Differential Equations and Mathematical Modeling

Chapter 6

6.1

Quick Review

3

2

3 ₃

Determine whether or not the function satisfies the differential equation.

1. 3

2. csc 5 cot 5 3 1

3.

x

x

y

dy _{y y e}

dx

dy _{x y x x}

dx

dy _y

y dx

 

     

Quick Review

4. 4 2 and 5 when 1 5. x tan and 2 when 0

y x x C y x

y e x C y x

     

Quick Review Solutions

3

2

Determine whether or not the function satisfies the differential equation.

1. 3 2. csc 5 cot 5 3

Yes

Yes

1

3.

x

y

dy _{y y e}

dx

dy _{x y x x}

dx dy

y dx

 

     

Quick Review Solutions

4

0

Let 5 and 1

5 4( 1) 2( 1)

5

Find the constant C.

4. 4 2 and 5 when 1

5. tan and 2 when

4 2 3

Let 2 and 0

2 tan(0)

2

3

0

1

x

y x x C y x

y e x C y x

What you’ll learn about



Differential Equations



Slope Fields



Euler’s Method

… and why

Differential equations have been a prime

Differential Equation

An equation involving a derivative is called a

differential equation

. The

order of a

Example

Solving a Differential Equation

2

Find all functions that satisfy y dy 3x cos .x dx  

2 3

The solution can be any antiderivative of 3 cos , which can be any function of the form sin .

x x

y x x C



First-order Differential Equation

If the general solution to a first-order differential

equation is continuous, the only additional

information needed to find a unique solution is

the value of the function at a single point, called

an

initial condition

. A differential equation

with an initial condition is called an

initial-value

problem

. It has a unique solution, called the

Example

Solving an Initial Value

Problem

2

Find the particular solution to the equation 3

1 whose graph passes through the point 1, .

2

x

dy

e x dx  

      2 2 2 2

2 2 2

1 3

The general solution is .

2 2

Applying the initial condition, we have

1 1 3

, from which we conclude that

2 2 2

1

2 . Therefore, the particular equation is 2

1 3 1

2 .

2 2 2

x

x

y e x C

e C

C e

y e x e

  

  

 

Example

Using the Fundamental Theorem to

Solve an Initial Value Problem

2

Find the solution to the differential equation '( ) cos for which (3) 5.

f x _x _ f

 

 

 

The solution is ( ) xcos 5.

Slope Field

Example

Constructing a Slope Field

Construct a slope field for the differential equation

cos .

dy

x dx 













The slope at any point 0, will be cos 0 1. The slope at any point

will be 0. 2

The slope is 1 along the lines 2 .

, or , will be -1.

The slope at all odd multiples of

y

x

y y











Euler’s Method for Graphing a Solution to

an Initial Value Problem

5. To construct the graph moving to the left from the initial point, repeat the process using negative values for .

3. Increase by . Increase by , wherex x y y

2. Use the differential equation to find the slope / at the point. 1. Begin at the point ( , ) specified by the initial condition.x y

dy dx

that lies along the linearization.

4. Using this new point, return to step 2. Repeating the process constructs the graph to the right of the initial point.

y dy dx x

( / ) . This defines a new point (x x, y y)

x

Example

Applying Euler’s Method

Use Euler's Method with increments of 0.1 to approximate the value of when 1.5

given 1 and 2 when 1.

x

y x

dy x y x

dx

  

   

(1, 2) 2 0.1 0.2 (1.1, 2.2)

(1.1, 2.2) 2.1 0.1 0.21 (1.2, 2.41)

(1.2, 2.41) 2.2 0.1 0.22 (1.3, 2.63)

(1.3, 2.63) 2.3 0.1 0.23 (1.4, 2.86)

(1.4, 2.86) 2.4 0.1 0.24 (1.5, 3.1) 1

dy x dx  



x y

,



_

_x

dx

 

 _{ }

6.2

Quick Review

2 2 0

1. Evaluate the definite integral.

x dx















t 2

3 2

2

Find given 2. 2

3. 2

4. cos 1 5. ln tan

x

dy dx

y dt

y x x

y x

y x







 

 

Quick Review Solutions







 















1 8 8

0

3 3 3

1. Evaluate the definite integral.

Find given

2. 2

3. 2

2

3 2 4 1

2 sin 1

1 ta

4. cos 1

5. ln tan

n x x x dx dy dx y dt

y x x

y

x

dy dx dy

x x x

dx dy x x dx dy dx x y x x                    







What you’ll learn about



Indefinite Integrals



Leibniz Notation and Antiderivatives



Substitution in Indefinite Integrals



Substitution in Definite Integrals

… and why

Indefinite Integral

 

The family of all antiderivatives of a function ( ) is the and is denoted

by .

f x

f x dx



indefinite integral of with respect to f x

 

 

 

 

If is any function such that ' , then

, where is an arbitrary constant

called the .

