Chabot Mathematics
§9.2 1st
Order ODEs
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Review §
9.1
 Any QUESTIONS About
• §9.1 → Variable Separable Ordinary
Differential Equations
 Any QUESTIONS
About HomeWork
• §9.1 → HW-13
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
§9.2 Learning Goals
 Solve first-order linear differential
equations and
• Initial Value Problems (IVP)
• Boundary Value Problems (BVP)
 Explore compartmental
analysis with applications
to finance, drug
administration, and
dilution models.
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
FirstOrder, Linear ODE
 The General form of a First Order,
Linear Ordinary Differential Equation
dy
 pt   y  qt 
dt
IVP
dy
 px   y  qx 
dx
BVP
 Solve the General Equation with
Integrating Factor
p u du

• Let I u   e

1
• Then the ODE
yu  
I u   qu   du  C

Solution
I u 
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx

Quick Example
dy
 Find Solution to ODE:
+y=x
dx
 The Integrating
1 dx
ò
x
I(x) = e
=e
Factor →
 Thus the
1é
x
y = x ë ò xe dx + C ùû
Solution
e
1é x x
= x ë xe - e + Cùû
e
 x  1  Ce
Chabot College Mathematics
5
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Solve dy

 Find the Particular dy
yt

t

Solution for ODE: dt
t 2 1
 Subject to Initial Value: yt  0  1.000
 SOLUTION:
 Note that this Eqn is NOT Variable
Separable, so ReWrite in General Form
dy
 p (u ) y  q (u )
du
dy
t
 2
y  t
dt t  1
pt 
Chabot College Mathematics
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
dt  t  yt t  1
2
qt 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Solve dy

p ( t ) dt
 Then the Integrating
 t 1 dt

I t   e
e
Factor:
 Now Let u  t 2  1
du
du
 t  dt
 Then  2t 
dt
2
 Using u and du in integrating Factor
t
2
I t   e
t dt
 t 2 1
e

du 2
u
e
1 du
2 u

e
1
ln u
2
e
1
ln t 2 1
2
 Now t2+1 is always positive so:
I t   e
1
ln t 2 1
2
Chabot College Mathematics
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e

1
ln t 2 1
2

e 
ln t 1
2
  e ln
1
2

dt  t  yt t  1
2
t 2 1
 t 2 1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Solve dy


dt  t  yt t  1
2
 Using this Integrating Factor find:
y


1
1
t2
t 2  1  t  dt  C

 Using u and du from before
y


1
1
t2
Chabot College Mathematics
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

1 
 du 
t  1  t  dt   C 
 u   2   C 
u
 

1 1


u
du

C

u  2 

2
1
 12
u
1 3/ 2

3 u  C


Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Solve dy


dt  t  yt t  1
2
 Then the General Solution by Back
SubStitution
1
y  12
u

 
 ReCall the
y0  1.00
Initial Condition (IC)
 Using IC 1 = 1 0 2 +1 + C 02 +1 -1/2
(
)
(
)
3
in Solution
2
C
3
Chabot College Mathematics
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
1 2
1 2
1 3/ 2
 1 1
1 2
2
 3 u  C   3 u  Cu  3 t  1  C t  1


Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Solve dy

 Finally the Full General Solution
-1/2
1 2
2 2
y = ( t +1) + ( t +1) .
3
3
MTH16 • Bruce Mayer, PE
8
7
y = f(t)
6
5
4
3
2
1
MTH15 Quick Plot BlueGreenBkGnd 130911.m
0
Chabot College Mathematics
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1
2
3
t

dt  t  yt t  1
2
4
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
I(x) → How Does it Work?
 Multiplication of Both Sides of the ODE
by I(x) changes ODE appearance
 dy

I x     p x   y  q x 
 dx

dy
I x    I x   px  y  I x   qx 
dx
 For Solution This must be of the form
d
I x   y   I x  qx 
dx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
I(x) → How Does it Work?
 So that by the PRODUCT Rule
d
dy dI  x 
I x  y   I x  
y
dx
dx
dx
 ReCall the I(x) multiplied ODE L.H.S.
dy
I x    I x   px  y  I x   qx 
dx
 Thus by Correspondence need
dI  x 
I x   px  
dx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
I(x) → How Does it Work?
 Then by Substitution
dy
dy dI x 
d
I x    I  x   p x  y  I x   
 y  I  x   y 
dx
dx
dx
dx
 Then the I(x) multiplied ODE
dy
d
d I  x   y 
I x    I x   p x  y  I x   qx   I x   y  
dx
dx
dx
 Which is VARIABLE SEPARABLE
d I x   y 
 d I x   y  I x   qx  dx
 I x   qx   


dx
1
 dx
 1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
I(x) → How Does it Work?
 Or
d I x   y   I x   qx   dx 
 1 d I x  y    I x  qx  dx
 Then Let: u  I x y
 Using u in the Variable Separated ODE
1 d u    I x  qx  dx
 BackSubbing for u
 u  C1   I x   qx   dx
u  C1  I x   y  C1   I x   qx   dx 
 Let −C1 = +C

I x   qx   dx  C

y
1
y
I x   qx   dx  C

I x 
Chabot College Mathematics
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I x 

