Lecture Note - Chapter 1

7/27/2019 Lecture Note - Chapter 1

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CHAPTER 1: FIRST ORDER ORDINARY DIFFERENTIAL EQUATION SSCE1793

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- Ordinary Differential Equations (ODE) Contains one or more dependent variables with respect to one independent variable =>

22 =>

22

+

+

=

=>

+

+

=

- Partial Differential Equations (PDE) involve one or more dependent variables and two or more independent variables

+

= 0 => + = 0

2

2+

2

+

=

=>

+

+

=

is the dependent variablewhile is the independent

variable

is the dependent variablewhile is the independentvariable

Can you determine which one is the DEPENDENT VARIABLE and which

one is the INDEPENDENT VARIABLES from the following equations ???

Classification by type

Dependent Variable:

Independent Variable:

Dependent Variable:


Dependent Variable:



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- Order of Differential Equation Determined by the highest derivative

- Degree of Differential Equation Exponent of the highest derivative

Examples:

a) 2

+ = sin

b) 2 2 + 2 = cos

c) 2 2 + 2

+ =

d) 3 34

+ 2

+ = 0

- Linear Differential Equations Dependent variables and their derivative are of degree 1 Each coefficient depends only on the independent variable A DE is linear if it has the form

+ 111 + + 1

+ 0 = ()

Examples:

1) + = sin 2) 22 + = sin 2

3) 55 + sin2 = tan2

- Nonlinear Differential Equations

Dependent variables and their derivatives are not of degree 1

Order : Degree:

Order : Degree:

Order : Degree:

Order : Degree:

Classification by order / degree

Classification as linear / nonlinear


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Examples:

1) + 2 = sin

2)

2 + = sin

3) 332

+ 223

+

2 + 1 =

Initial conditions : will be given on specified given point

Boundary conditions : will be given on some points

Examples :

1) 0 = 1 ; (0) = 2 Initial condition

2) 1 = 5 ; 2 = 2 Boundary conditionInitial Value Problems (IVP)

22 + 2

+ + sin

Initial Conditions:

0 = 0 ; 0 = 1Boundary Value Problems (BVP)

22

+ 2

+

+ sin

Boundary Conditions:

0 = 1 ; 1 = 2

- General Solutions Solution with arbitrary constant depending on the order of the equation

- Particular Solutions Solution that satisfies given boundary or initial conditions

Order : Degree:

Order : Degree:

Order : Degree:

Initial & Boundary Value Problems

Solution of a Differential Equation


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Examples:

= cos + sin (1)Show that the above equation is a solution of the following DE

+ = 0 (2)Solutions:

EXERCISE:

Show that = ( ) + ( ) is the solution of the following DE +

+

=


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Types of first order ODE

- Separable equation- Homogenous equation- Exact equation- Linear equation- Bernoulli Equation

1.2.1 Separable Equation

() =

1() =

How to identify?

Suppose, = = Hence this become a SEPARABLE EQUATION if it can be written as

Method of Solution : integrate both sides of equation

1.2 First Order Ordinary Differential Equations (ODE)


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Example 1:

Solve the initial value problem

=

cos1 + 22 , 0 = 1

Solution:

i) Separate the functions

ii) Integrate both sides

iii) Use the initial condition given, =

iv) Final answer

Note: Some DE may not appear separable initially but through

appropriate substitutions, the DE can be separable.


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Example 2:

Show that the DE = ( + )2 can be reduced to a separable equation by using

substitution = + . Then obtain the solution for the original DE.

Solutions:

i) Differentiate both sides of the substitution wrt

ii) Insert (2) and (1) into the DE

iii) Write (3) into separable form

iv)

Integrate the separable equation


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1.2.2 Homogenous Equation

Example 1:

Determine whether the DE is homogenous or not

a)

=2 + 2

( + )

b) = 2 + 2Solutions:

a) , = =2 + 2

( + ), =

b) , = = + 2 + 2

, =,

How to identify?

Suppose, = ,, is homogenous if

for every real value ofMethod of Solution :

i) Determine whether the equation homogenous or notii) Use substitution = and = + in the original DEiii) Separate the variable and iv) Integrate both sidesv) Use initial condition (if given) to find the constant value

Separable

equation

method


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Example 2:

Solve the homogenous equation

2 + 2 = 0Solutions:

i)

Rearrange the DE

ii) Test for homogeneity

iii) Substitute = and = + into (1)

iv) Solve the problem using the separable equation method

Note:

Non-homogenous can be reduced to a homogenous equation by usingthe ri ht substitution.


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Example 3:

Find the solution for this non-homogenous equation =

2 + 5

(1)

by using the following substitutions

=

+ 3,

=

+ 2 (2),(3)

Solutions:i) Differentiate (2) and (3)

and substitute them into (1),

ii) Test for homogeneity,, =(, )

iii) Use the substitutions = and = +

iv) Use the separable equation method to solve the problem


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1.2.3 Exact Equation

=

(, )(, )

=> ,

+ ,

= 0

=

, + , = 0

=

How to identify?