F F x f x

f x dx F x C C





 

Example

Evaluating an Indefinite

Integral

Evaluate 2_ xcosxdx.

2

2xcosxdx x sinx C

Properties of Indefinite Integrals





( ) ( ) for any constant

( ) ( ) ( ) ( )

kf x dx k f x dx k

f x g x dx f x dx g x dx

 

  



Power Functions

1

1

when 1 1

1

ln

n

n u

u du C n

n

u du du u C u







 

   



Trigonometric Formulas

2 2

cos sin sin cos

sec tan csc cot

sec tan sec csc cot csc

udu u C udu u C

udu u C udu u C

u udu u C u udu u C

 

 

 

    

    

Exponential and Logarithmic Formulas

ln

ln ln

ln ln

log

ln ln

u u

u u

a

e du e C

a

a du C

a

udu u u u C

u u u u

udu du C

a a







 

 

 

  



Example

Paying Attention to the

Differential

2 3

Let ( ) 1 and let . Find each

of the following antiderivatives in terms of . a. ( )

b. ( ) c. ( )

f x x u x

x f x dx

f u du f u dx

     

















3 3 9 3

2

2 3

6 7

2 1

a. ( ) 1

3 1

b. ( ) 1

3

1 1

3 3

c. ( ) 1 1

1 1

7

f x dx x dx x x C f u du u du u u C

x x C x x C

f u dx u dx x dx x dx x x C

Example

Using Substitution

3

2

Evaluate x e dxx .

 3 3 3 2 2 2

Let . Then 3 , from which we conclude that 1

. We rewrite the integral and proceed as follows 3 1 x 3 1 3 1 3 x u u x du

u x x

dx du x dx

e dx e du

Example

Using Substitution

2

Evaluate 6_ x 1 x dx.





2 2

3 2

3 2 ₂

Let 1 . Then 2 . Rewrite the integral in terms of :

6 1 3

2 3

3 2 1

u x du xdx u

x x dx udu

u C

x C

 

  

 

 

 _{ } 

 

Example

Setting Up a Substitution with a

Trigonometric Identity

3

Evaluate sin_ xdx.













3 2

2 2 3

3

let cos and - sin

sin sin sin

1 cos sin 1

3

cos cos

3

u x du xdx

xdx x xdx

x xdx u du

u

u C

x

x C

 

 

 



 

  

   

Example

Evaluating a Definite Integral

by Substitution

2 0 ₂

Evaluate .

9 x dx x  





0 ₂ 9

5 9

1 ln 2

Let 9 and 2 . Then (0) 0 9 9 and (2) 2 9 5. So,

1

9 2

= 1

ln 5 ln 9 2

1 5

ln

2 9

u

u x du xdx u

6.3

Quick Review

1

1 3

1. Find given tan 3 .

2. Solve for in terms of given tan 3 . 3. Solve the differential equation .

4. Solve the initial value problem cos , (0) 3.

x

dy

y x

dx

x y y x

dy e dx

dy

x x

dx y











 

Quick Review Solutions

1. Find given tan 3 .

2. Solve for in terms of given tan 3 .

3. Solve the differential equation .

4. Solve the initial value

3 1 9 1 tan 3 1 robl 3 p e x x dy dx x dy x y y e y x dx

x y y x

d e C y dx           2

m cos ,

(0) 3. 1 sin 3 2

dy

x x

d

y

y x x

x

 

 



What you’ll learn about



Product Rule in Integral Form



Solving for the Unknown Integral



Tabular Integration



Inverse Trigonometric and Logarithmic Functions

… and why

Integration by Parts Formula

udv uv vdu

Example

Using Integration by Parts

Evaluate sin_ x xdx.

Use - with

sin -cos

sin - cos cos - cos sin

udv uv vdu

u x dv xdx

du dx v x

x xdx x x xdx

x x x C

 

 



 

 

 

Example

Repeated Use of Integration by

Parts

2

Evaluate 2x e dxx .







