Topic 2 Part 2 - Solutions to 2nd Order ODEs Students

8/12/2019 Topic 2 Part 2 - Solutions to 2nd Order ODEs Students

1/91

Dr. Ganeshsree/AS102_Mathematics II/Topic 2_Part 2_Solutions to 2ndOrder ODEs/Jan 2014 Page 1

AS102MATHEMATICS II

TOPIC 2Part 2: SOLUTIONS TO SECOND ORDER DIFFERENTIAL EQUATIONS

SUBTOPICS/KEY QUESTIONS AND KEY IDEAS

Part 2 : Second Order Differential Equations

a. Definition of second order differential equationsb. The method of undetermined coefficientsc. Reduction of orders methodd. Variation of parameterse. The power series methods

Part 3 : Numerical Methods

a. Eulers Methodb.

Runge-Kutta Methodi. The Second-Order Runge-Kutta (RK2) Methodii. The Fourth-Order Runge-Kutta (RK4) Method


2/91


INTRODUCTION TO DIFFERENTIAL EQUATIONS (D.E.)

A second order differential equation is one containing a second derivative but no higherderivatives.

The focus in this chapter will be on linear second order DEs.

LINEAR SECOND ORDER DIFFERENTIAL EQUATIONS

General form:() () () ()

where () Rewriting it by dividing all the items by ()we have:

()() ()()

()() ()

() () () .Eq. 1where () ()() () ()() () ()()This is known as the second order linear differential equation.

Examples of second order linear DEs are:i. ii. iii.


3/91


THE STRUCTURE OF SOLUTIONS FOR THE SECOND ORDER LINEAR DEs

The second order linear homogeneous differential equation has the form () ()

In order to obtain the solutions to second order linear homogeneous DEs, we need to firststudy a few related principles and theorems.

THE PRINCIPLE OF LINEAR SUPERPOSITION

The principle of linear superposition states that:

If () () are solutions to a DE, then so is () () () where A andBare any constants.

In other words, this principle states that the linear combination of all the solutions of thelinear homogeneous differential equation is also a solution.

The point to taking the linear combinations of and are to generate new solutionsfrom and

However, if is itself a constant multiple of , say then the linearcombination that is produced

() ( )is just a constant multiple of , so does not contribute any new information to theprocess of finding solutions for the DE.

Therefore, it can be concluded that the principle of linear superposition can only beapplied if the two solutions to the 2

ndorder DE are linearly independent solutions.

LINEAR INDEPENDENCE AND LINEAR DEPENDENCE

The definitions of these two concepts are as given below:

Two functions are said to be linearly independenton an open interval (which can be the entirereal line) if neither function is a constant multiple of the other for all in the interval.If one of the functions is a constant multiple of the other on the entire interval , then thesefunctions are said to be linearly dependent.


4/91


To test for linear independence, a function called the Wronskian ( )is used.

THE WRONSKIAN

The Wronskian, denoted by ( )of two solutions and is the determinant of and and it is defined as given below:

Often we the Wronskian is denoted as just ()

Theorem : Properties of the Wronskian

Suppose and are the solutions of () () on an open interval Then the following holds true:

(a)Either () for all in or ( ) for all in (b)and are linearly independent on if and only if ( ) on

* Part (b) is called the Wronskian test for linear independence.

* The Wronskian must be either zero or non-zero for the entire interval of It cannot be zero forsome and be non-zero for some on

Example 1:

Consider the DE The solutions of the DE are and Arethese solutions linearly independent?


5/91


Summary of Procedure:

Let and be two linearly independent solutions of on an open interval Then every solution on is a linear combination of and Therefore, to find all the solutions of the homogeneous linear second order DEs:

1. Find the two linearly independent solutions and 2. The linear combination contains all the possible solutions.

The solution is called the general solution of the DE on 3. The particular solution can be found by solving an initial value problem (IVP) to

determine the values of and

THE NONHOMOGENEOUS CASE

A nonhomogeneous second order DE is of the form () () ()for() The main difference between the homogeneous and nonhomogeneous second order DE is

that, for the nonhomogeneous DE, the sums and constant multiples of solutions need not

be solutions.


6/91


THE CONSTANT COEFFICIENT CASE

To solve both second order linear homogeneous and nonhomogeneous DEs, we must begin with

finding two linearly independent solutions of a homogeneous equation.

When the coefficients are constants, the general solutions can be found easily.