Suppose, = ( ,)

(

,

),

Therefore the first order DE is given by

Condition for an exact equation.

Method of Solution (Method 1):

i) Write the DE in the formAnd test for the exactness

ii) If the DE is exact, then = , =

(1), (2)

To find (, ), integrate (1) wrt to get, = , + () (3)


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=

, + () =

, + , = 0

=

iii) To determine (), differentiate (3) wrt to get

iv) Integrate to get v) Replace

(

) into (3). If there is any initial conditions given, substitute the

condition into the solution.

vi) Write down the solution in the form, =, where A is a constant

Method of Solution (Method 2):

i) Write the DE in the form

And test for the exactness

ii) If the DE is exact, then = , =

(1), (2)

iii) To find (, ) from , integrate (1) wrt to get, = , + 1() (3)

iv) To find (, ) from , integrate (2) wrt to get, = , + 2() (4)

v) Compare (3) and (4) to get value for 1() and 2().vii) Replace 1() into (3) OR2() into (4).viii) If there are any initial conditions given, substitute the conditions into the

solution.ix) Write down the solution in the form

, =, where A is a constant


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Example 1:

Solve 2 + 3 + 2 1 = 0Solution (Method 1):

i) Check the exactness

ii) Find ,

To find ,, integrate either (1) or (2), lets say we take (1)

iii) Now we differentiate (3) wrt to compare with =

iv) Find

v) Now that we found , our new , should looks like this

vi) Write the solution in the form , =


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Exercise :

Try to solve Example 1 by using Method 2

Note:

Some non-exact equation can be turned into exact equation by

multiplying it with an integrating factor.


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Example 2:

+ 2 + + + 2 = 0Show that the following equation is not exact. By using integrating factor, , = , solve theequation.

Solution:

i) Show that it is not exact

ii) Multiply, into the DE to make the equation exact

iii) Check the exactness again

iv) Find ,

v) Find

vi) Write , =


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1.2.4 Linear First Order Differential Equation

+ = ()

()

+

() =

()

=

= () = ()

= 1()

How to identify?

The general form of the first order linear DE is given by

When the above equation is divided by ,

+ = () ( 1 )Where = () and = ()Method of Solution :

i) Determine the value of dan such the the coefficient of is 1.ii) Calculate the integrating factor, iii) Write the equation in the form of

iv) The general solution is given by

NOTE:

Must be + here!!


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Example 1:

Solve this first order DE

+

1 + =

Solution:

i) Determine and

ii) Find integrating factor, =

iii) Write down the equation

iv) Final answer

Note:

Non-linear DE can be converted into linear DE by using the right

substitution.


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Example 2:

Using = 2, convert the following non-linear DE into linear DE.2 2 = 1; 1 = 1

Solve the linear equation.

Solutions:

i) Differentiate = to get and replace into the non-linear equation.

ii) Change the equation into the general form of linear equation & determine and

iii) Find the integrating factor, =

iv) Find

v) Use the initial condition given, = .


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1.2.5 Equations of the form = ( + )

How to identify?

When the DE is in the form

= ( + ) ( 1 )

use substitution

= + to turn the DE into a separable equation

( 2 )

Method of Solution :

i) Differentiate ( 2 ) wrt ( to get ) = +

( 3 )

ii) Replace ( 3 ) into ( 1 )iii) Solve using the separable equation solution


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Example 1:

Solve = 1 + ( + 2)1

Solution:

i) Write the equation as a function of

ii) Let = and differentiate it to get

iii) Replace ( 2 ) into ( 1 )

iv) Solve using the separable equation solution


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1.2.6 Bernoulli Equation

1

1 + =

+ 1

(

)

= 1

(

)

How to identify?

The general form of the Bernoulli equation is given by

+ = ( 1 )

where 0, 1To reduce the equation to a linear equation, use substitution

= 1 ( 2 )Method of Solution :

iv) Divide ( 1 ) with + 1 =

( 3 )

v) Differentiate ( 2 ) wrt ( to get ) = 1

1

1 =

( 4 )

vi) Replace ( 4 ) into ( 3 )

vii) Solve using the linear equation solution Find integrating factor, = Solve = ()


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Example 1:

Solve

+

1

3 = 4 ( 1 )

Solutions:

i) Determine =ii) Divide ( 1 ) with

iii) Using substitution, =

iv) Replace ( 3 ) into ( 2 ) and write into linear equation form

v) Find the integrating factor

vi) Solve the problem

vii) Since

=

3 = 3


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1.3 Applications of the First Order ODE

The Newtons Law of Cooling is given by the following equation

= ( )Where is a constant of proportionality is the constant temperature of the surrounding medium

General Solution

Q1: Find the solution for ()It is a separable equation. Therefore

= ln = +

= + = + =

Q2: Find when = 0, = 00 = +1 => = 0

= + 0

Q3: Find . Given that = 70, 0 = 100, 6 = 8080 = 70 + 100 70

6

6 = 10

30

ln 6 = ln 13

= 16

ln1

3= 1.098612

= + 0 1.0986 = 70 + 301.0986

The Newtons Law of Cooling

Do you know

what type of DE

is this?


