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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
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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.
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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.
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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?
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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.
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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:
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Case 1: implies that the CE has
Case 2: implies that the CE has
Case 3: 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
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Examples:
Question 1
Find the general solution of the following differential equations:
(a) (b) (c) (d)
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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)
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Question 2
Find the solutions to the following:
(a) () () (b) () (c) () () (d) () () (e) () ()
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Question 3
Find the solutions to the following:
(a) () () (b)
() ()(c) ()
(d) () () (e) ( ) () ()
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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
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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:
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Examples:
Question 1
Find the general solution of for
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Question 2
Find the general solution of .
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Exercises:
Question 1
Find the general solution of the following nonhomogeneous differential equations using the
method of variation of parameters.
(a)
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(b) ()
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(c) ()
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(d)
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(e) ( )
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Question 2
Solve the following differential equations using the method of variation of parameters.
(a)
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(b)
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(c)
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(d)
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(e)
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(f)
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(g)
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(h) ()
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(i)
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(j)
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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
() ()
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Examples:
Question 1
Find the general solution of
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Question 2
Find the general solution of
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Question 3
Find the general solution of
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Question 5
Find the general solution of
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Question 6
Find the general solution of
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Exercises:
Question 1
Find the general solution of the following nonhomogeneous differential equations using the
method of undetermined coefficients.
(a)
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(b)
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(c)
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(d)
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(e)
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(f)
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(g)
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(h)
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(i)
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(j)
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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 () () .
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Exercises:
Question 1
Find the general solution to the following homogeneous differential equations using the
reduction of order method.
(a) ()
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(c) ()
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(d) ()
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(e) ()
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(f) ()
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(g) ( ) ()
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(h) ( ) ()
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(i) ( ) ()
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(j) ()
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Question 2
Find the general solution to the following nonhomogeneous differential equations using the
reduction of order method.
(a) ()
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(b) ()
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(c) ()
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(d) ()
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(e) ( ) ()
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(f) ( ) ( ) ()
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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.
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Examples:
Question 1
Use the power series method to solve the equation
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Question 2
Solve using the power series method.
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Exercises:
Question 1
Use the power series method to solve the following differential equations.
(a)
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(b)
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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:
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Examples/exercises:
Question 1
Use Eulers method to solve approximately the IVP
on [ ]with () Take the step size and find and
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Question 2
Use Eulers method with a step size of to solve approximately the IVP
( ) on
[] with
()
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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.
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A. Heuns method :If then and Then substituting these values into Eq. (1), we obtain
where
and and
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Example:
Question 1
Use Heuns method to solve approximately the IVP
on [ ]with () Take the step size and find and .
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B. Midpoint method ( ):If then and Then substituting these values into Eq. (1), we obtain
where
and
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Example:
Question 1
Use midpoint method to solve approximately the IVP
on [ ]with () Take the step size and find and .
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C. Ralstons method :If then and Then substituting these values into Eq. (1), we obtain
where
and
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Example:
Question 1
Use Ralstons method to solve approximately the IVP
on [ ]with () Take the step size and find and .
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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
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Example:
Question 1
Use the RK4 method to solve approximately the IVP
on [ ]with () Take the step size and find and .
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Exercises:
Question 1
Use Heuns method with a step size of to solve approximately the IVP on [ ]with ()
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Question 2
Use the RK4 method with a step size of to solve approximately the IVP on []with ()
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