Let 2 4

2 2 4

Now let

2 2 4

2 4 4

x

x

x x x

x

x

x x x x

x x x

u x dv e dx du xdx v e

x e dx x e xe dx u x dv e dx du dx v e

x e dx x e xe e dx x e xe e C

Example

Solving for the Unknown

Integral

Evaluate ex sin xdx.



Let sin -cos

sin cos cos

Now let cos sin

sin cos sin sin

2 sin cos sin

sin

x

x

x x x

x

x

x x x x

x x x

x

u e dv xdx du e dx v x

e xdx e x e xdx u e dv xdx du e dx v x

e xdx e x e x e xdx e xdx e x e x

e xdx                         cos sin 2 x x

e x e x

C

 

Example

Antidifferentiating ln

x

Find ln_ xdx.

 

Let ln 1

1

ln ln

ln ln

-u x dv dx

du dx v x x

xdx x x x dx x x x dx x x x C

 



 

 

 

 _{ }

  

Quick Quiz

Sections 6.1-6.3

Solve the following problems without using a graphing calculator.

1. Which of the following differential equations would produce the slope field shown below?

(A) 3

(B)

3 (C)

3 (D)

dy

y x dx

dy x

y dx

dy x

y dx

 

 

 

3 (E)

3

dy x

y dx

dy x x dx

 

Quick Quiz

Sections 6.1-6.3

Solve the following problems without using a graphing calculator.

1. Which of the following differential equations would produce the slope field shown below?

(A) 3

(B)

3 (C)

3 (D)

dy

y x dx

dy x

y dx

dy x

y dx

 

 

 

3

dy x

Quick Quiz

Sections 6.1-6.3

1/2 0

1/ 2 2 0

2 1/ 2 0

/ 4 2 0

/ 4 2 0

/ 6 2 0

2. If the substitution sin is made in the integrand

of , the resulting integral is 1

(A) sin sin (B) 2

cos (C) 2 sin (D) sin (E) 2 sin

Quick Quiz

Sections 6.1-6.3

1/2 0

1/ 2 2 0

2 1/ 2 0

/ 4 2 0

/

/ 4 2

0

0

6 2

2. If the substitution sin is made in the integrand

of , the resulting integral is 1

(A) sin sin (B) 2

cos

6.4

Quick Review

1. Rewrite the equation in exponential form or logarithmic form.

a. ln b.

2. Solve for . a. ln( 2) 3 b. 2 10 c. 1.5 2

3. Solve for given ln( 1) 2 - 3.

c

x

x

a b

e d

x x

e

y y x

 

 

 

Quick Review Solutions

3

ln

1. Rewrite the equation in exponential form or logarithmic form.

a. ln b. 2. Solve for .

a. ln( 2) 3 b. 2 10 c. 1.5 2

2 ln 5 ln 2 n l c x x b e a d c x a b e d x x e x e x             2 -3 3. Solve for given ln( 1) 2 - 3.

1.5

1

-x

y e

What you’ll learn about



Separable Differential Equations



Law of Exponential Change



Continuously Compounded Interest



Modeling Growth with Other Bases



Newton’s Law of Cooling

… and why

Understanding the differential equation

gives us new insight into exponential growth and decay.

dy ky

Separable Differential Equation

   

 

 

A differential equation of the form is

called . We by writing it

in the form

1

.

The solution is found by antidifferentiating each s

dy

f y g x dx

dy g x dx

f y





separable separate the variables

Example

Solving by Separation of

Variables

2 2

Solve for if y dy x y and y 3 when x 0.

dx   

2 2 2 2 2 2 3 1 3 1 3 3

Apply the initial conditions to find C.

1 1 3

So, and

3 3 3 1

This solution is valid for the continuous section

dy

x y dx

y dy x dx y dy x dx

x

y C

x

C y y

The Law of Exponential Change

If changes at a rate proportional to the amount present (that is, if ), and if when 0, then

The constant is the if 0 or the

O

kt O

y

dy

ky y y t

dt

y y e

k k

  





Continuously Compounded Interest

If the interest is added continuously at a rate proportional to the amount in the account, you can model the growth of the account with the initial value problem:

Differential equation: Ini

dA

rA dt 

tial condition: (0)

The amount of money in the account after years at an annual interest rate :

( ) .