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
1
No Need for Memorization


 Do Need to
1
y
I x   qx   dx  C

Memorize
I x 
 Only need to find a good I(x) to multiply
the ODE so that by the PRODUCT Rule
the L.H.S.:
dy
d I x   y 
 dy

I x     px   y   I x    I x   px  y 
dx
dx
 dx

 Then can Separate the Variables and
Integrate  1 d I x   y    I x   qx   dx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Key to Integrating Factor
 Need
dI  x 
I x   px  
dx
 Then ln I x    px   dx  e
ln I  x 
e
 p  x dx
• Assumes, withOUT loss of generality, that
the Constant of Integration is Zero
 So Finally the Integrating-Factor Formula
I x   e
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 p  x dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Key to Integrating Factor
dI  x 
 For Solution Need: I x   px  
dx
 Next Integrate this ODE
dI x 
1
 px   dx  
 dI x    px   dx
I x 
I x 
 Then ln I x    px   dx  elnI  x   e  p  x dx
• Assumes, withOUT loss of generality, that
the Constant of Integration is Zero
 So Finally the
Integrating-Factor
Formula
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I x   e
 p  x dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Dilution over Time
 A 60-gallon barrel containing 20 gallons
of simple syrup at 1:1 sugar-to-water
ratio is being stirred and filled with pure
sugar at a rate of 1 gallon per minute.
Unfortunately, a crack in the bottom of
the barrel is leaking solution at a rate
of 4 oz per minute.
 After how long will there be 40 gallons
of Pure Sugar in the barrel?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Dilution over Time
 SOLUTION:
 First to set up an equation to model the
quantity of sugar in the barrel over time,
 Next solve this eqn and find the time at
which the desired quantity occurs.
 A general Mass Balance for a “Control
Volume”
• Storage Rate = InFlow
InFlow − OutFlow
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Storage OutFlow
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Dilution over Time
 The Pure Sugar Mass Balance Statement
æ rate of
ö æ rate of ö æ rate of ö
ç
÷ ç
÷ ç
÷
ç change in ÷ = çsugar ÷ - çsugar ÷
ç
÷ ç
÷ ç
÷
èsugar amt. ø è inflow ø è outflowø
 The Model above accounts for modeling
the change in pure-sugar quantity, the
inflow is 1 Gallon per Minute (1 gpm) or
128 oz per minute.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Dilution over Time
 The outflow is of the mixed solution, so
it leaks at a rate of 4 oz/min, with total
quantity of sugar Q(t) and total quantity
of solution equal to:
V s 20 gal   inflow of solution   outflow of solution
 20 128 128t  4t
 So the concentration of
OutFlowing Syrup:
Chabot College Mathematics
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Q(t )
20  124t
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx

Example  Dilution over Time
 Now we express the differential equation
for the rate of change in sugar quantity:
æ rate of
ö æ rate of ö æ rate of ö
ç
÷ ç
÷ ç
÷
ç change in ÷ = çsugar ÷ - çsugar ÷
ç
÷ ç
÷ ç
÷
sugar
amt.
inflow
outflow
è
ø è
ø è
ø
dQ
Q
 128  4 
dt
20  124t
 This ODE is first-order and linear, so it
can solved using the general strategy.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Dilution over Time
 Calculate the Integrating dQ
4

Q  128
Factor for the ODE
dt 20  124t
I(t) = e
4
dt
ò 20+124t
=e
1
31 ln 5+31t
 (5  31t )
1
31
 Then the form of the solution
1
1
31
é
ù
y=
×128
dt
+
C
1 ë ò (5+ 31t)
û
(5+ 31t) 31
é128 31
ù
32
1
31
y=
× (5+ 31t) + Cú
1 ê
û
(5+ 31t) 31 ë 31 32
C
y  20  124t 
1
(5  31t ) 31
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Dilution over Time
 Use the IC to find the Constant Value
• Initially there is a 1:1 ratio of water to
sugar, so exactly half of the 20 gallons, or
10 gallons (1280 oz), is sugar. Use this
Data-Point to find the value of C:
C
y = 20 +124t +
1
(5 + 31t) 31
C
1280  20  124  0 
5  31 0
1
31
C  1260  5  1327.14
1
31
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Dilution over Time
 Finally, find the time at which there are
40 gallons of sugar in the barrel, which
happens when y = 40*128 = 5120 oz.
1327.14
y = 20 +124t +
1
(5 + 31t) 31
1327.14
5120 = 20 +124t +
1
(5+ 31t) 31
1327.14
0  5100  124t 
1
(5  31t ) 31
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Example  Dilution over Time
 This a transcendental (NonAlgebraic)
eqn for which there is NO exact solution
 Solve using the MuPAD Computer
Algebra System (CAS):
t  32.568
 In other words, after about 32.6 minutes
of pouring and mixing, there will be 20
gallons of pure sugar in the barrel.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
MuPAD Calculation
 tsoln := 124*t - 5100 +
1327.14/(5+31*t)^(1/31)
 numeric::solve(tsoln)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
WhiteBoard Work
 Problems From §9.2
• P51
Glacier Ice
Removal
Rate
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
All Done for Today
Linear
st
1 Order
ODEs
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Bruce Mayer, PE
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
–
Chabot College Mathematics
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BMayer@ChabotCollege.edu
2b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-14_Sp14_sec_9-2_1st_Linear_ODEs.pptx