Chapter 2-Differential Equation

7/30/2019 Chapter 2-Differential Equation

1/9

10/19/20

CHAPTER 2

DIFFERENTIALEQUATION1

2.1 INTRODUCTION TODIFFERENTIAL EQUATION

OBJECTIVES : At the end of the lesson, students will be

able to

1) define differential equations

2) understand degree, order and solution

3) define differential equations of separable

variable type

4) solve differential equations of separable variable

type2

Introduction

A simple type of differential equation is

We integrate it to produce the required solution.

In many practical situations such as population

growth, radioactive decay, chemical mixture,

temperature cooling, velocity and acceleration

problem and etc. we have much more

interesting differential equations

32

xdx

dy

3

1) Definitions

2) Differential equations with

separable variables

3) Linear first-order differential

equations

4

Definition 1 A differential equation (DE) is one which

relates an independent or dependent variables

with one or more derivatives.

Examples of differential equations:

a) dxxdy 32

b) 523 x

dx

dy

c) xydx

dysin2

2

d)0743

2

2

dx

dy

dx

yd

5

Definition 2

Orderis the highest derivative in a differential

equation.

Degree is the highest power of the highest

derivative which occurs in a differential

equation.

Examples: a) 823

3

2

2

yxdx

dy

dx

yd

This DE has order 2 (the highest derivative appearing is the second

derivative) and degree 1 (the power of the highest derivative is 1.)6


2/9

10/19/20

b)

This DE has order 1 (the highest derivative appearing

is the first derivative) and degree 5 (the power of the

highest derivative is 5.)

yxxdx

dysinsin32

5

35'2" 74 yyy

This DE has order 2 (the highest derivative appearing

is the second derivative) and degree 4 (the power of

the highest derivative is 4.)

c)

7

Definition 3 General solution The general solution

of a differential equation contains an

arbitrary constant c.

Particular solution - The particular

solution of a differential equation

contains a specified initial value and

containing no constant.8

Differential Equations with

Separable Variables

Definition 4

Differential equations with separable variables Differential

equations in which the variables can be algebraically separated.

Some differential equations can be solved by the method of

separation of variables. It is the technique used when integrals were

antiderivatives. A differential equation is separable if it can be

written in the form 0 dyyNdxxMThe method for solving separable differential equations is to

separate the variables so that the yN is with dy and the xMis with the dx, where yN is a function ofy only and xM

is a function of x only. Once we can write it in the above form, all

we do is integrate throughout, to obtain our general solution. 9

Below are examples ofgeneral solution for the

differential equations.

Example 1:

, where and

This is already in the required form, so we

simply integrate:

, c is constant

032 dxxdyy 3xxM

2yyN

cdxxdyy 32

cxy

43

43

10

In this chapter we only deal with first

order, first degree differential equations.

A solution for a differential equation is a

function whose elements and derivatives

may be substituted into the differential

equation. There are two types of solution

for differential equations.11

Example 2: 011

dx

dyyxxy

First we must separate the variables:

011 dyyxdxxy Multiply throughout by dx

01

1

y

dyyxdxx Divide throughout byy

0

11

y

dyy

x

dxxDivide throughout byx

12


3/9

10/19/20

This gives us: 01

11

1

dy

ydx

x

where

xxM

11 and

y

yN1

1

We now integrate,

constantis

11

11

c,cx

ylnyx

cxlnylnyx

cylnyxlnx

cdyy

dxx

13

Example 3 21 y

dx

dye

x


dxydye x 21 Multiply throughout by dx

dx

e

ydy

x

21

Divide throughout by xe

dx

edy

yx

1

1

12

Divide throughout by 21 y

14

This gives us: dxedyy x 21

We now integrate:

constantis,

1

1

11

1

1

1

2

ccey

cey

dxedyy

x

x

x

15

2.2 Separable Variables and

Integrating Factors

Example 1:

Solve the differential equation

012

dx

dyyexx

y


cdyyedxxx

dyyedxxxy

y

101

2

2

Consider dxxx 12 : 16

By using substitution,

Let 12 xt x

dtdx

xdx

dt

2

2

cx

ct

ct

dtt

x

dttxdxxx

2

32

2

3

2

3

2

1

2

13

13

1

2

32

1

2

1

21Thus,

17

Consider :