O

rt O

A A

t r

A t A e



Example

Compounding Interest

Continuously

Suppose you deposit $500 in an account that pays 5.3% annual interest. How much will you have 4 years later if

the interest is compounded continuously? compounded monthly?

(a) (b)

  

   0.053 4

12 4

Let 500 and 0.053. a. (4) 500 618.07

0.053

b. (4) 500 1 617.79 12

O

A r

A e

A

 

 

 

 _  _ 

Example

Finding Half-Life

-Find the half-life of a radioactive substance with decay equation .

kt O

y  y e

-1 The half-life is the solution to the equation .

2 1

Solve algebraically

2 1 - ln

2

1 1 ln 2 - ln

2 Note: The value

kt

O O

kt

y e y

e

kt

t

k k

 





 

is the half-life of the element. It depends only on the value of .

Half-life

The of a radioactive substance with rate constant ( 0) is

ln 2 half-life .

k k

k





Newton’s Law of Cooling

The rate at which an object's temerature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium.

If is the temperaT





ture of the object at time , and is the surrounding temperature, then

. (1)

Since ( - ), rewrite (1)

S

S

S

t T

dT

k T T dt

dT d T T

d d

  







Its solution, by the law of exponential change, is

S S

T T k T T

Example

Using Newton’s Law of Cooling

A temperature probe is removed from a cup of coffee and placed in water that has a temperature of T = 4.5 C.

Temperature readings T, as recorded in the table below, are taken after 2 sec, 5 sec, and every 5 sec thereafter.

Estimate

(a) the coffee's temperature at the time the temperature probe was removed. (b) the time when the temperature

probe reading will be 8 C.

o

S

Example

Using Newton’s Law of Cooling













According to Newton's Law of Cooling, - , where 4.5 and is the temperature of the coffee at 0. Use exponential regression to find that - 4.5 61.66 0.9277 is a model for the , - ,

kt

S O S

S O

t

S

T T T T e

T T t

T t T T t

      





















4.5 data. Thus, 4.5 61.66 0.9277 is a model of the , data.

(a) At time 0 the temperature was 4.5 61.66 0.9277 66.16 C. (b) The figure below shows the graphs of 8 and

4.5 61.66 0.9277

t

t

T

T t T

is a model for the





 4.5  61.66 0.9277

Example

Using Newton’s Law of Cooling





Use exponential regression to find that

According to Newton's Law of Cooling, T  T =





e

where TS O = 4.5 and T is the temperature of the coffee at t  0.

t

S

T



T  4.5 + 61.66 0.9277 is a model of the









(b) The figure below shows the graphs of y 8 and y  T  4.5 + 61.66





(a) At time t  0 the temperature was

6.5

Quick Review

2 3 2

1. Use polynomial division to write the rational ( )

function in the form ( ) , where the degree ( )

of is less than the degree of . a.

1 3 b.

1

R x Q x

D x

R D

x x x x x



  

Quick Review

-0.1

x x

30 2. Let ( ) .

1 5

a. Find where ( ) is continuous. b. Find lim ( ).

c. Find lim ( ).

d. Find the y-intercept of the graph of .

e. Find all horizontal asymptotes of the graph of .

x

f x

e f x f x f x

f

f





Quick Review Solution

2

3

2

1. Use polynomial division to write the rational ( )

function in the form ( ) , where the degree ( )

of is less than the degree of .

a. 2 2

1 3 b. 1 1 R x Q x D x R D x x x x x x x x x         





2. Let ( ) .

1 5

a. Find where ( ) is continuous. b. Find lim ( ).

c. Find lim ( ).

d. Find the y-intercept of the graph of .

3 1 , 30 0 5

e. Find all horizontal asym

x x f x e f x f x f x f        

What you’ll learn about



How Populations Grow



Partial Fractions



The Logistic Differential Equation



Logistic Growth Models

… and why

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 x

f x P Q

Q x

P Q Q x

f x



Example

Finding a Partial Fraction

Decomposition



 



2

2

6 8 4

Write the function ( ) as a sum of rational 4 1

functions with linear denominators. x x f x x x     



 

 





 





 





 





 

 





 





 





 



6 8 4

Since ( ) , we will find numbers A, B and C 2 2 1

so that ( ) . - 2 2 1

2 1 2 1 2 2

Note that ,

- 2 2 1 2 2 1

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

x x f x

x x x

A B C

f x

x x x

A x x B x x C x x

A B C

x x x x x x

A x x B x x C x x x x

                                     

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.

x A B C A

x A B C B

x A B C C

    

    

Example

Antidifferentiating with Partial

Fractions



 

 



2

6 8 4

Find .

