Introduction ODE

Differential Equations & Linear Algebra Dr. Gharib, December 23, 2010

Differential Equations

The Second Part of the MTH 2132/2311Course

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LECTURE 1

Introduction

Why should one be interested in differential equations? Many laws governing natural phenomena are relations (equations) involving rates at which things happen (derivatives).

Examples of fields using differential equations in their analysis include:

Solid mechanics & Motion

Heat transfer & Energy balances

Vibrational Dynamics & Seismology

Aerodynamics & Fluid Dynamics

Electronics & Circuit Design

Population Dynamics & Biological Systems

Climatology and Environmental Analysis

Options Trading & Economics

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Learning Objectives1. To be able to identify and classify an ordinary

differential equation

2. To understand what it means for a function to be a solution of an ordinary differential equation.

3. To be able to find the solution to certain simple ordinary differential equations

4. To be able to discover some properties of the solution of an ordinary differential equation Without actually finding the solution

5. To be able to derive an ordinary differential equation as the mathematical model for a physical phenomenon

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Introduction to Differential EquationsDefinitions and Terminology

Definition 1: Differential Equation (DE)

An equation containing the derivatives of one or more dependent variables, with respect to one or more independent variables, is said to be a differential equation.

Are differential equations easy to solve? Some are, but many are not

What do solutions look like? Solutions are functions.

If expressed symbolically they look like mathematical formulas.

Geometrically, they are curves.

Notation:1)Leibniz notation 2)Prime notation

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(a) Classification by typei) Ordinary Differential Equation (ODE)

Example 1

ii) Partial Differential Equation (PDE):

Example 2

(b) OrderThe order of a differential equation is the order of the highest derivative that appears in it.

Examples of first order differential equations:

Examples of second order differential equations:

Examples of third order differential equations:

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(c) DegreeIf all derivatives in the equation are raised to integer powers, then

The degree of a differential equation is the highest power to which the highest order derivative in the equation is raised.

Example of a second degree, first order differential equation:

Example of a third degree, second order differential equations:

(d) LinearityA linear differential equation has the following properties:

1) The dependent variable and its derivatives appear raised only to the first power: there are no terms like

.

2) Functions of y or of its derivatives are forbidden: there are no terms like .

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3) The equation must not contain products of derivatives of different order, or of y with its derivatives: no terms like

Notes

1. The independent variable x may appear in any form whatsoever.

2. A linear differential equation only contains derivatives raised to the first power so all linear differential equations are of the first degree.

Example:

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SOLVING DIFFERENTIAL EQUATIONSWhen we solve an algebraic equation, the aim is to find the unknown number or numbers that satisfy it.

When we solve a differential equation, the aim is to find the unknown function or functions that satisfy it.

Note:

A solution to a differential equation is a relation between the variables x and y in which no derivatives

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appear and when we substitute it into the differential equation, that equation becomes an identity (true for all allowed values of the variables).

Example: Show that is a solution of the differential equation .To do this, we substitute the solution into

the equation

This is an identity (true for all x), so is a solution for all x.

GENERAL SOLUTIONS AND PARTICULAR SOLUTIONS

is a solution of the differential equation , but this is not the only solution.

Let’s try where C is an arbitrary constant:

The solution is called a general solution because it

involves an arbitrary constant. For any choice of the

constant C, it is still a solution of the differential equation.

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The general solution represents a “family” of solutions

with similar properties.

A particular solution of the differential equation is given

by a specific choice of the arbitrary constant: it is

a solution that does not contain any arbitrary constants.

Examples: and are particular solutions corresponding to C = 0 and C = 2, respectively.

General nth Order Differential Equation: is given by

1. If g(x) = 0 then it is called Homogeneous

otherwise Non-homogeneous.

Notes

1. If

Subject to

Then it is called nth order Initial-Value Problem (IVP).

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2. If

Subject to: then it is called nth order Boundary Value Problem (BVP)

Theorem: EXISTENCE OF THE SOLUTION

Example: Determine whether or not subject to

has a unique solution over .

Sol: are all continuous on for every x. but

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Let are continuous on an interval I, for every x in I, and

Then, a solution of the IVP

Subject to: is exists on the interval I

and it is unique.


Hence IVP does not have a solution.

Example: Find an Interval I for which IVP

has a solution.

Sol: are all continuous every where.

&

Hence,

Note:

A BVP may have no solution, a unique solution, or many solutions.

Example: Let . Investigate the existence of the solution(s) under the following boundary condition.

i.

ii.

iii.

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Sol.

i.

Hence is a solution, i.e., infinite number of solutions for infinite values of b.

ii.

Hence is a solution.

iii.