By using integration by parts,

Let

dyye y

yu

dydudy

du

1

and yedy

dv

y

y

y

ev

dyedv

dyedv

1

Thus,

ye

eye

dyeye

vduuvdyye

y

yy

yy

y

18


4/9

10/19/20

So the general solution is:

23

21

3

1x 01 cyey

19

Example 2

Solve the differential equation ,


2 yxdx

dy

xdxy

dy

2

We now integrate: xdx

y

dy

2

cx

y 2

2ln2

20

We now use the information which means

at , to find c.

, gives

0x 5y

c2

025ln

2

3lnc

So the particular solution is:

3ln22ln2

xy

21

Separable Variables and Integrating Factors

Examples of DE with separable variables. These

equations can easily be solved after we separate the

variables

xydxdyxx

ydx

dyx

xydx

dy

11c)

1b)

1xa)

2

22

Examples of DE with non-separable variables.

These equations cannot be solved by separating

the variables, because the variables are unseparable.

These are called linear first-order DE.

xydx

dy

xydx

dyx

xx

y

dx

dy

3c)

02b)

1a)

23

Linear First-Order Differential Equations

A linear first-order differential equation is

one of the form .

The solution must be on an interval where

both and are continuous. The

usual solution to the differential equation

is to change it to an exact equation by

means of an integrating factor. This

integrating factor is

xQyxPdx

dy

xP xQ

.xV24


5/9

10/19/20

By multiplying both sides of the equation by the integrating

factorV(x), you change the left-hand side of the equation into

the derivative of the product V(x)y:

xQxVyxVxPdx

dyxV

Once we have chosen V(x), (we shall do so in a moment) and

carried out the multiplication ,

xVxQxVyxVdx

d

xQxVyxVxPdx

dyxV

ofChoice

we can solve the above equation by integrating both sideswith respect to x.

25

To find , we find the functions ofx that satisfies

the equations xV

ln

dintergrateEquation

separatedVariables

cancelledThe

cancelledtermsThe

sderivativeforRuleProduct

dxxPCexV

dxxPxV

dxxPxV

xdV

dxxPxV

xdV

yxVxPdx

xdV

dx

dyxVyxVxP

dx

xdVy

yxVxP

dx

dyxV

dx

xdVy

dx

dyxV

yxVxPdx

dyxVyxV

dx

d

26

Since we do not need the most general function V(x), we

may take Cto be 1 and we get dxxPexV

.

By multiplying both sides of the equation by the integrating

factor

dxxPexV , you obtain the exact differential

equation:

xQeyxPe

dx

dye

dxxPdxxPdxxP

xQe

dx

yeddxxP

dxxP

dxxQeyeddxxPdxxP

To solve this differential equation, integrate both sides andsolve for y. 27

Example 1:

Solve

This is already in the required form

with and

The integrating factor is

0 ydx

dy

xQyxPdx

dy 1xP 0xQ

dxxP

exV dx

e

Thus the integrating factor is xe . Multiplying bothsides of the equation by xe

28

0dxyed

x

0yed x

dxyed x 0

cyex

xe

cy

gives the solution: 0 yedx

dye

xx

29

Example 2:

Solve1

12

x

yxdx

dy

This is already in the required form xQyxPdx

dy

with x

xP2

and 1

1

x

xQ

.

The integrating factor is dx

xexV2

xe

ln22ln xe2x

30


6/9

10/19/20

Multiply by to both sides of the equation gives2x

1

22

2

x

xx

x

y

dx

dy

1

22

2

x

xyx

dx

dyx

1

22

x

x

dx

yxd

dxx

xyxd

1

22

dxxx

yxd1

2

2

31

Consider divide the denominator into

the numerator to obtain

dxxx

1

2

dxx

x

1

11

Let

dx

xxdx

xx

111

1

2

Thus

dxx

xyxd1

112

cxxx

yx 1ln2

22

32

Solve differential equations using integrating

factors (particular solutions)

Example 1:

Solve and that when .23 xydx

dyx

5

2y 1x

First we change the equation to the required form:

xQyxPdx

dy by multiplying it throughout by dx.

xx

y

dx

dy

3with

xxP

3 and xxQ

.