HOMOGENEOUS SECOND-ORDER DIFFERENTIAL EQUATIONS

Now consider the constant-coefficient linear homogeneous DE given by:

The behavior and pattern of Eq. (1) suggests an exponential function to be the solution,because derivatives of

are constant multiples of

- These different solutions are summarized in the next page:


7/91


Case 1: implies that the CE has



Summary:

If has CE 2+ a + b= 0, then the following cases are true:Case Roots of CE General solution

1

2

3


8/91


Examples:

Question 1

Find the general solution of the following differential equations:

(a) (b) (c) (d)


9/91


Exercises:

Question 1

Find the general solution of the following:

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o)


10/91

Dr. Ganeshsree/AS102_Mathematics II/Topic 2_Part 2_Solutions to 2ndOrder ODEs/Jan 2014 Page

10


11/91


11


12/91


12

Question 2

Find the solutions to the following:

(a) () () (b) () (c) () () (d) () () (e) () ()


13/91


13


14/91


14

Question 3

Find the solutions to the following:

(a) () () (b)

() ()(c) ()

(d) () () (e) ( ) () ()


15/91


15


16/91


16


17/91


17

SOLVING NONHOMOGENEOUS DEs

The method of finding the solutions for nonhomogeneous differential equations are:

1. Finding the linearly independent solutions of the associated homogeneous DEs.2.

Finding a particular solution for the nonhomogeneous DE.

Since finding the solutions in Step 1 has been already been studied in the previous section, this

section would focus on the various methods used to find There are 3 methods which are commonly used to determine the particular solution :

1. Variation of parameters2. Undetermined coefficients3. Reduction of orders


18/91


18

METHOD 1: VARIATION OF PARAMETERS

Suppose we know two linearly independent solutions and of the associated homogeneousDE.

One method of finding is called the method of variation of parameters.We need to look for functions and so that

The formula for finding () and ()are as given below:


19/91


19

Examples:

Question 1

Find the general solution of for


20/91


20

Question 2

Find the general solution of .


21/91


21

Exercises:

Question 1

Find the general solution of the following nonhomogeneous differential equations using the

method of variation of parameters.

(a)


22/91


22

(b) ()


23/91


23

(c) ()


24/91


24

(d)


25/91


25

(e) ( )


26/91


26

Question 2

Solve the following differential equations using the method of variation of parameters.

(a)


27/91


27

(b)


28/91


28

(c)


29/91


29

(d)


30/91


30

(e)


31/91


31

(f)


32/91


32

(g)


33/91


33

(h) ()


34/91


34

(i)


35/91


35

(j)


36/91


36

METHOD 2: UNDETERMINED COEFFICIENTS

This method applies only to solving nonhomogeneous differential equations of the constant

coefficient case given by ()The basic/general idea behind this method is that sometimes (most times) we can guess thegeneral form of ()from the appearance of()Summary of method/procedure:

Suppose we want to find the general solution of ()1. Write the general solution

() () ()of the associated homogeneous DE with

and

being the linearly independent solutions of the DE. This is similar to the

constant coefficient case.2. Find the particular solution ()of the nonhomogeneous DE using()and Table 1 as

a guide.

3. If any term of the first attempt is a solution of the associated homogeneous DE, multiplyby . If any term of this revised attempt is a solution of the homogeneous DE, multiplyby again. Substitute this final general form of a particular solution into the DE andsolve for the constants to obtain ()

4. The general solution is Table 1: Functions to try for

()in the Method of Undetermined Coefficients

() ()


37/91


37

Examples:

Question 1

Find the general solution of


38/91


38

Question 2



39/91


39

Question 3



40/91


41/91


41

Question 5



42/91


42

Question 6



43/91


43

Exercises:

Question 1

Find the general solution of the following nonhomogeneous differential equations using the

method of undetermined coefficients.

(a)


44/91


44

(b)


45/91


45

(c)


46/91


46

(d)


47/91


47

(e)


48/91


48

(f)


49/91


49

(g)


50/91


50

(h)


51/91


51

(i)


52/91


52

(j)


53/91


53

METHOD 3: REDUCTION OF ORDERS

Procedure/summary of method:

1. Set () ()() for some unknown function () and substitute it into thedifferential equation.Here()is a non-trivial solution of the differential equation.

2. Now we have a separable differential equation in terms of and . Use the integratingfactor (IF) to get and then integrate to get ()

3. Substitute ()back into the equation () ()()to get the complete solution ofthe DE.

The restriction of this method:

In order to use this method, we must be given one non-trivial solution ()to the differentialequation () () .


54/91


55/91


55

Exercises:

Question 1

Find the general solution to the following homogeneous differential equations using the

reduction of order method.