Hence, there is no solution.

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LECTURE 2

HOMOGENEOUS EQUATIONS

Given a linear nth-order differential equation of the form

Is called a homogeneous equation where as the equation

is called non-homogenous

1.In case of homogeneous equation;

y = 0 is always a solution.

A multiple ( y = c1y1) of a solution is always a solution.

A linear combination (y = c1y1+c2y2+---+cnyn) of solutions is also a solution Superposition Principle.

Example: Sol.

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Solving (i) and (ii) we get ,

Hence the solution is .

Linear Dependence/Independence

1.The set of functions f1, f2, ---, fn are Linearly Independent on the interval I if the equation

c1f1+c2f2+---+cnfn = 0 has only trivial solution

c1= c2= ---= cn = 0

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Otherwise linearly dependent

2.Equivalently; f1, f2, ---, fn are linearly independent if

Wronskian:

Otherwise linearly dependent

Example: Is L. ind or L. dep.?

Sol. Recall from Linear Algebra that any set containing 0 is L.dep.

Now,

Example: Is L. ind or L. dep.?

Sol. By Inspection

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Fundamental Set of Solution: A set {y1, y2,---,yn } of linearly independent solutions of a homogeneous equation is called fundamental set of solution.

In this situation y = c1y1+c2y2+---+cnyn is called General solution of homogeneous equation.

Example: Verify that is a fundamental set of

Sol. is a solution.

Similarly is a solution.

Now, .

Hence the set is a fundamental set.

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Example: Verify that is a fundamental set of

Sol. is a solution.

Similarly is a solution.

Now, .

Hence the set is a fundamental set.

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NON-HOMOGENEOUS EQUATIONS

Given non-homogeneous equation

1. is called a particular solution or particular integral and this can be any solution but free of arbitrary constants.

2. is called complementary solution or complimentary function and it is general solution of the homogeneous part

i.e., if a set {y1,

y2,---,yn } is linearly independent solutions of the homogeneous part then yc = c1y1+c2y2+---+cnyn .

3. is known as general solution of non-homogeneous equation.

Example: Verify that is a general solution of

Sol. Consider .

Need to verify that is a particular solution. .

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Now, need to verify that is a fundamental set for homogeneous part

and

. Hence verified.

Another Superposition Principle

Let { ypj } be a particular solutions of the equation

where .

Then the is a particular solution for the equation

Example: Verify that is a particular solution of

Let ,

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and then we can see that

, and are particular solutions of non-homogenous equations with right side defined by the above functions

SECTION 4.2: REDUCTION OF ORDER

Let be a linear second order differential equation. Suppose that is a nontrivial solution for this equation. We can find a second solution which is linearly independent from as

1.This method is called a reduction of order need to solve only first order equation to find .

2.This method can be used to find the general solution of a non-homogenous equation

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whenever a solution of the associated homogeneous equation is known.

Example: Let is a solution of . Find a second solution ?

Sol. Standard form

Reduction Formula:

Example: Let is a solution of Find a second solution

Sol. Standard form

Reduction Formula:

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Example: Let

Sol. Standard form

Reduction Formula:

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LECTURE 3Methods for solving First Order Differential Equations1. Separation of variables2. Integrating Factor 3. Bernoulli’s Equation

1. Separation of variables

Examples:

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Definition: Separable Equation

A first –order differential equation of the form

Is said to be separable or to have separable variables


, are separable and nonseparable, respectively.( explain )

From the definition there is no way of expressing as a product of a function of x times a function of y.

Another form of separable equation

where

General solution can be solved by directly integrating both the sides

Example:1

Example:2

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2. Integrating Factor Problem Statement: Given a first order linear differential equation and (possibly) an initial condition of the form

, .

Find the general solution for the given equation.

That is, find the one-parameter family of functions that are solutions of the given equation.

Notice that the terms involving and y are on the left-hand side of the equation and the only variable on the right-hand side is the independent variable t.

Step 0: Put the problem into standard form.

Divide the equation through by making the coefficient of

exactly one. This means that the problem has the (standard) form

, .

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Step 1: Determine the integrating factor, i.e.,

.

You will need to evaluate the integral by techniques from the calculus.

Step 2: Multiply the equation through by the integrating factor found in the previous step.

This step is straight forward algebra. Just remember to

multiply both sides of the equation by .

Step 3: Integrate both sides of the equation.

You should check (by mentally differentiating) that the left hand side is exactly the derivative of the product .

If it isn’t, you have made a mistake in Steps 1 to 3.

Don’t forget the constant of integration.

Step 4: Solve (algebraically) the result in Step 3 for y explicitly.