The integrating factor is dx

xexV

3xe ln3

3lnxe

3x33

Multiply by to both sides of the equation

gives

3x

333 xxxx

y

dx

dy

423 3 xyxdx

dyx

43x

dx

yxd

dxxyxd 43

dxxyxd43

cx

yx 5

53

34

We now use the information which means

at , to find c.1x5

2y

c

5

1

5

21

53 , gives

5

1c


5

1

5

53

xyx

35

Example 2

Solve 20,costan 2 yxyxdx

dy

xxQxxPyxdxdy

22 cosandtanx withcostan

The integrating factor is dxxPexV xdxe tan

x

e

e

e

e

x

x

dxx

x

dxx

x

sec

cos

1ln

cosln

cos

sin

cos

sin

36


7/9

10/19/20

Multiplying both sides of the equation by gives :xsec

x

cxy

cxyx

xdxyxd

xdxyxd

xdxxyxd

xxdx

yxd

xxyxxdx

dyx

sec

sin

sinsec

cossec

cossec

cossecsec

cossecsec

cossectansecsec

2

2

2

37

We now use the information which means at ,

to find c.

0x

2y

0sec

0sin2

c , gives 2c


x

xy

sec

2sin

xx cos2sin

38

Applications of Differential Equations Now we will apply the methods in this section to the solution

of some practical situations.

A) population growth model

The simplest growth model has a constant relative growth

rate. If we denote the population we are considering by ,

then the rate of change of the population is . To say that

the rate of change is proportional to the population is just

saying that there is a constant of proportionality ksuch that

ty

dt

dy

39

kydt

dy

Since kis constant, this can be immediately separated and

integrated to yieldkdt

y

dy

kdtydy

ckty lncktey ckt

eey ktAey

40

Example 1:

Assume that the population of the earth changes

at a rate proportional to the current population.

a) Write the differential equation satisfied bythe population

b) In 1650 the population is estimated to havebeen about 600 million and in 1950 about 2800million. By solving the differential equation andfitting this data, estimate the population at timet(years AD).

c) Using this fitted solution, and assuming thatthe greatest population the earth can support ispeople, in what year will this limit be reached?

41

Solution

a)

b)

42


8/9

10/19/20

43

c)

Radioactive decay models, on the other hand, are very

accurate over long periods of time. They are theprimary method for determining age of prehistoric

fossils and ancient artifacts. If we denote the decay we

are considering by , then the decreasing rate of

the decay is . To say that the decreasing rate is

proportional to the decay is just saying that there is a

constant of proportionality ksuch that

tC

B) Radioactive decay models

dt

dC

44

Since kis constant, this can beimmediately separated and integrated toyield

kCdt

dC

kdtC

dC

kdtC

dC

cktC lnckteC

ckt

eeC ktAeC

45

The rate of decay for a radioactive substance is

proportional to the remaining concentration. Write

down the differential equation satisfied by the decay.

If one third of the substance still remains within 12

years, how much left is still available after 3 years.

Solution:

concentration of radioactive substance at t

time

tC

46

Example :

47

C) Newton's Law of Cooling When an object has a temperature greater than the

ambient temperature, it cools according to Newton's

Law of cooling which states that the rate of cooling is

proportional to the difference in the temperatures, that

is: , where is the temperature of

the object at any time tand a is the ambient temperature.

The solution to this separable differential equation is:

akdt

d

t

48


9/9

10/19/20

dtakd

kdta

d

kdtad

ckta ln

ae ckt

ae ckt

aAe kt 49

Example :

A turkey is taken from the oven at 300 o F and

placed at room temperature of 70o F. In t= 2

minutes, the turkey's temperature is 200o F. Find

how long it takes the turkey to cool to 150 o F.

Solution

temperature of object at t time t

50

51

Chapter 2-Differential Equation

Documents

Transcript of Chapter 2-Differential Equation