(a) ()


56/91


57/91


57

(c) ()


58/91


58

(d) ()


59/91


59

(e) ()


60/91


60

(f) ()


61/91


61

(g) ( ) ()


62/91


62

(h) ( ) ()


63/91


63

(i) ( ) ()


64/91


64

(j) ()


65/91


65

Question 2

Find the general solution to the following nonhomogeneous differential equations using the

reduction of order method.

(a) ()


66/91


66

(b) ()


67/91


67

(c) ()


68/91


68

(d) ()


69/91


69

(e) ( ) ()


70/91


70

(f) ( ) ( ) ()


71/91


71

THE POWER SERIES METHOD FOR SOLVING DIFFERENTIAL EQUATIONS

Many differential equations cannot be solved explicitly in terms of finite combinations of simple

familiar functions. This situation is true even for a simple-looking DE such as

( )However, it is important to be able to solve equations such as Eq. (1) because they are applicablein a lot of real-life problems such as physical problems, quantum mechanics, finance and

economics.

In such cases, we use the method of power series to solve the DEs, that is we look for a solution

of the form

The method is to substitute this expression into the DE and determine the values of the

coefficients This technique resembles the method of undetermined coefficients discussed earlier.


72/91


72

Examples:

Question 1

Use the power series method to solve the equation


73/91


73

Question 2

Solve using the power series method.


74/91


74

Exercises:

Question 1

Use the power series method to solve the following differential equations.

(a)


75/91


75

(b)


76/91


76

NUMERICAL METHODS FOR FINDING SOLUTIONS OF DIFFERENTIAL EQUATIONS

There are numerous numerical methods that are available to find the values of the solutions to

differential equations.

However in this course, we will only be looking at two classical methods in calculating thevalues of differential equations. The two methods are the Eulers method and the Runge-Kutta

(RK) method.

Method 1: The Eulers method

In this section, we shall look at one of the simplest ways of calculating an approximate numerical

solution of a differential equation. However, Eulers method is rarely used in practice.

The Eulers method of generating approximate numerical values of the solution of an initial-

value problem

( ) () at selected points issummarized in the recursive formula given below.

Here is a positive integer which represents the number of iterations to be performed and is a(small) positive number which is called the step size.

Eulers formula:


77/91


77

Examples/exercises:

Question 1

Use Eulers method to solve approximately the IVP

on [ ]with () Take the step size and find and


78/91


78

Question 2

Use Eulers method with a step size of to solve approximately the IVP

( ) on

[] with

()


79/91


79

Method 2: The Runge-Kutta Method

There are several types of Runge-Kutta methods available and these types are classified by their

order.

The classical method is the Second-Order Runge-Kutta (RK2) method while the most useful andmost popular method is the Fourth-Order Runge-Kutta (RK4) method.

Part 1: The Second-Order Runge-Kutta method (RK2)

Although this case is not usually useful in practice, it is very useful in understanding the other

forms of Runge-Kutta methods.

The recursive formula for this method is as given below:

( ) ()

where

( ) ()

( )

( )

and

Since we can choose an infinite number of values for there are an infinite number of RK2methods.

However, we present three of the most commonly used and preferred versions.

It is to be noted that every version would yield the exact same results if the solution to the IVP

were to be quadratic, linear or a constant.

However, they yield different results when the solution is more complicated.


80/91


80

A. Heuns method :If then and Then substituting these values into Eq. (1), we obtain

where

and and


81/91


81

Example:

Question 1

Use Heuns method to solve approximately the IVP

on [ ]with () Take the step size and find and .


82/91


82

B. Midpoint method ( ):If then and Then substituting these values into Eq. (1), we obtain

where

and


83/91


83

Example:

Question 1

Use midpoint method to solve approximately the IVP



84/91


84

C. Ralstons method :If then and Then substituting these values into Eq. (1), we obtain

where

and


85/91


85

Example:

Question 1

Use Ralstons method to solve approximately the IVP



86/91


86

Part 2: The Fourth-Order Runge-Kutta method (RK4)

As with the second-order approaches, there are an infinite number of versions for this method.

The most commonly used form is called the classical Fourth-Order Runge-Kutta (RK4) method.

The recursive formula for this method is as given below:

where

and and .Take note that in each step, we have to first compute the four auxiliary quantities andand then the new value


87/91


87

Example:

Question 1

Use the RK4 method to solve approximately the IVP



88/91


88

Exercises:

Question 1

Use Heuns method with a step size of to solve approximately the IVP on [ ]with ()


89/91


89

Question 2

Use the RK4 method with a step size of to solve approximately the IVP on []with ()


90/91


90


91/91

Topic 2 Part 2 - Solutions to 2nd Order ODEs Students

Documents

Transcript of Topic 2 Part 2 - Solutions to 2nd Order ODEs Students