The result is the general solution for the equation.

Step 5: (only if an IVP is given) Apply the initial conditions and solve (algebraically) for the value of the constant of integration.

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This is accomplished by substituting for t and for y. The constant of integration is the unknown in the resulting equation.

Example:

Problem statement: Solve the IVP

, , t > 0.

Step 0: Put the problem in standard form.

The problem is given in standard form with and

.

Step 1: Determine the integrating factor, i.e., .

Step 2: Multiply the equation through by the integrating factor.

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Check (mentally) that the left-hand side is correct:

.

Then

Integrating,


Divide both sides of the equation by :

Step 5: (only if an IVP is given) Apply the initial conditions and solve (algebraically) for the value of the constant of integration.

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The initial condition is . Substitute

and into the equation and solve for C.

Multiplying through by :

Substituting for C gives the solution

.

Notice that this solution exists only for the interval .

Furthermore , since the numerator is

bounded between 1 and -1while the denominator is arbitrarily

large numerically. Finally, does not

exist, not even as an infinite limit.

Example:

Find the general solution of .


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Put all the terms involving the variable y on the right side of the equation and all the terms involving only the variable t on the left side of the equation.

Make the coefficient of the term involving exactly one.

The equation is now in standard form with and

.


.


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Then

Integrating



or

, where .

Step 5: (only if an IVP is given) apply the initial conditions and solve (algebraically) for the value of the constant of integration.

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There is no IVP. So there is nothing to do in this step.

Notice that no matter the choice of constant K, these solutions exist on the interval . Also , since both the terms and are numerically large and positive for large values of t. So their product is numerically large and positive. On the other hand is different since the term

is close to zero while the term is numerically large when t is numerically large and negative. This is an indeterminate form. So one must use L’Hopital’s Rule to evaluate this limit.Accordingly,

.

Example: Find the general solution of the equation

.


Put all the terms involving the variable y on the right side of the equation and all the terms involving only the variable t on the left side of the equation.

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Since the coefficient of the term involving is exactly one, the

equation is in standard form with and

.


.




Then

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Integrating

Use Integration by parts twice:


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,

Step 5: (only if an IVP is given) apply the initial conditions and solve (algebraically) for the value of the constant of integration.

There is no IVP. So there is nothing to do in this step.

Notice that no matter the choice of the constant C, these solutions exist on the interval . Also

, this follows by the following reasoning: For t numerically large

and positive, the terms and are both

numerically large and positive and so their product is too. The

term is very close to zero. Thus the limit is numerically

large and positive. On the other hand,

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is different

because when t is numerically large and negative, the term

is very close to zero. This limit actually depends on the

value of C!

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LECTURE 43. Bernoulli’s Equation

BERNOULLI EQUATION ( non-linear First Order D.E ) Any differential equation of the form

where P and Q are functions of x, and n is any real number is called Bernoulli’s equation. It is solved by making the substitution: ,

thereby converting the original non-linear D.E. into a linear one.

Example 2.11: Solve the following differential equations:

(a)

(b)

Solution: (a) Let (n=3)

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or

Substituting these values into the given differential equation, we

get

or

This equation is of the standard form, Integrating factor

where Therefore (x) =x-2

Solution is given by

v.x-2 = -6x-2 dx +c or v.x-2 =6x-1 +c or v = 6x + cx2

Since v = y-2 we get y-2 =6x +cx2 or

(b) Let w = y-1, then the equation

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takes the form

Integrating factor (x) = = eP(x)dx , where P(x) = 1

or (x) = eP(x)dx =ex

Solve the following Bernoulli-type equations:(a) . Ans: (b) . Ans: (c) . Ans: (d) Ans: (e) ; given y(0) = 4. Ans:

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SECTION 4.3: HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT

COEFFICIENTS

Consider the special case of second order equation

.

1. Auxiliary equation:

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Case 1. If of auxiliary equation then there will be two real and distinct solution m1 and m2 and corresponding solution of differential equation will be

Example:

Sol. m2-3m +2 = 0 or (m-1)(m-2) = 0 implies m = 1 , 2

Therefore, y = c1 ex + c2 e2x is the solution.

Case 2:

Solution will be

Example:

Solution is

Case 3:

Solution will be

Example:

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The previous procedure can be generalized for the nth-order linear differential equation with constant coefficients

The Auxiliary equation:

If all roots of the auxiliary equation are distinct the solution is

If one of the roots (m1) of the auxiliary equation has multiplicity k then solution is

Note: Let are roots of an auxiliary equation then solution will be

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Example: Find general solution;

Sol.


Sol.


Sol.

