two Equations in
Two Unknowns
 Using V(0) = 0.5 mm3
0.5 =14 - Ae-k(0)
0.5 =14 - A
A =13.5
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Solve Mouse ODE
 Now use the other Time Point: V 6  4 mm 3
4 =14 -13.5e-k(6)
-10
= e-6k
-13.5
ln(20 / 27) = -6k
k   1 6  ln(20 / 27)  0.05
 Thus the particular solution for the
volume of the cell after t days is


V t   14 mm 3  13.5 mm 3 e
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0.05
t
day
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Verify ODE Solution
 Verify ODE↔Solution Pair
dB
• ODE
 1 2t B
dt
t t 2
• Solution Bt   2e
 Take Derivative of Proposed Solution
(
)
dB d
t-t 2
=
2e
dt dt
t-t 2 d
= 2e × ( t - t 2 )
dt
= 2e
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t-t 2
× (1- 2t )
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Verify ODE Solution
 Sub into ODE the dB/dt relation
dB
 1 2t B
dt
2e
t t 2
dB
 1  2t  
dt
 Which by Transitive Property Suggests
1  2t B  2e
t t 2
 1  2t 
 Thus by Calculus and
t t 2
Algebra on the ODE B  2e
 Which IS the ProPosed Solution for B
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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
WhiteBoard Work
 Problems From §9.1
• P52 → Work Efficiency
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
All Done for Today
GolfBall
FLOW
Separation
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx