Non-Homogeneous

Differential equation

Wuletaw Desalew

10/16/2007

Advisor: Ato Biadgelign Asmare

MiliyonNew Stamp

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Contents Page

Acknowledgments ................................................................................................................................... 3

Introduction ............................................................................................................................................ 4

Objective of study ................................................................................................................................... 5

Chapter one: Non homogenous system of differential equation ......................................................... 6

Basic Definitions .................................................................................................................................. 6

Method of solutions for characteristics equation .............................................................................. 6

Methods of finding a particular solution ............................................................................................ 7

Variation of parameters .................................................................................................................. 7

Method of undetermined coefficient ............................................................................................. 9

Chapter two: Application of non homogeneous differential equation ................................................ 11

Oscillations ........................................................................................................................................ 11

Electric circuits .................................................................................................................................. 13

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Acknowledgments

First and above all I thanks to almighty god for his provision and protection in all aspects in

my life. Next, I would like to thanks my family who bring up to this stage of my life. At last I

would like to express my heartfelt appreciation to my advisor Ato Biadglign Asmare for his

constrictive comments and materials on preparing my project.

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Introduction

Differential equation is an equation relating an unknown function and one or more of its

derivatives. First, we discuss the concept of differential equation and its classification that is

ordinary differential partial differential equation and next differential equation can be

classified as homogeneous and non-homogenous. And in this project we discuss on the

solution of ordinary differential equation on non-homogenous system of differential

equation by the method of variation of parameters and undetermined coefficients. Finally,

we discuss about many physical process can be modelled by linear differential equation. For

example oscillation, electrical circuits, which is an application of differential equations.

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Objective of study

To find the solution of differential equations.

To be aware on definitions, solutions and properties of system of non-homogeneous

differential equations.

Present on the application of differential equation in a real situations.

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Chapter one: Non homogenous system of differential equation

Basic Definitions

Definition 1: an equation involving derivative of one or more dependent variables with

respect to one or more independent variables is called differential equations.

Definition 2: A differential equation is either non-homogeneous or homogeneous. A

differential equation of the form () +

() + + + = (). If g(x)=0, a

differential equation is called homogeneous differential equation. If g(x)0 ,a differential

equation is called non-homogeneous differential equation.

A general solution of () +

() + + + = () if g(x) 0 is the

form,

= +

Where Yh is the solution of homogenous or characteristic equation and Yp is a particular

solution depending on g(x).

Example 1: Given a differential equation 6 + 9 =

Therefore, g(x)0. Hence 6 + 9 = is a non-homogeneous differential

equation.

Example 2: Given+ 3 + = 0. Then, g(x)=0. Hence, the equation is homogeneous.

Method of solutions for characteristics equation

Let + = = 0;be characteristic equation .Let = be a solution of

+ + = .Then, = and = substitute the value of

, in the equation + + = 0.

It becomes; + + = 0

= [ + + ]=

+ + = 0 (which is a quadratic equation). Therefore, it has three different

cases.

Case 1: If 4 > 0.Then, the quadratic equation has two distinct real solutions, say

and .Thus, and are linear independent solutions. Hence,

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= +

is a solution of = + = 0.

Case 2: If 4 = 0;the equation has exactly one real solution. Hence, = and = .

Case 3: If 4 < 0.,the equation has two complex solutions. Say =+i and =-i. Then,=

( ) and = ().hence, = + =

( ) + ().

Example: Solve the differential equation + + = 0

Solution: Let = be a solution of + + = 0, = =

.Then, substitute in the equation it becomes; + + = 0

+ + 1 = 0, form quadratic. Hence, 4 0.

Solution: and x is the solution of 2 + = 0.Then,= and =

and let

= + by the formula: =()

(,)=

(,),

(,) = (,)=

+ = .

Therefore, =

=

=

.

=

=

=

= .

Hence, = + =-

+ .Therefore,the general solution of

2 + =

is

+

+ .

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Example 2: solve the differential equation + = tan

Solution: the characteristics equation + = 0 + 1 = 0 = .then

= ccos + csin.To find the characteristics equation; replace by

to produces = cos + sin .

The result system of equation is :

cos + sin=0..........(1)

sin + cos = tan......(2)

Multiplying the first equation by sin cos to get

sin cos + sin = 0

sin cos + cos =

sin

coscos = sin

Adding these to equations produces ; = sin = sin = cos and

=

=

= cos sec.Then, integrate to get the value of .Therefore,

= (cos sec) = sin ln |sec + tan| .

= sin cos cos |sec + tan| sin cos

= cos |sec + tan|

= + = cos + csin cos |sec + tan| .

Method of undetermined coefficient

Consider a non homogeneous differential equation() +

() + + = ()

.Recall that the general solution of such equation of the form: = + , , the

general solution of homogeneous equation and is the particular solution. The method of

undetermined coefficient determines the particular solution in the following cases.

Case 1:() is a product of a polynomial and exponential.

() = () , Then, =

[ +

+ + ]

Where s is the number of times p appears on the list of zeros of the characteristic equation

and ,, are undetermined coefficients.

Case 2:() a product of a polynomial, exponential and trigonometric function.

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If () = cos () =

sin,then,the particular solution guess

as; = [

+ + + ]

cos + [ +

+ +

] sin.

Where s is the number of times and ,, are undetermined coefficients.

Case 3:() is trigonometric function.