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Example 1:

According to Newtons Law of Cooling, the rate of change of the temperature satisfies = ( )

Where is the ambient temperature, is a constant and is time in minutes. When objectis placed in room with temperature 10C, it was found that the temperature of the object

drops from 90C to 30C in 30 minutes. Then determine the temperature of an object after

20 minutes.

Solution:

i) Determine all the information given.Room temperature = = 10CWhen = 0, 0 = 90CWhen

= 30,

30= 30C

Question: Temperature after 20 minutes, = 20, = ?ii) Find the solution for

= + iii) Use the conditions given to find and

When = 0, 0 = 90C , = 10C90 = 10 + => = 80 = 10 + 80

When = 30, 30 = 30C30 = 10 + 8030

30 = 2080

= 130

ln1

4= 0.04621

= 10 + 800.04621 iv)

=

,

= ?

20 = 10 + 800.04621 20 = 41.75


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The differential equation =

Where

is a constant

is the size of population / number of dollars / amount of radioactiveThe problems:

1. Population Growth2. Compound Interest3. Radioactive Decay4. Drug Elimination

Example 1:

A certain city had a population of 25000 in 1960 and a population of 30000 in 1970. Assumethat its population will continue to grow exponentially at a constant rate. What populations

can its city planners expect in the year 2000?

Solution:

1) Extract the information = 0, 0 = 25000 = 10, 10 = 30000 = 40, 40 = ?2) Solve the DE

= Separate the equation and integrate

= => = 3) Use the initial & boundary conditions = 0, 0 = 25000

= 25000 =>

= 25000

= 10, 10 = 3000030000 = 25000 10 10 = ln 30000

25000 => = 0.01823

= 25000 0.01823 = 40, 40 = ?40 = 25000

0.01823

40

= 51840In the year 2000, the population size is expected to be 51840

Natural Growth and Decay

Do you know

what type of DE

is this?


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Given that the DE for an RL-circuit is

+ = ()Where

(

) is the voltage source

is the inductance is the resistanceCASE 1 : = (constant)

+ = 0(1)

i) Write in the linear equation form

+ = 0 = , =

0 ii) Find the integrating factor,

= =

iii)

Multiply the DE with the integrating factor

= 0 iv) Integrate the equation to find

= 0

=

1

0

+

=

0

+

Electric Circuits - RC

Do you know

what type of DE

is this?


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CASE 2 : = or = Consider = 0 sin , the DE can be written as

+ = 0 sin i) Write into the linear equation form and determine

and

+ = 0 = , =

0 sinii) Find integrating factor,

= =

iii) Multiply the DE with the integrating factor

= 0

iv) Integrate the equation to find = 1

0 (1)

= 1 0

1w cos R

w2L sin

+ 2 1

w2 sin

= 0w

cos + 0R

wL2 sin R

wL2 0

sin (2)From (1)

Tabular Method

Differentiate Sign Integrate

sin

1

w cos

2

1

w

2

sin

+

+

-


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= 0 (3)

Replace (3) into (2)

= 0 cos + 02 sin

2

1 + RwL

2 = 0w cos + 02 sin

= 11 + R

wL2

0w cos +

02 sin

The Newtons Second Law of Motion is given by

= Where is the external force is the mass of the body is the velocity of the body with the same direction with

is the time

Example 1:

A particle moves vertically under the force of gravity against air resistance 2, where is aconstant. The velocity at any time is given by the differential equation

= 2.

If the particle starts off from rest, show that

= 2

1

2 + 1 Such that = . Then find the velocity as the time approaches infinity.

Vertical MotionNewtons Second Law of Motion


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Solution:

i) Extract the information from the questionInitial Condition = 0, = 0

ii) Separate the DE1

2 = 1

2

2 =

Let = ,1

2

2

=

iii) Integrate the above equation 12 2 =

1

21

+ +1

= + 1

2 ln + ln = + ln +

= 2 + 2

iv) Use the initial condition, = , = ln1 = 2 => = 0

ln + = 2

v) Rearrange the equation + = 2

+ = 2 2 + = 2

1

2 2 =1

+

1 + = 12 + + 12 Using Partial Fraction


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2 + 1 = 2 1 = 2 12 + 1

vi) Find the velocity as the time approaches infinity.

= 1

1

2 1 + 12

When , 12 0 => =

Lecture Note - Chapter 1

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