CHAPTER 6:DIFFERENTIAL EQUATIONS AND

MATHEMATICAL MODELINGSECTION 6.5:

LOGISTIC GROWTH

AP CALCULUS AB

What you’ll learn about

How Populations GrowPartial FractionsThe Logistic Differential EquationLogistic Growth Models

… and whyPopulations in the real world tend to grow logistically over extended periods of time.

Section 6.5 – Logistic Growth

Partial Fraction Decomposition

1. If is an improper fraction

(degree of N(x) > degree of D(x), divide the denominator into the numerator to obtain

xDxN

xD

xN 1polynomial

xDxN

Section 6.5 – Logistic Growth

Example

112

43

4411

111

82

1324

4

43112

4

132

2

2

2

x

x

x

xx

xxx

xx

x

xx

Section 6.5 – Logistic Growth

Partial Fraction Decomposition (cont)2. If the degree of the numerator is smaller than

the degree of the denominator, factor the denominator into factors of the form

a) For each linear factor the partial fraction

decomposition would be

. and 2 nm cbxaxqpx

mm

qpx

A

qpx

A

qpx

A

...2

21

Section 6.5 – Logistic Growth

Example:

23

32

23

32

23

3223

23

7

6

72

xx

BAxBA

xx

BBxAAx

xx

xBxAx

B

x

A

xx

x

xx

x

2

1

3

2

6

7

2

11

1

55

7322

7312 1

732 and 1

2

xxxx

x

And

A

A

B

B

BB

BBBA

BABA

So

Section 6.5 – Logistic Growth

Partial Fraction Decomposition (cont)b) For each quadratic factor of the form

where is non-reducible, the partial fraction decomposition must include

mcbxax 2 cbxax 2

nnn

cbxax

CxB

cbxax

CxB

cbxax

CxB

222

222

11 ...

Partial Fraction Decomposition with Distinct Linear Denominators

( )

If ( ) , where and are polynomials with the ( )

degree of less than the degree of , and if ( ) can be written

as a product of distinct linear factors, then ( ) can be written

as a sum of r

P xf x P Q

Q x

P Q Q x

f x

ational functions with distinct linear denominators.

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational

4 1

functions with linear denominators.

x xf x

x x

2

2

6 8 4Since ( ) , we will find numbers A, B and C

2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.

S

x xf x

x x x

A B Cf x

x x xA x x B x x C x xA B C

x x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.

Setting -2 : (0) (-4)(-3) (0) 36, so 3.

Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .

2 2 1 2 2 1

x A B C A

x A B C B

x A B C C

x xf x

x x x x x x

Example Antidifferentiating with Partial Fractions

26 8 4

Find .2 2 1

x xdx

x x x

2

3 2

We know from the last example that

6 8 4 1 3 2

2 2 1 - 2 2 1

ln - 2 3ln 2 2ln 1

ln - 2 2 1

x xdx dx

x x x x x x

x x x C

x x x C

Link to Purple Math Partial Fractions

Stop and do partial fraction drill sheet

Logistic Differential Equation

Exponential growth can be modeled by the differential equation

for some 0.

If we want the growth rate to approach zero as approaches a

maximal carrying capacity , we can introduce a limitin

dPkP k

dtP

M

g factor

of - : .

This is the .

dPM P kP M P

dt

logistic differential equation

Example Logistic Differential Equation

The growth rate of a population of bears in a newly established

wildlife preserve is modeled by the differential equation

0.008 100 - , where is measured in years.

a. What is the carrying capac

P

dPP P t

dt

ity for bears in this wildlife preserve?

b. What is the bear population when the population is growing the fastest?

c. What is the rate of change of the population when it is growing the fastest?

a. The carrying capacity is 100 bears.

b. The bear population is growing the fastest when it is half

the carrying capacity, 50 bears.

c. When 50, 0.008 50 100 50 20 bears per year.dP

Pdt

The General Logistic Formula

-

The solution of the general logistic differential equation

is

1

where is a constant determined by an appropriate

initial condition. The and

Mk t

dPkP M P

dt

MP

AeA

M

carrying capacity the

are positive constants.kgrowth constant