() = cos or sin.Then, = cos + sin

Case 4:() is a product of exponential and trigonometric function.

() = cos sin.Then, = [ cos + Bsin].

Example: solve the differential equation of

=

Solution: The differential equation family of is{}and it is assumed that

,then, =

= and substituting in to = leads to the equation

= = = 0.which is impossible ,Because the solution of

homogeneous equation corresponding to = .

And hence, the preceding method does not apply .In this particular instance it is helpful to

note that if one assume that = has solution of the form particularly (this

is ,use the family {}) instead of {} ,then,

= + ,= + 2And substituting for in to

= =

+ 2 = 2 = =

.

And it follow that the particular solution =

and the solution of the characteristics

equation is = and =

.

Therefore, the general solution of = is

= + = +

+

.

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Chapter two: Application of non homogeneous differential equation

Many physical processes can be modelled by linear differential equation .For example,

oscillation, electric circuits and more.

Oscillations

Consider a mass m on a spring. let () denotes the position at time t.the following forces

acting on the spring with mass .

The gravitational force

The spring force this is proportional to the total elongation plus (),

= ( + ()) ,

= , ,

=

The damping or resistive force: that may arise because resistance from the air,

internal energy dissipation, friction between the masses. It is proportional to the

speed of the mass = () .

The constant depending on the

direction of motion of the mass.

Possible external force ()

The product of mass and acceleration is equal to the sum of all four forces. =

+ () () + (). = 0,we have the differential equation

describing the motion:

+ = ()

If = 0, the oscillation are said to be undammed oscillation, otherwise they are damped. If

() = ;the oscillation are said to be free, otherwise they are forced.

Undammed free oscillation :the equation of motion for undamped free oscillation is

+ = 0.

Note: The characteristics equation of this system + = 0 and has the solution

=

.If we denote

by ,the general solution of this equation is

= cos sin + sin

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The constant is called the natural frequency and the constant

the period of the

motion .If we put. = Rcos = sin,R is the amplitude is the phase angle and

u= cos cos + sin sin = cos( ).

Damped free oscillation: The equation of motion for damped free oscillation is

+ + = 0.The solution of the characteristic equation are:

, =

.

There are three cases

Case 1: If 4 > 0, the equation has two different real solutions. Because 4 < ,so the solution is =

+ .

Case2: 4 = 0,

= +

.

Case 3: 4 < 0,the equations has two complex solutions.

, =

.

so the solution is =

[ cos + sin].

, =

.

Note :1. frequency ( )=

,where T is a period.

2 . ( ) =

.

3. Amplitude(R)= +

4. Phase angle ()=cos(

)=sin(

)

Example 1:A mass of 0.1 kg stretches aspiring 0.05 .If the mass is set in motion from its

equilibrium position with a downward velocity of 10cm/sec, and if there is no damping

,determine the position u as a function of t.

Solution: given m=0.1kg ,k=

=

(.)(.)

.= 19.6.since,therwe is no damping and oscillation

are free.Thus,0.1u+19.6u=0 + 196 = 0.

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Thus, the general solution is = cos14 + sin14.since, the mass set in motion from

the equilibrium position u(0)=0.The initial velocity is 10m/sec so u(0)=10.From the initial

condition =0 and =5/7.Hence,u=5/7sin14 . These are un damped free oscillation.

Example 2: a mass of 20kg is oscillating on aspiring with the spring constant of 3920 in

a medium with the damping constant of 400kg/sec .If the mass is pulled down additional 2m

and then, released, determine the position uas the function of t.

Solution:

given Required

M=20k u(t) as a function of t

=400kg/sec

K=3920N/m

The equation of motion is 20400 + 3920 = 0.the roots of characteristics equation

20 + 400 + 3920 = 0 are -1046.so the general solution is u=[ cos46 +

46].The initial condition are u(0)=2 and u(0)=0.Thus, = 2 =

and the

solution is = [2 cos46 +

46].The oscillation is damped free oscillation.

Electric circuits

Consider an electric circuit with resistance R the capacitance C and the inductance

containing producing the voltage E(t) at time t.The current and the charge Q are related

by =

.

The voltage drops across the resistor is =

The voltage drops across the capacitor is =

The voltage drops across the inductance is =

The 2 Kirchhoffs law tells us the applied voltage E(t) is equal to the sum of voltage drops

in the rest of the circuit.i.e

+ +

= ().since, =

,

=

and we have a

second order differential equation. That is:

+ +

= ().

Example: a series circuit has capacitor of capacitance C=0.2510 farad and the inductance

of the inductor L=1H,if the initial charge on the capacitor is 10 coulomb and there is no

initial charge. Find Q as a function of t.

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Solution: given Required

C=0.2510 Q(t)=?

L=1H

Q(0)=10

The equation (circuit equation) is " + +

= (),() = 0," +

. =

0.Implies " + 4 10 = 0,Te general solution of " + 4 10 = 0 is = cos200 +

sin200.The initial condition Q(0)=10.Implies

Q (0) =0.It gives, = 10 = 0., = cos200 solution.

References:

[Dennis G. Zill] A first course in differential equation.

[Martin Braun] Differential equation and their application, Springer-

verlag 1993

[R.L. Ross] Introduction to ordinary differential equation,4

ed.wiley,1989

[Erwin Kreyzing] Advanced engineering mathematics10,.Wiley,

2000.