(i)

(ii)

(iii)

Solving (i),(ii) & (iii) we get

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Example: Why the given graph is a graph of a

particular solution of .

Sol. The graph indicates that auxiliary equation must have

one positive and one negative root. The given equation has this property.

Q. Determine the values of λ for which BVP has

(a) Trivial solution

(b) Non-trivial solution

Sol. Auxiliary equation is

If is the solution, applying boundary conditions

. Hence is the trivial solution.

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If , is the solution..

Applying boundary conditions . Hence is the trivial solution.

If is the solution .

For or , there will be non-trivial solution otherwise trivial solution.

LECTURE 5

SECTION 4.4

UNDETERMINED COEFFICIENT

The Solution to non-homogeneous equation

is y = yc + yp

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yc is complimentary solution to the homogeneous part

yp is the particular solution of the non-homogeneous equation

The undetermined coefficients technique is based on the idea that the particular solution will have the same form of the function . The method is limited to:

The coefficients are constants

The function is a polynomial (including constant functions), exponential function, a sine or cosine functions, or finite sums and product of these functions. The given table helps in finding yp

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1(any constant)

Case I No function in the assumed particular solution is a solution of the associated homogenous differential equation.

Example: Use undetermined coefficients method to solve

Sol. For

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Hence

For yp we assume . Now, need to determine A, B, & C. being a solution, it must satisfy the differential equation.

Comparing coefficient we get

Hence,

Case II If any yp, contains terms that duplicated terms in yc, then that yp , must be multiplied by xn, where n is the smallest positive integer that eliminates that duplication.


Sol.

First assume , but these terms appear in yc so we assume

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Sol. For

Hence

For yp we assume . Now, need to determine A, B, & C. being a solution, it must satisfy the differential equation.

Now, comparing co-efficient we get

Now,


Sol. . Assume

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Now,

Hence,

LECTURE 6

SECTION 4.6

VARIATION OF PARAMETERS

In the last section we solved non-homogeneous differential equations using the method of undetermined coefficients. This method fails to find a solution when the functions g(x) does not generate a UC-Set. For example if g(x) is sec(x), x -1, ln x, etc, we must use another approach. The approach that we will use is similar to reduction of order and it is called variation of parameters and the main idea is to assume that the particular solution is in the form

where the parameters and are functions in x after some calculation we can get the solution. Finally we can summaries the method in the following procedure.

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Step 1. Put the equation in standard form

Step 2. Find

Step3. Compute

Step 4. After computing .

Example:

Sol. .

Hence,

Example:

Sol. .

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Hence,

Example:

Sol.

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Solve IVP

Solution: Auxiliary equation

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LECTURE 7

SECTION 4.7:

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CAUCHY-EULER EQUATION

Cauchy-Euler equation is

Where as the coefficients are constant and the interval of solution is

General form of 2nd equation:

Where and are constant

Assume is the solution

Dropping

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Factor

1.For auxiliary equation we have

Case 1 distinct real roots:

and

General solution:

Ex. 1

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and General solution:

Example: Solve

Sol. Assume is the solution

Real but repeated then General solution:

Example: Solve

Sol. Let

Complex conjugate then

General solution:

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Example: Solve

Sol. Let

Example: Solve

Sol: Let

Example: Solve by variation of parameter.

Sol. Let

Auxiliary equation

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equation in standard form

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LECTURE 8

4.8: SYSTEM OF LINEAR EQUATIONS

Elimination technique:

Differential operatorWriting DE as differential operator that operates on

Ex. 1 or

Higher derivative: or

Second order DE operator:

The operator is linear:

Homogeneous differential equation:

Since

Since is solution

Case 1: 2 different real roots

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Case 2: double real root need second independent solution

Start with

Differentiating:

Where

For double root:

is second solution

Since is a polynomial is polynomial operator

Ex. 2 Factorize and solve

Since

Similarly:

Solutions: and

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Example: Solve

To eliminate x multiply equation (i) by 1 and (ii) by D.

Auxiliary equation of (v) is

. For , assume

Hence,

Similarly eliminating y we get;

Now, substituting (vi) and (vii) in either of (i) or (ii), say (ii) we get;

Hence,

and is the solution.

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Method 2:

To eliminate x multiply equation (i) by 1 and (ii) by D.


. For , assume

Hence,

Now, substitute the value of y in either (i) or (ii), say (ii) we get;

Example: Solve

To eliminate x multiply equation (i) by 3 and (ii) by (D+1).

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. For , assume

Hence,

Now, substitute the value of y in either (i) or (ii), say (ii) we get;

Example:

Sol: Now,

substituting these values of x and y in either (i) or (ii) say (ii),

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Introduction ODE