2 2 1

x x

dx

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 x

dx dx

x x x x x x

x x x C

x x x C

    _   _

   _   _

     

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

dP

kP k

dt

P M

 





g factor

of - : .

This is the .

dP

M P kP M P

dt  

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

dP

P 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 P 50, dP 0.008 50 100 50 20 bears per year.

dt

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

dP

kP M P dt

M P

Ae A

M

 

 

carrying capacity the

are positive constants.k

Quick Quiz

Sections 6.4 and 6.5

You may use a graphing calculator to solve the following problems.

1. The rate at which acreage is being consumed by a plot of kudzu is proportional to the number of acres already consumed at time .

If

t

there are 2 acres consumed when 1 and 3 acres consumed when 5, how many acres will be consumed when 8?

(A) 3.750 (B) 4.000 (C) 4.066 (D) 4.132 (E) 4.600

t

t t



Quick Quiz

Sections 6.4 and 6.5

You may use a graphing calculator to solve the following problems.

1. The rate at which acreage is being consumed by a plot of kudzu is proportional to the number of acres already consumed at time . If

t

there are 2 acres consumed when 1 and 3 acres consumed when 5, how many acres will be consumed when 8?

(A) 3.

(C) 4.066

750 (B) 4.000

(D) 4.132 (E) 4.600

t

t t



Quick Quiz

Sections 6.4 and 6.5

 

2. Let ( ) be an antiderivative of cos . If (1) 0, then (5)

(A) -0.099 (B) -0.153 (C) -0.293 (D) -0.992 (E) -1.833

F x x F

F

Quick Quiz

Sections 6.4 and 6.5

 

2. Let ( ) be an antiderivative of cos . If (1) 0, then (5)

(A) -0.099 (B) -0.153

(D) -0.992 (E) -1.8

(C) -0.293

33

F x x F

F

Quick Quiz

Sections 6.4 and 6.5



 





 





 





 



(A) ln

4 3

1 3

(B) ln

4 1

1

(C) ln 1 3

2

1 2 2

(D) ln

2 1 3

(E) ln 1 3

dx x x x C x x C x

x x C

x

C

x x

x x C

Quick Quiz

Sections 6.4 and 6.5



 





 





 





 



(B) ln

4 1

1

(C) ln 1 3

2

1 2 2

(D) ln

2 1 3

(E

1 1

) ln 1 3

(A) ln

4 3 dx x x x C x

x x C

x C x x C x x

x x C

Chapter Test





1. Evaluate . 5

2. Evaluate sin . 25

3. Evaluate . 25

4. Solve the initial value problem analytically: 1 , (0) 0.1.

5. Find an integral equation ( ) such that sin

x tdt t e xdx dx x dy

y y y dx

dy

y f t dt x

dx        

  and 5

when 4.

x

a y

x

 

Chapter Test

6. The intensity ( ) of light feet beneath the surface of the ocean satisfies the differential equation , where is a constant. As a diver you know from experience that diving to 18 ft in

L x x

dL

kL k

dx  

the Caribbean Sea cuts the intensity in half. You cannot work without

artificial light when the intensity falls below a tenth of the surface value. About how deep can you expect to work without artificial light?

Chapter Test Solutions









5

2. Evaluate sin . 25

3. Evaluate . 25

4. Solve the initial value problem analyticall

1 ln 5 2 3sinx cos 10 10 5 5 ln

y: 1 , (

5 0 2 ) x x t C x e C x C x tdt t e xdx dx x dy

y y y dx       _  _       _      -3 3 4 0.1.

5. Find an integral equation ( ) such that sin and

1 1 9

5 when 4. sin 5

x

x

x

a

dy

y f t dt x y

dx

y

x

e

y _ tdt

Chapter Test Solutions

6. The intensity ( ) of light feet beneath the surface of the ocean satisfies the differential equation , where is a constant. As a diver you know from experience that diving to 18 ft in

L x x

dL

kL k

dx  

the Caribbean Sea cuts the intensity in half. You cannot work without

artificial light when the intensity falls below a tenth of the surface value. About how deep can you expect to work without artificial light?

7. Find the amount of time required for $10,000 to double if the 6.3% annual interest is compounded annually,

about 59.8 feet

a. about 11.3 years

continuously.